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Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in Fig. 10.1.
A polaroid is placed in the path of the emergent ray at point \(P\) and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.
(c) The correct answer is the option c
Explanation:
Brewster’s law states that when a beam of unpolarized light is reflected in a transparent medium, the resultant reflected light is completely polarized at a certain angle of incidence. It is given by:
From the figure, it is clear that
\(
\theta p+\theta r=90^{\circ}
\)
When the Polaroid is aligned such that its transmission axis is perpendicular to the plane of polarization of the incident light (which is polarized at Brewster’s angle), no light will pass through the Polaroid. This will result in complete darkness when observed through the Polaroid.
Also, \(n=\tan \theta p\) (Brewster’s law)
(i)For \(i<\theta p\) or \(i>\theta p\)
Both of the reflected and refracted light rays become partially polarized.
(ii) Forglass \(\theta p=51^{\circ}\) for water \(\theta p=53^{\circ}\)
Consider sunlight incident on a slit of width \(10^4 \mathrm{~Ã…}\). The image seen through the slit shall
(a) The wavelength of the visible light lies in the range \(4000 \mathrm{~Ã…}-7000 \mathrm{~Ã…}\) whereas the width of the slit is about \(10^4 \mathrm{~Ã…}\). Since the width of the slit is comparable to that of wavelength. Diffraction occurs with maxima at centre.
The maxima consist of all the colour of lights present in the wave. Therefore, At the centre all colours appear i.e, mixing of colours from the white patch at the centre.
Consider a ray of light incident from air onto a slab of glass (refractive index \(n\) ) of width \(d\), at an angle \(\theta\). The phase difference between the ray reflected by the top surface of the glass and the bottom surface is
(a) Now consider the diagram, the ray \((P)\) is incident at an angle \(\theta\) and gets reflected in the direction \(P^{\prime}\) and refracted in the direction \(P^{\prime \prime}\). Due to reflection from the glass medium, there is a phase change of \(\pi\).
Time taken to travel along \(O P^{\prime \prime}\)
\(
\Delta t=\frac{O P^{\prime \prime}}{v}=\frac{d / \cos r}{c / n}=\frac{n d}{c \cos r} .
\)
From Snell’s law, \(n=\frac{\sin \theta}{\sin r}\)
\(
\Rightarrow \quad \sin r=\frac{\sin \theta}{n}
\)
\(
\cos r=\sqrt{1-\sin ^2 r}=\sqrt{1-\frac{\sin ^2 \theta}{n^2}}
\)
Phase difference, \(\Delta \phi=\frac{2 \pi}{T} \times \Delta t \Rightarrow \Delta \phi=\frac{2 \pi n d}{\lambda}\left(1-\frac{\sin ^2 \theta}{n^2}\right)^{-1 / 2}\)
So, net phase difference \(=\Delta \phi+\pi\)
\(
\Rightarrow \quad \Delta \phi_{\text {net }}=\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)+\pi
\)
In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case
(c) Key concept:
Condition for Observing Interference
The initial phase difference between the interfering waves must remain constant. Otherwise the interference will not be sustained.
The frequency and wavelengths of two waves should be equal. If not the phase difference will not remain constant and so the interference will not be sustained.
The light must be monochromatic. This eliminates overlapping of patterns as each wavelength corresponds to one interference pattern.
Here in this problem of Young’s double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filtration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen. Hence, in this case there shall be no interference fringes.
The wave front emitted by a narrow source is divided in two parts by reflection, refraction or diffraction. The coherent sources so obtained are imaginary.
Figure 10.2 shows a standard two slit arrangement with slits \(\mathrm{S}_1, \mathrm{~S}_2 . \mathrm{P}_1, \mathrm{P}_2\) are the two minima points on either side of P (Fig. 10.2).
At \(P_2\) on the screen, there is a hole and behind \(P_2\) is a second 2- slit arrangement with slits \(\mathrm{S}_3, \mathrm{~S}_4\) and a second screen behind them.
(d) Every point on the given wave front acts as a source of new disturbance called secondary wavelets which travel in all directions with the .velocity of light in the medium.
A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wave front at that instant. This is called secondary wave front. In the given question, there is a hole at point which is a maxima point. From Huygen’s principle, wave will propagate from the sources \(S_1\) and \(S_2\). Each point on the screen will act as secondary sources of wavelets.
For two interfering waves we get Minima and Maxima
For Minima,
\(
\Delta \mathrm{x}=\left(n+\frac{1}{2}\right) \lambda
\)
For Maxima,
\(
\Delta=\mathrm{n} \lambda
\)
where n is an integer.
Explanation:
From the figure,
At \(P_2\) Intensity is minimum.
\(\rightarrow\) Resultant Intensity is given by
\(
I_{\text {resultant }}=4 I_o \cos ^2\left(\frac{\theta}{2}\right)
\)
At \(P_2\) Intensity is minimum.
Conditions for Interference are-
\(\rightarrow\) Coherent
\(\rightarrow\) Monochromatic
Both the above conditions are fulfilled.Thus Interference must be fulfilled.
The wave front from the slit \(S\) will spread and interfere at \(P_2\). From there they will again pass through \(\mathrm{S}_3\) and \(\mathrm{S}_4\).
\(\rightarrow\) Thus There would be a regular two slit pattern on the second screen.
Two source \(\mathrm{S}_1\) and \(\mathrm{S}_2\) of intensity \(I_1\) and \(I_2\) are placed in front of a screen [Fig. 10.3 (a)]. The patteren of intensity distribution seen in the central portion is given by Fig. 10.3 (b). In this case which of the following statements are true.
(a, b, d) Key concept:
For getting the sustained interference the initial phase difference between the interfering waves must remain constant, i.e., sources should be coherent.
For two coherent sources, the resultant intensity is given by \(I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \varphi\)
Resultant intensity at the point of observation will be maximum.
\(
\begin{gathered}
I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2} \\
I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2
\end{gathered}
\)
Resultant intensity at the point of observation will be minimum.
\(
\begin{gathered}
I_{\min }=I_1+I_2-2 \sqrt{I_1 I_2} \\
I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
\end{gathered}
\)
Consider sunlight incident on a pinhole of width \(10^3 \mathrm{~Ã…}\). The image of the pinhole seen on a screen shall be
(b, d) Key concept: Diffraction of Light can be observed only if the size of obstacle/aperture is less than the wavelength of light.
Given, width of pinhole \(=103 Ã…=1000 Ã…\)
We know that wavelength of sunlight ranges from \(4000 Ã…\) to \(8000 Ã…\). Clearly, wavelength \(X\) < width of the slit.
Hence, light is diffracted from the hole. Due to diffraction from the slit the image formed on the screen will be different from the geometrical image.
Consider the diffraction patern for a small pinhole. As the size of the hole is increased
(a, b) Key concept: The “shadow” of hole of diameter \(d\) is spread out over an angle \(\Delta \theta=1.22 \frac{\lambda}{D} \Rightarrow \Delta \theta \propto \frac{1}{D}\)
\(
\theta=1.22 \lambda / \mathrm{D}
\)
\(
\therefore \theta \propto \frac{1}{D}
\)
As the size of the pinhole (D) increases, the angular width of central maximum decreases. when size of hole increases, then more light enters through hole, which results in increase of intensity.
For light diverging from a point source
(a, b) Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical.
As intensity of the source will be
\(
I \propto \frac{1}{r^2}
\)
where, \(r\) is radius of the wavefront at any time.
Hence the intensity decreases in proportion to the distance squared.
In a Young’s double slit experiment, the separation between the slits is 0.10 mm, the wavelength of light used is 600 nm and the interference pattern is observed on a screen 1.0 m away. What is the separation between the successive bright fringes?
(d) The separation between the successive bright fringes is
\(
\begin{aligned}
w & =\frac{D \lambda}{d}=\frac{1.0 \mathrm{~m} \times 600 \times 10^{-9} \mathrm{~m}}{0.10 \times 10^{-3} \mathrm{~m}} \\
& =6.0 \times 10^{-3} \mathrm{~m}=6.0 \mathrm{~mm}
\end{aligned}
\)
The wavelength of light coming from a sodium source is 589 nm. What will be its wavelength in water? Refractive index of water \(=1.33\).
(b) The wavelength in water is \(\lambda=\lambda_0 / \mu\), where \(\lambda_0\) is the wavelength in vacuum and \(\mu\) is the refractive index of water. Thus,
\(
\lambda=\frac{589 \mathrm{~nm}}{1.33}=443 \mathrm{~nm}
\)
Find the minimum thickness of a film which will strongly reflect the light of wavelength 589 nm. The refractive index of the material of the film is \(1 \cdot 25\).
(c) For strong reflection, the least optical path difference introduced by the film should be \(N 2\). The optical path difference between the waves reflected from the two surfaces of the film is \(2 \mu d\). Thus, for strong reflection,
\(
\begin{aligned}
2 \mu d & =\lambda / 2 \\
d & =\frac{\lambda}{4 \mu}=\frac{589 \mathrm{~nm}}{4 \times 1 \cdot 25}=118 \mathrm{~nm}
\end{aligned}
\)
A parallel beam of monochromatic light of wavelength 450 nm passes through a long slit of width 0.2 mm. What is the angular divergence in which most of the light is diffracted?
Most of the light is diffracted between the two first order minima. These minima occur at angles given by \(b \sin \theta= \pm \lambda\)
\(
\begin{aligned}
\sin \theta & = \pm \lambda / b \\
& = \pm \frac{450 \times 10^{-9} \mathrm{~m}}{0.2 \times 10^{-3} \mathrm{~m}}= \pm 2.25 \times 10^{-3}
\end{aligned}
\)
\(
\theta \approx \pm 2 \cdot 25 \times 10^{-3} \mathrm{rad}
\)
The angular divergence \(=4.5 \times 10^{-3} \mathrm{rad}\).
A beam of light of wavelength 590 nm is focussed by a converging lens of diameter 10.0 cm at a distance of 20 cm from it. Find the diameter of the disc image formed.
The angular radius of the central bright disc in a diffraction pattern from circular aperture is given by
\(
\begin{aligned}
\sin \theta & \approx \frac{1.22 \lambda}{b} \\
& =\frac{1.22 \times 590 \times 10^{-9} \mathrm{~m}}{10.0 \times 10^{-2} \mathrm{~m}}=0.7 \times 10^{-5} \mathrm{rad}
\end{aligned}
\)
The radius of the bright disc is
\(
0.7 \times 10^{-5} \times 20 \mathrm{~cm}=1.4 \times 10^{-4} \mathrm{~cm}
\)
The diameter of the disc image \(=2.8 \times 10^{-4} \mathrm{~cm}\).
White light is a mixture of light of wavelengths between 400 nm and 700 nm. If this light goes through water ( \(\mu=1.33\) ) what are the limits of the wavelength there ?
(a) When a light having wavelength \(\lambda_0\) in vacuum goes through a medium of refractive index \(\mu\), the wavelength in the medium becomes \(\lambda=\lambda_0 / \mu\).
For \(\quad \lambda_0=400 \mathrm{~nm}, \lambda=\frac{400 \mathrm{~nm}}{1.33}=300 \mathrm{~nm}\)
and for \(\lambda_0=700 \mathrm{~nm}, \lambda=\frac{700 \mathrm{~nm}}{1.33}=525 \mathrm{~nm}\).
Thus, the limits are 300 nm and 525 nm.
The optical path of a monochromatic light is the same if it goes through 2.00 cm of glass or 2.25 cm of water. If the refractive index of water is 1.33 , what is the refractive index of glass?
(c) When light travels through a distance \(x\) in a medium of refractive index \(\mu\), its optical path is \(\mu x\). Thus, if \(\mu\) is the refractive index of glass,
\(
\mu(2.00 \mathrm{~cm})=1.33 \times(2.25 \mathrm{~cm})
\)
\(
\mu=1.33 \times \frac{2.25}{2.00}=1.50
\)
White light is passed through a double slit and interference pattern is observed on a screen 2.5 m away. The separation between the slits is 0.5 mm . The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringe. What is the wavelengths of the violet and the red light?
(b) For the first bright fringe, the distance from the centre is
\(
y=\frac{D \lambda}{d}
\)
For violet light, \(y=2.0 \mathrm{~mm}\). Thus,
\(
\begin{aligned}
2.0 \mathrm{~mm} & =\frac{(2.5 \mathrm{~m}) \lambda}{0.5 \mathrm{~mm}} \\
\lambda & =\frac{(0.5 \mathrm{~mm})(2.0 \mathrm{~mm})}{2.5 \mathrm{~m}}=400 \mathrm{~nm}
\end{aligned}
\)
Similarly, for red light, \(y=3.5 \mathrm{~mm}\). Thus,
\(
\begin{aligned}
3.5 \mathrm{~mm} & =\frac{(2.5 \mathrm{~m}) \lambda}{0.5 \mathrm{~mm}} \\
\lambda & =700 \mathrm{~nm} .
\end{aligned}
\)
A double slit experiment is performed with sodium (yellow) light of wavelength 589.3 nm and the interference pattern is observed on a screen 100 cm away. The tenth bright fringe has its centre at a distance of 12 mm from the central maximum. What is the separation between the slits?
(d) For the \(n\)th maximum fringe, the distance above the central line is
\(
x=\frac{n \lambda D}{d} .
\)
According to the data given,
\(
x=12 \mathrm{~mm}, n=10, \lambda=589 \cdot 3 \mathrm{~nm}, D=100 \mathrm{~cm} .
\)
Thus, the separation between the slits is
\(
\begin{aligned}
d=\frac{n \lambda D}{x} & =\frac{10 \times 589.3 \times 10^{-9} \mathrm{~m} \times 100 \times 10^{-2} \mathrm{~m}}{12 \times 10^{-3} \mathrm{~m}} \\
& =4.9 \times 10^{-4} \mathrm{~m}=0.49 \mathrm{~mm}
\end{aligned}
\)
The intensity of the light coming from one of the slits in a Young’s double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed.
(c) The intensity of the light originating from the first slit is double the intensity from the second slit. The amplitudes of the two interfering waves are in the ratio \(\sqrt{ } 2: 1\), say \(\sqrt{ } 2 A\) and \(A\).
At the point of constructive interference, the resultant amplitude becomes \((\sqrt{2}+1) A\). At the points of destructive interference, this amplitude is \((\sqrt{ } 2-1) A\). The ratio of the resultant intensities at the maxima to that at the minima is
\(
\frac{(\sqrt{ } 2+1)^2 A^2}{(\sqrt{ } 2-1)^2 A^2}=34
\)
The width of one of the two slits in a Young’s double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit-width, find the ratio of the maximum to the minimum intensity in the interference pattern.
(c) Suppose the amplitude of the light wave coming from the narrower slit is \(A\) and that coming from the wider slit is \(2 A\). The maximum intensity occurs at a place where constructive interference takes place. Then the resultant amplitude is the sum of the individual amplitudes. Thus,
\(
A_{\max }=2 A+A=3 A
\)
The minimum intensity occurs at a place where destructive interference takes place. The resultant amplitude is then difference of the individual amplitudes. Thus,
\(
A_{\min }=2 A-A=A
\)
As the intensity is proportional to the square of the amplitude,
\(
\frac{I_{\max }}{I_{\min }}=\frac{\left(A_{\max }\right)^2}{\left(A_{\min }\right)^2}=\frac{(3 A)^2}{A^2}=9
\)
Two sources \(S_1\) and \(S_2\) emitting light of wavelength 600 nm are placed a distance \(1.0 \times 10^{-2} \mathrm{~cm}\) apart. A detector can be moved on the line \(S_1 P\) which is perpendicular to \(S_1 S_2\). (a) What would be the minimum and maximum path difference at the detector as it is moved along the line \(S_1 P\)? (b) Locate the position of the farthest minimum detected.
(a)
(a) The situation is shown in the figure (below). The path difference is maximum when the detector is just at the position of \(S_1\) and its value is equal to \(d=1.0 \times 10^{-2} \mathrm{~cm}\). The path difference is minimum when the detector is at a large distance from \(S_1\). The path difference is then close to zero.
(b) The farthest minimum occurs at a point \(P\) where the path difference is \(\lambda / 2\). If \(S_1 P=D\),
\(
\begin{array}{r}
S_2 P-S_1 P=\frac{\lambda}{2} \\
\sqrt{D^2+d^2}-D=\frac{\lambda}{2}
\end{array}
\)
\(
\begin{gathered}
D^2+d^2=\left(D+\frac{\lambda}{2}\right)^2 \\
d^2=D \lambda+\frac{\lambda^2}{4} \\
D=\frac{d^2}{\lambda}-\frac{\lambda}{4} \\
=\frac{\left(1 \cdot 0 \times 10^{-4} \mathrm{~m}\right)^2}{600 \times 10^{-9} \mathrm{~m}}-150 \times 10^{-9} \mathrm{~m}=1 \cdot 7 \mathrm{~cm}
\end{gathered}
\)
A beam of light consisting of two wavelengths, \(6500 Ã…\) and \(5200 Ã…\) is used to obtain interference fringes in a Young’s double slit experiment \(\left(1 Ã…=10^{-10} \mathrm{~m}\right)\). The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. (a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength \(6500 Ã…\). (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
(a)
(a) The centre of the \(n\)th bright fringe is at a distance \(y=\frac{n \lambda D}{d}\) from the central maximum. For the 3rd bright fringe of \(6500 Ã…\),
\(
\begin{aligned}
y & =\frac{3 \times 6500 \times 10^{-10} \mathrm{~m} \times 1.2 \mathrm{~m}}{2 \times 10^{-3} \mathrm{~m}} \\
& =0.117 \mathrm{~cm} \approx 0.12 \mathrm{~cm}
\end{aligned}
\)
(b) Suppose the \(m\) th bright fringe of \(6500 Ã…\) coincides with the \(n\)th bright fringe of \(5200 Ã…\).
Then,
\(
\frac{m \times 6500 \AA \times D}{d}=\frac{n \times 5200 \AA \times D}{d}
\)
\(
\frac{m}{n}=\frac{5200}{6500}=\frac{4}{5}
\)
The minimum values of \(m\) and \(n\) that satisfy this equation are 4 and 5 respectively. The distance of the 4th bright fringe of \(6500 Ã…\) or the 5th bright fringe of \(5200 Ã…\) from the central maximum is
\(
\begin{aligned}
y & =\frac{4 \times 6500 \times 10^{-10} \mathrm{~m} \times 1.2 \mathrm{~m}}{2 \times 10^{-3} \mathrm{~m}} \\
& =0.156 \mathrm{~cm} \approx 0.16 \mathrm{~cm}
\end{aligned}
\)
Monochromatic light of wavelength 600 nm is used in a Young’s double slit experiment. One of the slits is covered by a transparent sheet of thickness \(1.8 \times 10^{-5} \mathrm{~m}\) made of a material of refractive index \(1 \cdot 6\). How many fringes will shift due to the introduction of the sheet?
(d) When the light travels through a sheet of thickness \(t\), the optical path travelled is \(\mu t\), where \(\mu\) is the refractive index. When one of the slits is covered by the sheet, air is replaced by the sheet and hence, the optical path changes by \((\mu-1) t\). One fringe shifts when the optical path changes by one wavelength. Thus, the number of fringes shifted due to the introduction of the sheet is
\(
\frac{(\mu-1) t}{\lambda}=\frac{(1.6-1) \times 1.8 \times 10^{-5} \mathrm{~m}}{600 \times 10^{-9} \mathrm{~m}}=18
\)
White light is incident normally on a glass plate of thickness \(0.50 \times 10^{-6}\) and index of refraction 1.50. Which wavelengths in the visible region ( \(400 \mathrm{~nm}-700 \mathrm{~nm}\) ) are strongly reflected by the plate?
(c) The light of wavelength \(\lambda\) is strongly reflected if
\(
2 \mu d=\left(n+\frac{1}{2}\right) \lambda \dots(i)
\)
where \(n\) is a nonnegative integer.
Here,
\(
\begin{aligned}
2 \mu d & =2 \times 1.50 \times 0.5 \times 10^{-6} \mathrm{~m} \\
& =1.5 \times 10^{-6} \mathrm{~m} \dots(ii)
\end{aligned}
\)
Putting \(\lambda=400 \mathrm{~nm}\) in (i) and using (ii),
\(
\begin{aligned}
1.5 \times 10^{-6} \mathrm{~m} & =\left(n+\frac{1}{2}\right)\left(400 \times 10^{-9} \mathrm{~m}\right) \\
n & =3.25
\end{aligned}
\)
Putting \(\lambda=700 \mathrm{~nm}\) in (i) and using (ii),
\(
\begin{aligned}
1.5 \times 10^{-6} \mathrm{~m} & =\left(n+\frac{1}{2}\right)\left(700 \times 10^{-9} \mathrm{~m}\right) \\
n & =1.66
\end{aligned}
\)
Thus, between 400 nm and 700 nm the integer \(n\) can take the values 2 and 3. Putting these values of \(n\) in (i), the wavelength become
\(
\lambda=\frac{4 \mu d}{2 n+1}=600 \mathrm{~nm} \text { and } 429 \mathrm{~nm}
\)
Thus, light of wavelengths 429 nm and 600 nm are strongly reflected.
A parallel beam of green light of wavelength 546 nm passes through a slit of width 0.40 mm. The transmitted light is collected on a screen 40 cm away. Find the distance between the two first-order minima.
The minima occur at an angular deviation \(\theta\) given by \(b \sin \theta=n \lambda\), where \(n\) is an integer. For the first order minima, \(n= \pm 1\) so that \(\sin \theta= \pm \frac{\lambda}{b}\). As the fringes are observed at a distance much larger than the width of the slit, the linear distances from the central maximum are given by
\(
\begin{aligned}
x & =D \tan \theta \\
& \approx D \sin \theta= \pm \frac{\lambda D}{b} .
\end{aligned}
\)
Thus, the minima are formed at a distance \(\frac{\lambda D}{b}\) from the central maximum on its two sides. The separation between the minima is
\(
\frac{2 \lambda D}{b}=\frac{2 \times 546 \times 10^{-9} \mathrm{~m} \times 40 \times 10^{-2} \mathrm{~m}}{0.40 \times 10^{-3} \mathrm{~m}}=1.1 \mathrm{~mm}
\)
Light is
(c) both a particle and a wave phenomenon.
Explanation:
Light exhibits characteristics of both waves and particles.
Wave nature:
Light can be diffracted (bent around obstacles) and can interfere with itself, creating patterns of bright and dark spots, similar to water waves.
Particle nature:
Light can also behave as a stream of particles called photons. For example, the photoelectric effect, where light can eject electrons from a metal surface, is explained by the particle nature of light. This dual nature of light is known as wave-particle duality, meaning light can behave as both a wave and a particle depending on the situation.
The speed of light depends
(d) neither on elasticity nor on inertia.
We are going to use the following given formula in order to solve the problem:\(\mu=\frac{c}{v}\). \(\text { where } \mu \text { is the absolute refractive index, } c \text { is the speed of light in vacuum and } v \text { is the speed of light in the medium.}\)Â
Explanation: The speed of light in a vacuum is a fundamental constant, approximately \(3.0 \times 10^8\) meters per second. When light travels through a medium, its speed is affected by the medium’s refractive index, not its elasticity or inertia. The refractive index is a property of the medium related to its electrical and magnetic properties, which determines how much light bends (refracts) when entering the medium.
The equation of a light wave is written as \(y=A \sin (k x-\omega t)\). Here, \(y\) represents
(d) Electric field. In the equation of a light wave, \(y=A \sin (k x-\omega t)\), \(y\) represents the electric field in the wave. Light is an electromagnetic wave, and its wave nature is described by oscillating electric and magnetic fields. The equation describes how the electric field varies with position ( \(x\) ) and time ( \(t\) ).
Which of the following properties show that light is a transverse wave?
(d) Polarization.
Explanation: Polarization is the ability of light waves to vibrate in a specific plane, which is only possible with transverse waves.
Why other options are incorrect:
(a) Reflection: Reflection is the bouncing of light off a surface, and it can occur with both transverse and longitudinal waves.
(b) Interference: Interference is the interaction of overlapping waves, and it can also occur with both transverse and longitudinal waves.
(c) Diffraction: Diffraction is the bending of light around an obstacle, and it too can happen with both transverse and longitudinal waves.
When light is refracted into a medium,
(c) When light is refracted into a medium, its wavelength decreases but frequency remains unchanged.
Explanation: Refraction occurs when light enters a medium with a different density, causing it to bend. The speed of light changes in the new medium, but the frequency, which is determined by the source of the light, stays the same. Since speed is related to wavelength ( \(v=f \lambda\) ), when the speed decreases in a denser medium, the wavelength must also decrease to maintain the constant frequency.
Formula used: The formula of the velocity of the light is given by
\(\)
v=\lambda f
\(\)
Where \(\)v\(\) is the velocity of the light, \(\)\lambda\(\) is the wavelength of the light and \(\)f\(\) is the frequency of the light.
Complete answer: It is known that the velocity of the light decreases when the light moves from the rarer medium to the denser medium. The rarer medium is the medium that is made up of the less dense medium. The rarer medium is the medium that is made up of the denser medium.
\(\)
\begin{aligned}
& \mu=\frac{v_{\text {rarer }}}{v_{\text {denser }}}>1 \\
& v_{\text {rarer }}>v_{\text {denser }}
\end{aligned}
\(\)
Substituting the formula of the velocity in the above step, we get
\(\)
\lambda_1 f_1>\lambda_2 f_2 \dots(1)
\(\)
It is known that the frequency of the wave depends on the nature of the wave and it is independent of the moving medium. For both the denser and the rarer medium, the value of the frequency remains the same.
\(\)
\text { So } f=f_1=f_2
\(\)
Substitute the above value in the equation (1)
\(\)
\lambda_1 f>\lambda_2 f \lambda_1>\lambda_2
\(\)
The above step shows that when the light moves from the rarer medium to the denser medium, the wavelength decreases.
When light is refracted, which of the following does not change?
(b) Frequency does not change.
Explanation: Refraction occurs when light enters a new medium, causing it to bend. This change in direction is due to a change in the light’s speed, but the frequency, which is related to the energy of the light wave, remains constant.
Why other options are incorrect:
(a) Wavelength: Wavelength is related to the speed of light by the equation \(v=\lambda f\), where \(v\) is velocity, \(\lambda\) is wavelength, and \(f\) is frequency. Since the velocity changes when light is refracted, the wavelength must also change to maintain the constant frequency.
(c) Velocity: As mentioned before, the velocity of light changes when it enters a new medium.
(d) Amplitude: While some sources suggest that amplitude might change during refraction, others indicate that it remains the same. The change in amplitude is more related to the process of reflection, which often accompanies refraction. The focus of refraction is primarily on the change in wavelength and velocity due to the change in medium.
The amplitude modulated (AM) radio wave bends appreciably round the corners of a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) board but the frequency modulated (FM) wave only negligibly bends. If the average wavelengths of AM and FM waves are \(\lambda_a\) and \(\lambda_f\),
(a)Â The phenomenon described, where AM radio waves bend significantly around obstacles like a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) board while FM waves bend negligibly, is known as diffraction. Diffraction is the bending or spreading of waves as they pass around an obstacle or through an opening.
The key principle governing diffraction is that diffraction is more pronounced when the wavelength of the wave is comparable to or larger than the size of the obstacle or opening.
In this scenario:
AM waves are bending appreciably around a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) board.
FM waves are bending negligibly.
This indicates that the wavelength of the AM wave \(\left(\boldsymbol{\lambda}_{\boldsymbol{a}}\right)\) is larger relative to the size of the board ( 1 m ) compared to the wavelength of the FM wave ( \(\boldsymbol{\lambda}_f\) ). Since longer wavelengths diffract more, and AM waves are diffracting more significantly than FM waves in this situation, it can be concluded that the wavelength of the AM wave is greater than the wavelength of the FM wave.
Therefore, the correct relation between the wavelengths is \(\lambda_a>\lambda_f\).
Which of the following sources gives best monochromatic light?
(d) The source that gives the best monochromatic light is a laser.
Explanation: A laser produces light with a very specific wavelength, meaning it is essentially pure color, making it the most monochromatic light source among the options.
Why other options are incorrect:
(a) A candle: Candlelight is a mixture of wavelengths, producing a broad spectrum of colors, not considered monochromatic.
(b) A bulb: Similar to a candle, a regular light bulb emits a mixture of wavelengths, resulting in white light.
(c) A mercury tube:
While mercury tubes emit light with more defined wavelengths than a candle or bulb, they still produce a mixture of wavelengths compared to a laser.
The wavefronts of a light wave travelling in vacuum are given by \(x+y+z=c\). The angle made by the direction of propagation of light with the \(X\)-axis is
(d) \(\cos ^{-1}(1 / \sqrt{3})\).
Explanation:
The wavefronts of a light wave are surfaces of constant phase. The direction of propagation of the light wave is always perpendicular to these wavefronts.
The equation of the wavefronts is given by \(\boldsymbol{x}+\boldsymbol{y}+\boldsymbol{z}=\boldsymbol{c}\), where \(\boldsymbol{c}\) is a constant. This is the equation of a plane in three-dimensional space. The normal vector to this plane gives the direction of propagation of the light wave.
From the equation of the plane \(\boldsymbol{A} \boldsymbol{x}+\boldsymbol{B} \boldsymbol{y}+\boldsymbol{C} \boldsymbol{z}=\boldsymbol{D}\), the normal vector is given by \(\overrightarrow{\boldsymbol{n}}=(\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C})\). In our case, the equation is \(\boldsymbol{x}+\boldsymbol{y}+\boldsymbol{z}=\boldsymbol{c}\), so the normal vector is \(\vec{n}=(1,1,1)\).
To determine the angle between the direction of propagation (the normal vector) and the X-axis, calculate the angle between the vector \(\vec{n}=(1,1,1)\) and the X-axis, which is represented by the vector \(\vec{i}=(1,0,0)\).
The dot product formula can be used to find the cosine of the angle between two vectors:
\(
\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}
\)
Here, \(\vec{a}=\vec{n}=(1,1,1)\) and \(\vec{b}=\vec{i}=(1,0,0)\).
The dot product is \(\vec{n} \cdot \vec{i}=(1)(1)+(1)(0)+(1)(0)=1\).
The magnitude of \(\vec{n}\) is \(|\vec{n}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}\).
The magnitude of \(\vec{i}\) is \(|\vec{i}|=\sqrt{1^2+0^2+0^2}=1\).
So, \(\cos \theta=\frac{1}{\sqrt{3} \cdot 1}=\frac{1}{\sqrt{3}}\).
Therefore, the angle made by the direction of propagation of light with the \(x-\) axis is \(\theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\).
The wavefronts of light coming from a distant source of unknown shape are nearly
(a)Â The wavefronts of light coming from a distant source of unknown shape are nearly plane.
Explanation: As the light travels a large distance from the source, the curvature of the wavefronts becomes negligible, making them appear flat or plane.
Why other options are incorrect:
(b) elliptical:Â Elliptical wavefronts are associated with sources that are elongated in one direction, like a laser beam. A distant source of unknown shape is unlikely to have such a specific geometry.
(c) cylindrical:Â Cylindrical wavefronts originate from a line source, not a point source. While a distant source might appear elongated from our perspective, it’s still considered a point source for the purpose of wavefront analysis.
(d) spherical:Â Spherical wavefronts are characteristic of light emanating from a point source close by. As the source becomes distant, the wavefronts flatten out into plane wavefronts.
The inverse square law of intensity (i.e., the intensity \(\propto \frac{1}{r^2}\) ) is valid for a
(a) Firstly the inverse square law: Any point source which spreads its influence equally in all directions without limits to its range will obey the inverse square law.
Actually this comes from the strictly geometrical consideration. The intensity of the influence at any given radius \(r\) is the source strength divided by the area of the sphere. Being strictly geometric in its origin; the inverse square law is used in a diverse way.
Point source of the gravitational force, electrostatic field, light, sound or radiation obeys the inverse square law.
Explanation: The inverse square law, which states that intensity is inversely proportional to the square of the distance, is only valid for a point source. This is because a point source emits energy equally in all directions, and as the distance from the source increases, the energy is spread over a larger area, resulting in a decrease in intensity proportional to the square of the distance.
Note: Intensity of a point source obeys the inverse square law.
Intensity of light at distance r from the point source is given by
\(
I=S /\left(4 \pi r^2\right)
\)
where \(S[latex] is the source strength.
Line or plane or cylindrical source, it varies with the distance i.e. Intensity [latex]\propto \frac{1}{r}\)
Two sources are called coherent if they produce waves
(d) Two sources are called coherent if they produce waves with a constant phase difference. This means the phase difference between the waves from the two sources does not change over time.
When a drop of oil is spread on a water surface, it displays beautiful colours in daylight because of
(d) The bright colors seen in an oil film or in the soap bubble are caused by interference. The brightest colors are those that interfere constructively. This interference is between light reflected from upper and lower surfaces of the film; thus, the effect is understood as thin film interference. As noticed before, interference effects are most prominent when light interacts with something having a size almost like its wavelength.
When a drop of oil is spread on a water surface, it displays beautiful colors that appear in daylight because of interference of light. Here some of the light is reflected off the top surface and some the bottom surface. Because the thickness of the oil film is about an equivalent because the wavelength of the sunshine the 2 reflected rays interfere with one another.
Here interference of light is taking place.
Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is 25. The intensities of the sources are in the ratio
(c)
\(
\begin{aligned}
&\text { The ratio of maximum intensity and minimum intensity is given by, }\\
&\begin{aligned}
& I_{\max }=\left(\sqrt{I}_1+\sqrt{I}_2\right)^2, I_{\min }=\left(\sqrt{I}_1-\sqrt{I}_2\right)^2 \\
& \therefore \frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I}_1+\sqrt{I}_2}{\sqrt{I}_1-\sqrt{I}_2}\right)^2=25 \\
& \frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}=5 \\
& \Rightarrow \frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}=5 \\
& \Rightarrow \sqrt{\frac{I_1}{I_2}}+1=5 \sqrt{\frac{I_1}{I_2}}-5 \\
& \Rightarrow 6=4 \sqrt{\frac{I_1}{I_2}} \\
& \Rightarrow \sqrt{\frac{I_1}{I_2}}=\frac{3}{2} \\
& \Rightarrow \frac{I_1}{I_2}=\frac{9}{4}
\end{aligned}
\end{aligned}
\)
The slits in a Young’s double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is \(I_0\). If one of the slits is closed, the intensity at this point will be
(b)Â Step 1: In a Young’s double slit experiment, we have two slits (let’s call them S1 and S2) that are equally wide. The source of light is placed symmetrically with respect to these slits. The intensity at the central fringe (where the two waves from the slits constructively interfere) is given as \(I_0\).
Hint: Remember that when both slits are open, the intensity at the central maximum is the result of constructive interference.
Step 2: Expression for Resultant Intensity
The resultant intensity \(I\) at any point due to two coherent sources can be expressed as:
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi
\)
where \(I_1\) and \(I_2\) are the intensities from each slit, and \(\phi\) is the phase difference between the two waves.
Hint: The phase difference \(\phi\) at the central maximum is zero because the path difference is zero.
Step 3: Applying Symmetry
Since the source is symmetrically placed with respect to the slits, the intensities from both slits are equal, i.e., \(I_1=I_2=I_c\). Thus, the expression for intensity simplifies to:
\(
I=I_c+I_c+2 \sqrt{I_c I_c} \cos 0=2 I_c+2 I_c=4 I_c
\)
At the central maximum, this resultant intensity is given as \(\boldsymbol{I}_{\mathbf{0}}\). Therefore, we have:
\(
4 I_c=I_0
\)
From this, we can solve for \(I_c\) :
\(
I_c=\frac{I_0}{4}
\)
Step 4: Closing One Slit
Now, if one of the slits (let’s say S2) is closed, the intensity from that slit becomes zero: \(I_2=0\). The resultant intensity when one slit is closed will be:
\(
I^{\prime}=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi
\)
Substituting \(I_2=0\) :
\(
I^{\prime}=I_1+0+2 \sqrt{I_1 \cdot 0} \cos \phi=I_1
\)
Since \(I_1=I_c\) (because the source is still symmetrically placed), we have:
\(
I^{\prime}=I_c
\)
Hint: When one slit is closed, the intensity at the central maximum is simply the intensity from the open slit.
Step 5: Final Calculation
Substituting the value of \(I_c\) that we found earlier:
\(
I^{\prime}=\frac{I_0}{4}
\)
A thin transparent sheet is placed in front of a Young’s double slit. The fringe-width will
(c) When a thin transparent sheet is placed in front of one of the slits in a Young’s double-slit experiment, the fringe width will remain the same.
Explanation:
Key concept: The fringe width in a Young’s double-slit experiment is determined by the wavelength of light, the distance between the slits, and the distance between the slits and the screen.
Effect of the transparent sheet: Placing a transparent sheet in front of one slit introduces a slight optical path length difference, causing the interference pattern to shift slightly.
No change in fringe width: However, the transparent sheet doesn’t change the fundamental parameters (wavelength, slit separation, screen distance) that govern the fringe width, so the fringe width remains the same.
Note:
The expression for the fringe width in Young’s double slit experiment is
\(
\beta=\frac{\lambda D}{d} \ldots(1)
\)
Here, \(\beta\) is the fringe width, \(\lambda\) is the wavelength, \(d\) is the distance between the slits and \(D\) is the distance between the slit and the screen.
Complete answer:
When a thin transparent sheet is placed in front of the double slit, no physical quantities affecting the fringe width are changed.Therefore, the fringe width remains the same.Hence, the options A, B and D are incorrect.
From equation (1), it can be concluded that the fringe width depends on the wavelength \(\)\lambda\(\) of the light, distance \(\)d\(\) between the slits and the distance \(\)D\(\) between the slit and screen.
Although a thin transparent sheet is placed in front of the double slit, the distance between the slits and the distance between the slit and screen remains the same.Since the sheet is transparent, the light will pass through the sheet and hence, the wavelength of the light will also remain unchanged.As all the quantities in the equation for fringe width remains the same, the fringe width will also remain the same.
If Young’s double slit experiment is performed in water,
(a)Â If Young’s double slit experiment is performed in water, the fringe width will decrease.Â
Explanation: Fringe width formula:
The fringe width in Young’s double slit experiment is given by the formula \(\beta=\frac{\lambda D}{d}\), where \(\lambda\) is the wavelength of light, \(D\) is the distance between the slits and the screen, and \(d\) is the distance between the two slits.
Wavelength change in water: When light passes from air into water, its wavelength decreases because the refractive index of water is greater than that of air.
Resulting fringe width decrease: Since the fringe width is directly proportional to the wavelength, if the wavelength decreases, the fringe width will also decrease.
Alternate explanation:
Fringe width is given by:
\(
\beta=\frac{\lambda D}{d}
\)
But in our question Young’s double slit experiment is performing under water.
And we know that water has a refractive index higher than that of air.
Refractive index: The ratio of speed of light in one medium with respect to another medium. In general the first medium is air. Refractive Index of air is 1. Medium other than air has a refractive index greater than one.
Refractive Index is a unit less quantity and generally represented by \(n\).
\(
\begin{aligned}
& n=\frac{v_{\text {air }}}{v_{\text {medium }}} \\
& n=\frac{\lambda_{\text {air }}}{\lambda_{\text {medium }}} \\
& \lambda_{\text {medium }}=\frac{\lambda_{\text {air }}}{n}
\end{aligned}
\)
Hence wavelength in the medium other than air will decrease.
Also, \(\beta=\frac{\lambda D}{d}\).
Here, distance between the slit and screen ( \(D\) ) and distance between two slits ( \(d\) ) remains unchanged under water. Therefore,
\(
\begin{aligned}
& \beta \propto \lambda \\
& \beta_{\text {medium }} \propto \lambda_{\text {medium }} \\
& \beta_{\text {medium }}=\frac{\lambda_{\text {air }}}{n} \\
& \beta_{\text {medium }}=\frac{\beta_{\text {air }}}{n}
\end{aligned}
\)
It is clear from the above equation, Fringe width will decrease when we perform Young’s Double Slit Experiment under water.
A light wave can travel
(a, c) First of all let us take a look at the basic concepts of a light wave. Light is referred to as a transverse wave as well as an electromagnetic wave that can be visible by the typical human eye. Like all electromagnetic waves, light can travel through vacuum too. Light waves are able to travel through a vacuum, which do not need a medium to travel. In empty space or vacuum, the wave does not dissipate or grow smaller no matter how far it moves, since the wave is not making any interaction with anything else. As a wave gets travelled from a source, it will propagate outward in all directions. As it is not requiring any extra medium to be travelled, the media has no significance in the case of this propagation. Therefore even if it is a material medium, the light wave can travel. The only difference is that the velocity with which the propagation occurs will be different in different mediums. Therefore we can conclude that light can travel through a material medium and vacuum also.
Which of the following properties of light conclusively support wave theory of light?
(b, c) Snell’s Law, which states that the speed of light reduces on moving from a rarer to a denser medium, can be concluded from Huygens’ wave theory and interference of light waves is based on the wave properties of light.
When light propagates in vacuum there is an electric field and a magnetic field. These fields
(b, c, d) When light propagates in a vacuum, the electric and magnetic fields have a zero average value, are perpendicular to the direction of propagation, and are mutually perpendicular.
Explanation:
Zero Average Value: The electric and magnetic fields in a light wave oscillate sinusoidally. When averaged over a complete cycle, these oscillations result in a zero average value for both fields.
Perpendicular to Propagation: Light is an electromagnetic wave, and both the electric and magnetic fields are perpendicular to the direction the wave is traveling. This is a characteristic of transverse waves.
Mutually Perpendicular: The electric and magnetic fields themselves are also perpendicular to each other.
Huygens’ principle of secondary wavelets may be used to
(c, d) Huygen’s wave theory explains the origin of points for the new wavefront proceeding successively. It also explains the variation in speed of light on moving from one medium to another, i.e. it proves Snell’s Law.
Explanation: Huygens’ principle states that every point on a wavefront acts as a source of secondary wavelets, and the envelope of these wavelets determines the new position of the wavefront at a given time. The word interference is used to describe the superposition of two waves, whereas diffraction is interference produced by several waves. By this principle, we can find the exact position of the next ripple/wavefront that is to be produced.
Three observers \(A, B\) and \(C\) measure the speed of light coming from a source to be \(v_A, v_B\) and \(v_C\). The observer \(A\) moves towards the source and \(C\) moves away from the source at the same speed. The observer \(B\) stays stationary. The surrounding space is vacuum everywhere.
(c, d) According to the principle of relativity, all observers, regardless of their motion relative to the light source, will measure the same speed of light in a vacuum. Therefore, regardless of whether observer A is moving towards the source, observer C is moving away from the source, or observer B is stationary, they will all measure the same speed of light, denoted as “c”.
Since the speed of light is a universal constant, \(v_A=v_B=v_C=3 \times 10^8\) \(v_B=\frac{1}{2}\left(v_A+v_C\right)\)
This expression also implies that \(v_{v{A}}=v_{v{B}}=v_{v{C}}\).
Three observers \(A, B\) and \(C\) measure the speed of light coming from a source to be \(v_A, v_B\) and \(v_C\). The observer \(A\) moves towards the source and \(C\) moves away from the source at the same speed. The observer \(B\) stays stationary. Consider the medium as water. Select the correct option(s) from the list given in that
question.
(a, d) Step 1: Understanding the Setup:
Observer A is moving towards the light source.
Observer B is stationary.
Observer C is moving away from the light source.
The medium is water.
Step 2: Speed of Light in Different Mediums:
The speed of light in a vacuum is constant, but in a medium like water, it is affected by the motion of the observer relative to the medium.
The speed of light in water is less than in vacuum, but the perceived speed will change based on the observer’s motion.
Step 3: Analyzing Observer A:
Since observer A is moving towards the source, the speed of light measured by A (\(v_A\)) will be greater than the speed measured by B (\(v_B\)), who is stationary. Thus, we can write:
\(
v_A>v_B
\)
Step 4: Analyzing Observer C:
Observer C is moving away from the source. Therefore, the speed of light measured by C (\(v_C\)) will be less than the speed measured by B (\(v_B\)). Thus, we can write:
\(
v_B>v_C
\)
Step 5: Combining the Relationships:
From the above two inequalities, we can combine them to get:
\(
v_A>v_B>v_C
\)
Step 6: Equal Magnitudes of \(v_A\) and \(v_C\):
The problem states that the magnitudes of the speeds of \(A\) and \(C\) are equal but in opposite directions. This means:
\(
|v_A|=|v_C|
\)
Since they are in opposite directions, this does not affect the inequality derived earlier.
Step 7: Finding the Average Speed:
We can express the relationship mathematically. Since \(v_A\) and \(v_C\) are equal in magnitude:
\(
v_B=\frac{v_A+v_C}{2}
\)
This shows that the speed of \(B\) is the average of the speeds of \(A\) and \(C\).
Conclusion: From the analysis, we conclude that:
The correct relationship is:
\(
v_A>v_B>v_C
\)
Additionally, we have:
\(
v_B=\frac{v_A+v_C}{2}
\)
Light waves travel in vacuum along the \(X\)-axis. Which of the following may represent the wavefronts?
(a) \(x}=c}\).
Explanation:
Wavefronts are perpendicular to the direction of wave propagation. Since light waves are traveling along the X -axis, the wavefronts must be planes perpendicular to the X -axis. The equation \({x}={c}\) represents a plane perpendicular to the X -axis.
Why other options are incorrect:
(b) \(y=c\) : This represents a plane parallel to the XZ-plane, which is not perpendicular to the Xaxis.
(c) \(z=c\) : This represents a plane parallel to the XY-plane, which is also not perpendicular to the \(X\)-axis.
(d) \(x+y+z=c\) : This represents a three-dimensional surface, not a plane perpendicular to the X axis.
If the source of light used in a Young’s double slit experiment is changed from red to violet,
(b) Consecutive fringes will come closer.
Explanation:
In a Young’s double-slit experiment, the fringe width is directly proportional to the wavelength of light used. Since violet light has a shorter wavelength than red light, the fringe width will decrease when switching from red to violet, meaning the fringes will come closer together.
Why other options are incorrect:
(a) the fringes will become brighter: Changing the wavelength of light does not affect the brightness of the fringes. The brightness depends on the intensity of the light source, not the wavelength.
(c) the intensity of minima will increase: The intensity of the minima is related to the contrast between the bright and dark fringes, not the wavelength. Changing the wavelength affects the fringe spacing, not the intensity of the minima.
(d) the central bright fringe will become a dark fringe: The central bright fringe is always the result of constructive interference and will remain bright regardless of the wavelength used. Changing the wavelength only affects the spacing of the other fringes, not the central one.
A Young’s double slit experiment is performed with white light.
(a, b, d) The superposition of all the colours at the central maxima gives the central band a white colour. As we go from the centre to corner, the fringe colour goes from violet to red. There will not be a completely dark fringe, as complete destructive interference does not take place.
Explanation:
Step 1: Understanding White Light in Young’s Double Slit Experiment: In a Young’s double slit experiment, when white light is used instead of monochromatic light, the light consists of multiple wavelengths corresponding to different colors (red, orange, yellow, green, blue, indigo, violet).
Step 2: Formation of Fringes: Each color in white light will create its own interference pattern due to the different wavelengths. The central bright fringe will be formed by the superposition of all colors, resulting in a white light fringe.
Step 3: Wavelength Order: The wavelengths of the colors vary, with violet having the shortest wavelength and red having the longest. This means that the fringe positions will vary based on the wavelength:
Violet (shortest wavelength) will be closest to the central white fringe.
Red (longest wavelength) will be farthest from the central white fringe.
Step 4: Arrangement of Colors: As we move away from the central white fringe, the colors will appear in the following order:
Closer to the center: Violet, Indigo, Blue, Green, Yellow, Orange, and finally Red.
Step 5: Conclusion: The central fringe will be white, and the colors will spread out on either side, with violet being closest to the center and red being the farthest. There will not be a completely dark fringe since white light combines all colors.
Four light waves are represented by
(i) \(y=a_1 \sin \omega t\)
(ii) \(y=a_2 \sin (\omega t+\varepsilon)\)
(iii) \(y=a_1 \sin 2 \omega t\)
(iv) \(y=a_2 \sin 2(\omega t+\varepsilon)\).
Interference fringes may be observed due to superposition of
(a, d) Interference fringes are observed due to the superposition of coherent light waves. Coherent light refers to light waves with a constant phase relationship and the same frequency.
Explanation:
(i) \(y=a_1 \sin \omega t\) : This represents a wave with angular frequency \(\omega\).
(ii) \(y=a_2 \sin (\omega t+\varepsilon)\) : This represents a wave with the same angular frequency \(\omega\) and a phase difference of \(\boldsymbol{\varepsilon}\).
(iii) \(y=a_1 \sin 2 \omega t\) : This represents a wave with angular frequency \(2 \omega\).
(iv) \(y=a_2 \sin 2(\omega t+\varepsilon)\) : This represents a wave with angular frequency \(2 \omega\) and a phase difference of \(2 \varepsilon\).
(i) and (ii): Both have the angular frequency \(\boldsymbol{\omega}\). They are coherent as they have the same frequency and a constant phase difference \((\boldsymbol{\varepsilon})\).
(iii) and (iv): Both have the angular frequency \(2 \omega\). They are coherent as they have the same frequency and a constant phase difference ( \(2 \varepsilon\) ).
Based on the conditions for interference, interference fringes can be observed due to the superposition of waves with the same frequency.
Two sources with intensity \(I_0\) and \(4 I_0\) respectively, interfere at a point in a medium. Find the ratio of (i) maximum and minimum intensities, (ii) amplitudes.
(a) (i) Maximum intensity, \(I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)
\(
\begin{array}{ll}
\text { Here, } & I_1=I_0 \text { and } I_2=4 I_0 \\
\therefore & I_{\max }=\left(\sqrt{I_0}+\sqrt{4 I_0}\right)^2=9 I_0
\end{array}
\)
and minimum intensity, \(I_{\text {min }}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=I_0\)
\(
\Rightarrow \quad \frac{I_{\max }}{I_{\min }}=\frac{9 I_0}{I_0}=9: 1
\)
(ii) Ratio of intensities, \(\frac{I_{\text {max }}}{I_{\text {min }}}=\left(\frac{r+1}{r-1}\right)^2\)
where, \(r=\) ratio of amplitudes.
\(
\begin{array}{ll}
\Rightarrow & \frac{9}{1}=\left(\frac{r+1}{r-1}\right)^2 \Rightarrow 3=\frac{r+1}{r-1} \\
\Rightarrow & 3 r-3=r+1 \Rightarrow 2 r=4 \Rightarrow r=2: 1
\end{array}
\)
According to corpuscular theory of light, the different colours of light are due to
(c) According to the corpuscular theory of light, different colors are due to different sizes of the corpuscles.
Explanation: The corpuscular theory, proposed by Isaac Newton, envisioned light as tiny particles called corpuscles. This theory suggested that different colors of light correspond to different sizes of these corpuscles.
Why other options are incorrect:
(a) different electromagnetic waves: The wave theory of light, which superseded the corpuscular theory, explains color in terms of different wavelengths of electromagnetic waves.
(b) different force of attraction among the corpuscles: The corpuscular theory didn’t explain color based on forces of attraction.
(d) None of the above: Since the corpuscular theory explains color based on corpuscle size, option (d) is incorrect.
Two identical light sources \(S_1\) and \(S_2\) emit light of same wavelength \(\lambda\). These light rays will exhibit interference, if
(a)Â For two light rays to exhibit interference, their phase differences must remain constant.
Explanation: Interference occurs when waves overlap and their phases interact, creating a resulting wave with a combined amplitude. For this to happen, the phase difference between the waves needs to be consistent over time.
Why the other options are incorrect:
(b) Their phases are distributed randomly:
Randomly distributed phases would not lead to a constant phase difference, resulting in incoherent light and no interference.
(c) Their light intensities remain constant:
While constant intensity might be a condition for observing interference patterns, it is not the defining factor. The crucial condition is a constant phase difference.
(d) Their light intensities change randomly:
Changing light intensities would also disrupt the necessary constant phase difference, preventing interference.
When interference of light takes place
(c) Conservation of energy holds good and energy is redistributed.
Explanation: In interference, the energy of light waves is not created or destroyed, but rather it is redistributed across different regions, resulting in areas of higher intensity (constructive interference) and lower intensity (destructive interference). This redistribution is consistent with the principle of conservation of energy.
Why other options are incorrect:
(a) Energy is created in the region of maximum intensity:
This is not true. While the intensity is higher at regions of constructive interference, the total energy remains constant. Energy is not being created.
(b) Energy is destroyed in the region of minimum intensity:
Energy is not destroyed in any region during interference. The energy is simply shifted from areas of lower intensity to areas of higher intensity.
(d) Conservation of energy does not hold good:
This is incorrect. Interference is a phenomenon that demonstrably adheres to the principle of conservation of energy. The total energy of the light waves involved remains constant.
If two waves represented by \(y_1=4 \sin \omega t[latex] and [latex]y_2=3 \sin \left(\omega t+\frac{\pi}{3}\right)\) interfere at a point, then the amplitude of the resulting wave will be about
(b) Here, phase difference, \(\phi=\pi / 3\)
So, amplitude of resulting wave,
\(
\begin{aligned}
A & =\sqrt{a_1^2+a_2^2+2 a_1 a_2 \cos \phi} \\
& =\sqrt{4^2+3^2+2 \times 4 \times 3 \cos \pi / 3} \approx 6 \mathrm{~m}
\end{aligned}
\)
Two coherent sources of intensities \(I_1\) and \(I_2\) produce an interference pattern. The maximum intensity in the interference pattern will be
(d) Resultant intensity for two interfering waves is given by
\(
I_R=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi .
\)
For maximum, \(\cos \phi=\) maximum \(=1\)
\(
\therefore \quad I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2
\)
What is the path difference in case of destructive interference?
\(
\begin{aligned}
&\text { (d) For destructive interference, }\\
&\begin{aligned}
\cos \phi & =\text { minimum }=-1 \\
\phi & =\pi, \quad 3 \pi, \quad 5 \pi \\
\phi & =(2 n+1) \pi \\
n & =0,1,2,3, \ldots \\
\Delta x & =\frac{\lambda}{2 \pi} \times \phi=\frac{(2 n+1) \lambda}{2}
\end{aligned}
\end{aligned}
\)
If the amplitude ratio of two sources producing interference is \(3: 5\), then the ratio of intensities at maxima and minima is
\(
\begin{aligned}
&\text { (c) In interference, }\\
&\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}
\end{aligned}
\)
\(
\begin{array}{ll}
\text { Given, } & \frac{A_1}{A_2}=\frac{3}{5} \\
\therefore & \frac{I_{\max }}{I_{\min }}=\frac{(3+5)^2}{(3-5)^2}=\frac{16}{1}
\end{array}
\)
Two sources of light of the same frequency produce intensities \(I\) and \(4 I\) at a point \(P\) when used individually. If they are used together such that the light from them reach \(P\) with a phase difference of \(2 \pi / 3\), then the intensity at \(P\) will be
(b)
\(
\begin{aligned}
I & =I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi \\
& =I+4 I+2 \sqrt{I \times 4 I} \times \cos \frac{2 \pi}{3} \\
& =5 I+4 I \times\left(-\frac{1}{2}\right)=5 I-2 I=3 I
\end{aligned}
\)
Two incoherent sources of intensities \(I\) and \(4 I\) superpose, then the resultant intensity is
(a) When two incoherent sources of light superpose, the resultant intensity is simply the sum of the individual intensities. This is because incoherent sources do not have a constant phase difference, meaning the phase relationship between the waves from the sources varies randomly.
Given that the intensities of the two incoherent sources are \(\boldsymbol{I}\) and \(4 \boldsymbol{I}\), the resultant intensity \(\left(\boldsymbol{I}_{\boldsymbol{R}}\right)\) will be:
\(
\begin{aligned}
& I_R=I_1+I_2 \\
& I_R=I+4 I \\
& I_R=5 I
\end{aligned}
\)
Which of the following is not an essential condition for interference?
(c)Â The condition that is not essential for interference is (c) The amplitudes of the two waves must be equal.
Explanation: While having similar amplitudes can lead to more visible interference patterns, it’s not a strict requirement for interference to occur.
Key points about interference conditions:
Coherence:
The waves must be coherent, meaning they have a constant phase relationship. This often means they come from the same source.
Same wavelength/frequency:
The waves must have the same wavelength (or equivalent frequency) to interfere constructively or destructively.
Similar direction:
Although not perfectly aligned, the waves should propagate in roughly the same direction for significant interference to be observed.
In a YDSE, green light of wavelength 500 nm is used. Where will be the second bright fringe formed for a set up in which separation between slits is 4 mm and the screen is placed 1 m from the slits?
(b) Position of second bright fringe,
\(
\begin{aligned}
y_2 & =\frac{2 \lambda D}{d} \\
& =\frac{2 \times 500 \times 10^{-9} \times 1}{4 \times 10^{-3}} \\
& =0.25 \mathrm{~mm}
\end{aligned}
\)
Note While reporting our answer we should report \(y_2= \pm 0.25 \mathrm{~mm}\) as fringe will be formed both, above and below central bright fringe.
Bichromatic light is used in YDSE having wavelengths \(\lambda_1=400 \mathrm{~nm}\) and \(\lambda_2=700 \mathrm{~nm}\). Find minimum order of \(\lambda_1\) which overlaps with \(\lambda_2\).
(d) Let \(n_1\) bright fringe of \(\lambda_1\) overlaps with \(n_2\) bright fringe of \(\lambda_2\).
Then, their position will be same, i.e.
\(
\begin{aligned}
\frac{n_1 \lambda_1 D}{d} & =\frac{n_2 \lambda_2 D}{d} \\
\frac{n_1}{n_2} & =\frac{\lambda_2}{\lambda_1} \\
& =\frac{700}{400}=\frac{7}{4}
\end{aligned}
\)
The ratio \(\frac{n_1}{n_2}=\frac{7}{4}\) implies that 7th bright fringe of \(\lambda_1\) will overlap with 4th bright fringe of \(\lambda_2\).
Similarly, 14 th bright fringe of \(\lambda_1\) will overlap with 8 th bright fringe of \(\lambda_2\) and so on.
So, the minimum order of \(\lambda_1\) which overlaps with \(\lambda_2\) is 7.
Young’s double slit experiment is carried out using microwaves of wavelength \(\lambda=3 \mathrm{~cm}\). Distance between the slits is \(d=5 \mathrm{~cm}\) and the distance between the plane of slits and the screen is \(D=100 \mathrm{~cm}\). Find the number of maximas and their positions on the screen.
(c) The maximum path difference that can be produced \(=\) distance between the sources or 5 cm.
Thus, in this case, we can have only three maximas, one central maxima and two on its either side for a path difference of \(\lambda=3 \mathrm{~cm}\).
For maximum intensity at \(P\),
\(
S_2 P-S_1 P=\lambda
\)
\(
\sqrt{(y+d / 2)^2+D^2}-\sqrt{(y-d / 2)^2+D^2}=\lambda
\)
\(
\begin{aligned}
&\begin{aligned}
& \text { Here, } d=5 \mathrm{~cm}, D=100 \mathrm{~cm} \\
& \text { and } \\
& \therefore \quad \lambda=3 \mathrm{~cm} \\
& \therefore \quad y= \pm 75 \mathrm{~cm}
\end{aligned}\\
&\text { Thus, the three maximas will be at }\\
&y=0 \text { and } y= \pm 75 \mathrm{~cm}
\end{aligned}
\)
In YDSE, the two slits are separated by 0.1 mm and they are 0.5 m from the screen. The wavelength of light used is \(5000 Ã…\). Find the distance between 7th maxima and 11 th minima on the screen.
(d) Given, \(d=0.1 \mathrm{~mm}=10^{-4} \mathrm{~m}, D=0.5 \mathrm{~m}\)
and \(\lambda=5000 Ã…=5.0 \times 10^{-7} \mathrm{~m}\)
The distance between 7th maxima and 11th minima is
\(
\begin{aligned}
\Delta y & =\left(y_{11}\right)_{\text {dark }}-\left(y_7\right)_{\text {bright }} \\
& =\frac{(2 \times 11-1) \lambda D}{2 d}-\frac{7 \lambda D}{d} \\
\Delta y & =\frac{7 \lambda D}{2 d}=\frac{7 \times 5.0 \times 10^{-7} \times 0.5}{2 \times 10^{-4}}=8.75 \times 10^{-3} \mathrm{~m} \\
& =8.75 \mathrm{~mm}
\end{aligned}
\)
In YDSE, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the first dark fringe and fourth bright fringe is obtained to be 0.6 cm. Determine the wavelength of the light used in the experiment.
(a) Also, separation between first dark fringe and fourth bright fringe is 0.6 cm.
Now, position of 1st dark fringe from central maximum be
\(
y_1=\frac{\lambda D}{2 d}
\)
Position of 4th bright fringe will be \(y_2=\frac{4 \lambda D}{d}\)
Separation between them is equal to \(\frac{4 \lambda D}{d}-\frac{\lambda D}{2 d}=0.6 \mathrm{~cm}\)
\(
\frac{7 \lambda D}{2 d}=0.6 \mathrm{~cm}
\)
Wavelength of the light,
\(
\begin{aligned}
\lambda & =\frac{0.6 \times 10^{-2} \times 2 \times 0.28 \times 10^{-3}}{7 \times 1.4} \\
& =3.428 \times 10^{-7} \mathrm{~m}
\end{aligned}
\)
Fringe width in a particular YDSE is measured to be \(\beta\). What will be the fringe width, if the wavelength of the light is doubled, separation between the slits is halved and separation between the screen and slits is tripled?
(b) We know that, \(\beta=\frac{\lambda D}{d}\)
Now, the changed values are \(\lambda^{\prime}=2 \lambda, d^{\prime}=\frac{d}{2}[latex] and [latex]D^{\prime}=3 D\)
The new fringe width will be
\(
\begin{aligned}
& \beta^{\prime}=\frac{\lambda^{\prime} D^{\prime}}{d^{\prime}} \\
\Rightarrow \quad & \beta^{\prime}=\frac{(2 \lambda)(3 D)}{\left(\frac{d}{2}\right)}=12 \frac{\lambda D}{d}=12 \beta
\end{aligned}
\)
The fringe width increases to 12 times of the original.
Whose fringe width will be larger, the one for red light or the one for yellow light, all other fringes be the same?
(a) We know that, fringe width, \(\beta=\frac{\lambda D}{d}\)
Now, for same value of \(D\) and \(d, B \propto \lambda\).
\(
\lambda_{\text {red }}>\lambda_{\text {yellow }}
\)
Hence,
\(
\beta_{\text {red }}>\beta_{\text {yellow }}
\)
Hence, fringe width will be greater when red coloured light is used.
Two slits in YDSE are placed 1 mm from each other. The interference pattern is observed on a screen placed 1 m from the plane of slits. What is the angular fringe width for a light of wavelength 400 nm?
(c) Given, \(d=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)
\(
D=1 \mathrm{~m}, \lambda=400 \mathrm{~nm}=400 \times 10^{-9} \mathrm{~m}
\)
Angular fringe width, \(\theta=\frac{\lambda}{d}=\frac{400 \times 10^{-9}}{10^{-3}}=400 \times 10^{-6} \mathrm{rad}\)
Two slits are separated by 0.32 mm . A beam of 500 nm strikes the slits producing an interference pattern. Determine the number of maxima observed in the angular range \(-30^{\circ}<\theta<30^{\circ}\).
(b) Fringe width, \(\beta=\frac{\lambda D}{d}\) and \(y=\frac{D}{\sqrt{3}}\)
Therefore, number of fringe width in a distance \(y\),
\(
n=\frac{y}{\beta}=\frac{d}{\sqrt{3} \lambda}=\frac{0.32 \times 10^{-3}}{(\sqrt{3})\left(500 \times 10^{-9}\right)}=369.5
\)
Therefore, total number of maxima obtained in the angular range \(-30^{\circ}<\theta<30^{\circ}\) (including the central one) is
\(
N=2 n+1=2 \times 369.5+1=739+1=740
\)
Two coherent sources each emitting light of intensity \(I_0\) interfere, in a medium at a point, where phase difference between them is \(\frac{2 \pi}{3}\). Then, find the resultant intensity at that point.
(b) Given, phase difference, \(\phi=\frac{2 \pi}{3}\) and resultant intensity, \(I=4 I_0 \cos ^2\left(\frac{\phi}{2}\right)=4 I_0 \cos ^2(\pi / 3)=I_0\)
In a Young’s double slit set up using monochromatic light of wavelength \(\lambda\), the intensity of light at a point, where path difference is \(2 \lambda\) is found to be \(I_0\). What will be the intensity at a point where path difference is \(\lambda / 3\)?
(c) We know that, path difference is related to the phase difference according to the relation, \(\phi=\frac{2 \pi x}{\lambda}\)
Now, phase difference, where path difference ( \(\lambda / 3\) ) is equal to \(\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}\).
Also, at the point, where path difference \(=2 \lambda\)
Phase difference will be \(\frac{2 \pi}{\lambda} \times 2 \lambda=4 \pi\)
Let individual source intensity be \(I\).
Then,
\(
I_0=4 I \cos ^2 \frac{4 \pi}{2}=4 I
\)
Hence,
\(
I=\frac{I_0}{4}
\)
Now, at the point, where phase difference is \(\phi=\frac{2 \pi}{3}\)
\(
\begin{aligned}
& I^{\prime}=4 I \cos ^2 \frac{2 \pi}{3 \times 2} \\
& I^{\prime}=4 I \cos ^2\left(\frac{\pi}{3}\right)=I=\frac{I_0}{4}
\end{aligned}
\)
Maximum intensity in YDSE is \(I_0\). Find the intensity at a point on the screen, where
(i) the phase difference between the two interfering beams is \(\frac{\pi}{3}\).
(ii) the path difference between them is \(\frac{\lambda}{4}\).
(c)
(i) We know that, intensity,
\(
I=I_{\max } \cos ^2\left(\frac{\phi}{2}\right)
\)
Here, \(I_{\text {max }}\) is \(I_0\) (i.e. intensity due to independent sources is \(I_0 / 4\) ).
Therefore, at phase difference, \(\phi=\frac{\pi}{3}\) or \(\frac{\phi}{2}=\frac{\pi}{6}\)
\(\therefore \quad\) Intensity, \(I=I_0 \cos ^2\left(\frac{\pi}{6}\right)=\frac{3}{4} I_0\)
(ii) Phase difference corresponding to the given path difference, \(\Delta x=\frac{\lambda}{4}\) is
\(
\begin{array}{lll}
& \phi=\left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{4}\right) & \left(\because \phi=\frac{2 \pi}{\lambda} \Delta x\right) \\
\text { or } & \frac{\phi}{2}=\frac{\pi}{4} & \\
\therefore & I=I_0 \cos ^2\left(\frac{\pi}{4}\right)=\frac{I_0}{2} &
\end{array}
\)
In a Young’s double slit set up the wavelength of light used is 546 nm. The distance of screen from slits is 1 m. The slit separation is 0.3 mm.
(i) Compare the intensity at a point \(P\) distant 10 mm from the central fringe, where the intensity is \(I_0\).
(ii) Find the number of bright fringes between \(P\) and the central fringe.
(b) Given,
\(
\begin{aligned}
\lambda & =546 \mathrm{~nm}=5.46 \times 10^{-7} \mathrm{~m}, \\
D & =1 \mathrm{~m} \text { and } d=0.3 \mathrm{~mm}=0.3 \times 10^{-3} \mathrm{~m}
\end{aligned}
\)
(i) At a distance, \(y=10 \mathrm{~mm}=10 \times 10^{-3} \mathrm{~m}\) from the central fringe, the path difference will be
\(
\begin{aligned}
\Delta x=\frac{y \cdot d}{D} & =\frac{\left(10 \times 10^{-3}\right)\left(0.3 \times 10^{-3}\right)}{1} \\
& =3.0 \times 10^{-6} \mathrm{~m}
\end{aligned}
\)
The corresponding phase difference between the two beams will be
\(
\begin{aligned}
\phi & =\frac{2 \pi}{\lambda} \cdot \Delta x \\
& =\left(\frac{2 \pi}{5.46 \times 10^{-7}}\right)\left(3.0 \times 10^{-6}\right) \mathrm{rad} \\
& =34.5 \times \frac{180^{\circ}}{\pi} \\
& \approx 1977^{\circ}
\end{aligned}
\)
\(
\therefore \quad I=I_0 \cos ^2 \frac{\phi}{2}=I_0 \cos ^2\left(988^{\circ}\right)=3.0 \times 10^{-4} I_0
\)
(ii) Fringe width,
\(
\begin{aligned}
\beta=\frac{\lambda D}{d} & =\frac{\left(5.46 \times 10^{-7}\right)(1)}{0.3 \times 10^{-3}} \mathrm{~m} \\
& =1.82 \mathrm{~mm}
\end{aligned}
\)
Since, \(\frac{y}{\beta}=\frac{10}{1.82}=5.49\)
Therefore, number of bright fringes between \(P\) and central fringe will be 5 (excluding the central fringe).
Light travels in a mica sheet of refractive index 1.4 and length 10 cm. Find the optical path equivalent to length of the sheet.
(d) Given, length of the sheet, \(t=10 \mathrm{~cm}\)
Refractive index, \(\mu=1.4\)
\(\therefore\) Optical path \(=\mu t=(1.4) \times(10)=14 \mathrm{~cm}\)
Interference fringes are produced by a double slit arrangement and a piece of plane parallel glass of refractive index 1.5 is interposed in one of the interfering beam. If the fringes are displaced through 30 fringe widths for light of wavelength \(6 \times 10^{-5} \mathrm{~cm}\), then find the thickness of the plate.
(d) Path difference due to the glass slab, \(\Delta x=(\mu-1) t \dots(i)\)
Thirty fringes are displaced due to the slab.
Hence, \(\Delta x=30 \lambda \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
\begin{aligned}
(\mu-1) t & =30 \lambda \\
\therefore \quad t & =\frac{30 \lambda}{\mu-1}=\frac{30 \times 6 \times 10^{-5}}{1.5-1}=3.6 \times 10^{-3} \mathrm{~cm}
\end{aligned}
\)
In YDSE using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.5 and thickness 2 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the plane of slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the light.
(a) Shift \(=\frac{(\mu-1) t D}{d}\)
New fringe width \(=\frac{2 D \lambda}{d} \Rightarrow \frac{(\mu-1) t D}{d}=\frac{2 D \lambda}{d}\)
Therefore, wavelength of light,
\(
\lambda=\frac{(\mu-1) t}{2}=\frac{(1.5-1) \times 2 \times 10^{-6}}{2} \mathrm{~m}=0.5 \times 10^{-6} \mathrm{~m}
\)
In YDSE, find the thickness of a glass slab ( \(\mu=1.5\) ) which should be placed before the upper slit \(S_1\), so that the central maximum now lies at a point, where 5 th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is \(5000 Ã…\).
According to the question, shift \(=5 \times\) (fringe width)
\(
\begin{aligned}
& \Rightarrow \quad \frac{(\mu-1) t D}{d}=\frac{5 \lambda D}{d} \\
& \therefore \text { Thickness of a glass slab, } t=\frac{5 \lambda}{\mu-1}=\frac{25000}{1.5-1}=50000 Ã…
\end{aligned}
\)
Calculate the minimum thickness of a soap bubble film \((\mu=1.33)\) that results in constructive interference in the reflected light, if the film is illuminated with light whose wavelength in free space is \(\lambda=600 \mathrm{~nm}\).
(c) For constructive interference in case of soap bubble film,
\(
2 \mu t=\left(n-\frac{1}{2}\right) \lambda \text { (where, } n=1,2,3, \ldots \text { ) }
\)
For minimum thickness \(t, n=1\)
\(
\Rightarrow \quad 2 \mu t=\lambda / 2
\)
Thus, the minimum thickness of a soap bubble film is
\(
\begin{aligned}
t=\frac{\lambda}{4 \mu} & =\frac{600}{4 \times 1.33} \\
& =112.78 \mathrm{~nm}
\end{aligned}
\)
In solar cells, silicon \(\mathrm{Si}(\mu=3.5)\) is coated with a thin film of silicon monoxide \(\mathrm{SiO}(\mu=1.45)\) to minimise reflective losses from the surface. Determine the minimum thickness of SiO that produces the least reflection at a wavelength of 550 nm , near the centre of the visible spectrum.
(b)Â The reflected light is minimum when ray 1 and ray 2 (shown in figure) meet the condition of destructive interference.
Both rays undergo a \(180^{\circ}\) phase change upon reflection. The net change in phase due to reflection is therefore zero and the condition for a minimum reflection requires a path difference of \(\lambda_\mu / 2\).
For minimum thickness \(t, n=1\)
\(
\Rightarrow \quad 2 t=\frac{\lambda}{2 \mu}
\)
Thus, the minimum thickness of silicon monoxide is
\(
t=\frac{\lambda}{4 \mu}=\frac{550}{4(1.45)}=94.8 \mathrm{~nm}
\)
Note: When we observe this film fringes through refracted light waves, then condition for maxima is \(2 t=n \lambda \mu\) and for maxima, \(2 t=(2 n-1) \frac{\lambda \mu}{2}\) (where, \(\left.n=1,2,3, \ldots\right)\).
Light source of wavelength 400 nm used in a Lloyd’s mirror, lies 1 mm above the plane mirror. Find the distance of first minima above the mirror. Distance of the screen from the plane of the sources is 2 m.
(d) Consider the formation of minimum at point \(P\) as shown in the figure. Path difference between the rays reaching at point \(P\) is
\(
\Delta x=\frac{d}{2} \sin \theta=\lambda \quad \text { (for first minimum) }
\)
\(
d \times \tan \theta \approx 2 \lambda \quad(\because \sin \theta \approx \tan \theta)
\)
\(
\begin{aligned}
&\begin{array}{ll}
\Rightarrow & d \times \frac{y}{D}=2 \lambda \\
\Rightarrow & y=\frac{D}{d}(2 \lambda)
\end{array}\\
&\therefore \text { Distance of first minima above the mirror, }\\
&\begin{aligned}
y & =\frac{2}{2 \times 10^{-3}} \times 2 \times 400 \times 10^{-9} \\
& =0.8 \mathrm{~mm}
\end{aligned}
\end{aligned}
\)
In Young’s double slit experiment, an interference pattern is obtained for \(\lambda=6000 Ã…\), coming from two coherent sources \(S_1\) and \(S_2\). At certain point \(P\) on the screen third dark fringe is formed. Then, the path difference \(S_1 P-S_2 P\) (in micron) is
\(
\begin{aligned}
&\text { (b) For dark fringe at point } P \text {, }\\
&\begin{aligned}
S_1 P-S_2 P & =\frac{5 \lambda}{2} \\
& =5 \times \frac{6000}{2}=15000 Ã… \\
& =1.5 \mathrm{micron}
\end{aligned}
\end{aligned}
\)
In Young’s double slit experiment, when two light waves form third minimum intensity, they have
(d) For minima, path difference should be an odd multiple of half wavelength.
i.e. \(\quad \Delta x=(2 n-1) \frac{\lambda}{2}\)
For third minima \((n=3), \Delta x=(2 \times 3-1) \frac{\lambda}{2}=\frac{5 \lambda}{2}\)
Two slits are separated by a distance of 0.5 mm and illuminated with light of \(\lambda=6000 Ã…\). If the screen is placed 2.5 m from the slits. The distance of the third bright image from the centre will be
(d) Distance of \(n\)th bright fringe from the centre, \(y_n=\frac{n D \lambda}{d}\)
So,
\(
\begin{aligned}
y_3 & =\frac{3 \times 6000 \times 10^{-10} \times 2.5}{0.5 \times 10^{-3}} \\
& =9 \times 10^{-3} \mathrm{~m}=9 \mathrm{~mm}
\end{aligned}
\)
Fringe width decreases with increase in
(c) \(\beta=\frac{\lambda D}{d} \Rightarrow \beta \propto \frac{1}{d}\)
Hence, fringe width decreases with increase in \(d\).
In Young’s double slit experiment, the fringe width will remain same, if ( \(D=\) distance between screen and plane of slits, \(d=\) separation between two slits and \(\lambda=\) wavelength of light used)
(b) Fringe width, \(\beta=\frac{D \lambda}{d}\)
If both \(d\) and \(D\) are doubled, then \(\beta^{\prime}=\frac{(2 D) \lambda}{2 d}=\frac{D \lambda}{d}=\beta\)
Interference was observed in interference chamber when air is present. Now, the chamber is evacuated and if the same light is used, then for the same arrangement
(d) When chamber is evacuated, the refractive index \((\mu)\) decreases. Therefore, wavelength \(\left(\lambda \propto \frac{1}{\mu}\right)\) increases.
Now, fringe width is given by
\(
\begin{aligned}
& \beta=\frac{\lambda D}{d} \\
& \Rightarrow \beta \propto \lambda
\end{aligned}
\)
Therefore, as wavelength increases, the fringe width will increase.
Monochromatic green light of wavelength \(5 \times 10^{-7} \mathrm{~m}\) illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2 m away is
\(
\begin{aligned}
&\text { (c) Separation of bright lines or fringe width }\\
&\beta=\frac{D \lambda}{d}=\frac{5 \times 10^{-7} \times 2}{10^{-3}}=10^{-3} \mathrm{~m}=1.0 \mathrm{~mm}
\end{aligned}
\)
In double slit experiment, for light of which colour, the fringe width will be minimum?
(a) Fringe width in double slit experiment is given by
\(
\beta=\frac{D \lambda}{d}
\)
It implies that \(\beta \propto \lambda\), so in given options \(\lambda_{\text {violet }}=\) minimum.
In Young’s double slit experiment, green light \((\lambda=5461 Ã…)\) is used and 60 fringes were seen in the field view. Now, sodium light is used \((\lambda=5890 Ã…)\), then number of fringes observed are
(d) We have,
\(
n_1 \lambda_1=n_2 \lambda_2
\)
Number of fringes, \(n_2=\frac{n_1 \lambda_1}{\lambda_2}=\frac{60 \times 5461 \times 10^{-10}}{5890 \times 10^{-10}}=55\)
In Young’s double slit experiment, the ratio of intensities of bright and dark fringes is 9. This means that
(b) Given, \(\frac{I_{\text {bright }}}{I_{\text {dark }}}=\frac{I_{\max }}{I_{\min }}=9\)
or \(\quad\left(\frac{a_1+a_2}{a_1-a_2}\right)^2=9\)
or \(\frac{a_1+a_2}{a_1-a_2}=3\)
\(
\Rightarrow \quad \frac{a_1}{a_2}=\frac{3+1}{3-1} \Rightarrow \frac{a_1}{a_2}=2
\)
So,\(\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\frac{4}{1}\)
What happens to the fringe pattern, if in the path of one of the slits a glass plate which absorbs 50 % energy is interposed?
\(
\begin{aligned}
&\text { (d) In normal Young’s double slit apparatus, }\\
&\begin{aligned}
& & I_1 & =I_2=I_0 \\
& \therefore & I_{\max } & =\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=4 I_0 \\
& \text { and } & I_{\min } & =\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=0
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { When one of the slit is covered by a glass plate, }\\
&\begin{array}{ll}
& I_1<I_0, I_2=I_0 \\
\therefore & I_{\max }<4 I_0 \text { and } I_{\min }>0
\end{array}
\end{aligned}
\)
In Young’s double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern
(a) In interference between waves of equal amplitudes \(a\), the minimum intensity is zero and the maximum intensity is proportional to \(4 a^2\) (because slit width \(\propto\) amplitude). For waves of unequal amplitudes \(a\) and \(A(A>a)\), the minimum intensity is non-zero and the maximum intensity is proportional to \((a+A)^2\), which is greater than \(4 a^2\). So, the intensities of both the maxima and the minima increases.
A thin mica sheet of thickness \(2 \times 10^{-6} \mathrm{~m}\) and refractive index ( \(\mu=1.5\) ) is introduced in the path of the light from upper slit. The wavelength of the wave used is \(5000 Ã…\). The central bright maximum will shift
(a) On introducing mica sheet, fringe shift \(=\frac{\beta}{\lambda}(\mu-1) t\) where, \(t\) is the thickness of sheet.
So, shift \(=\frac{\beta}{\left(5000 \times 10^{-10}\right)} \times(1.5-1) \times\left(2 \times 10^{-6}\right)=2 \beta\)
So, central bright maximum will shift two fringes upward.
In Young’s double slit experiment, the aperture screen distance is 2 m . The fringe width is 1 mm. Light of 600 nm is used. If a thin plate of glass \((\mu=1.5)\) of thickness 0.06 mm is placed over one of the slits, then| there will be a lateral displacement of the fringes by
(b) When thin plate of glass is introduced, then lateral displacement of fringes
\(
\begin{aligned}
& =\frac{\beta}{\lambda}(\mu-1) t \\
& =\frac{1 \times 10^{-3}}{600 \times 10^{-9}}(1.5-1) \times 0.06 \times 10^{-3}=\frac{1}{20} \mathrm{~m}=5 \mathrm{~cm}
\end{aligned}
\)
Interference fringes were produced in Young’s double slit experiment using light of wavelength \(5000 Ã…\). When a film of thickness \(2.5 \times 10^{-3} \mathrm{~cm}\) was placed in front of one of the slits, then the fringe pattern shifted by a distance equal to 20 fringe widths. The refractive index of the material of the film is
\(
\begin{aligned}
& \text { (c) Shift }=\frac{(\mu-1) t D}{d} \text { or } 20\left(\frac{\lambda D}{d}\right)=\frac{(\mu-1) t D}{d} \\
& \therefore \quad \mu=1+\frac{20 \lambda}{t}=1+\frac{20\left(5.0 \times 10^{-7}\right)}{2.5 \times 10^{-5}}=1.4
\end{aligned}
\)
What is the minimum thickness of a thin film \((\mu=1.2)\) that results in constructive interference in the reflected light? If the film is illuminated with light whose wavelength in free space is \(\lambda=500 \mathrm{~nm}\)?
(a) For constructive interference in case of thin film,
\(
2 \mu t=\left(n-\frac{1}{2}\right) \lambda \quad(\text { where }, n=1,2,3, \ldots)
\)
For minimum thickness \(t, n=1\) or \(2 \mu t=\lambda / 2\)
The minimum thickness of a thin film,
\(
t=\frac{\lambda}{4 \mu}=\frac{500}{4 \times 1.2}=104.17 \approx 104 \mathrm{~nm}
\)
A screen is placed 50 cm from a single slit, which is illuminated with \(6000 Ã…\) light. If distance between first and third minima in the diffraction pattern is 3 mm . What is the width of the slit?
(d) Given, \(D=50 \mathrm{~cm}\) and \(\lambda=6000 Ã…=600 \mathrm{~nm}\)
Distance between first and third minima,
\(
\begin{aligned}
y_3-y_1 & =\frac{3 D \lambda}{a}-\frac{D \lambda}{a}=\frac{2 D \lambda}{a} \\
\Rightarrow \quad 3 \times 10^{-3} & =\frac{2 \times 0.5 \times 600 \times 10^{-9}}{a}
\end{aligned}
\)
Therefore, width of the slit,
\(
a=200 \times 10^{-6} \mathrm{~m}=0.2 \mathrm{~mm}
\)
Note: Distance of \(n\)th secondary minima from central maxima,
\(
\begin{gathered}
y_n=D \cdot \theta \Rightarrow \frac{n \lambda D}{a} \\
y_n=\frac{n \lambda f}{a}
\end{gathered}
\)
where, \(D=\) distance between slit and screen, \(f \approx D=\) focal length of converging lens.
A beam of light of wavelength 400 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen \(2 m\) away from the slit. It is observed that 2 nd order minima occurs at a distance of 2 mm from the position of central maxima. Find the width of the slit.
(a)
\(
\begin{aligned}
&\text { Given, } \lambda=400 \mathrm{~nm}=400 \times 10^{-9} \mathrm{~m} \text {, }\\
&D=2 \mathrm{~m}, y=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}
\end{aligned}
\)
Let point \(P\) is the point of second-order minima.
Now, \(\sin \theta=\tan \theta \approx \theta=\frac{2 \times 10^{-3}}{2}=10^{-3}\)
Let the slit separation be \(a\).
Now, for second order minima, \(\theta=\frac{2 \lambda}{a}=10^{-3}\)
\(
\Rightarrow \quad \frac{2 \times 400 \times 10^{-9}}{a}=10^{-3}
\)
Thus, we obtain \(a=0.8 \mathrm{~mm}\).
A beam of light of wavelength 600 nm from a distant source falls on a single slit 2 mm wide and the resulting diffraction pattern is observed on a screen \(2 m\) away. What is the distance between the first dark fringe on either side of central bright fringe?
(c) Given, \(\lambda=600 \mathrm{~nm}, a=2 \times 10^{-3} \mathrm{~m}\) and \(D=2 \mathrm{~m}\)
Distance of first dark fringe from centre, \(y_1=\frac{D \lambda}{a}\)
Distance between the first dark fringe on either side of the central bright fringe,
\(
\begin{aligned}
2 y_1=\frac{2 D \lambda}{a} & =\frac{2 \times 2 \times 600 \times 10^{-9}}{2 \times 10^{-3}}=1200 \times 10^{-6} \mathrm{~m} \\
& =1.2 \times 10^{-3} \mathrm{~m}=1.2 \mathrm{~mm}
\end{aligned}
\)
A slit of width \(d\) is illuminated by red light of wavelength \(6500 Ã…\). For what value of \(d\) will the first maximum fall at angle of diffraction of \(30^{\circ}\)?
(b) For first secondary maximum of the diffraction pattern,
\(
\begin{aligned}
d \sin \theta & =\frac{3 \lambda}{2} \\
\therefore \quad d & =\frac{3 \lambda}{2 \sin \theta} \\
\therefore \quad & =\frac{3 \times 6500 \times 10^{-10}}{2 \times \sin 30^{\circ}} \\
& =1.95 \times 10^{-6} \mathrm{~m}
\end{aligned}
\)
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.0 mm wide and the resulting diffraction pattern is observed on a screen \(2 m\) away. What is the distance between the second bright fringe on either side of the central bright fringe?
(b) For the diffraction at a single slit, the position of maxima is given by
\(
a \sin \theta=(2 n+1) \frac{\lambda}{2}
\)
For small value of \(\theta, \sin \theta \approx \theta=\frac{y}{D}\)
For second bright fringe ( \(n=2\) ),
\(
\therefore \quad a \times \frac{y}{D}=\frac{5 \lambda}{2} \text { or } y=\frac{5 \lambda}{2 a} D
\)
Substituting the values, we have
\(
y=\frac{5 \times 2 \times 6 \times 10^{-7}}{2 \times 1 \times 10^{-3}}=3.0 \times 10^{-3} \mathrm{~m}=3.0 \mathrm{~mm}
\)
Distance between second maxima on either side of central maxima \(=2 y=6.0 \mathrm{~mm}\).
In a single slit diffraction experiment first minima for \(\lambda_1=660 \mathrm{~nm}\) coincides with first maxima for wavelength \(\lambda_2\). Calculate the value of \(\lambda_2\).
(a) Position of minima in diffraction pattern is given by
\(
a \sin \theta=n \lambda
\)
For first minima of \(\lambda_1\), we have
\(
a \sin \theta_1=(1) \lambda_1 \text { or } \sin \theta_1=\frac{\lambda_1}{a} \dots(i)
\)
The first maxima approximately lies between first and second minima. For wavelength \(\lambda_2\), its position will be
\(
\begin{aligned}
& a \sin \theta_2 & =\frac{3}{2} \lambda_2 \\
\therefore & \sin \theta_2 & =\frac{3 \lambda_2}{2 a} \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
&\text { The two will coincide, if }\\
&\begin{array}{rlrl}
& & \theta_1 & =\theta_2 \text { or } \sin \theta_1=\sin \theta_2 \\
\therefore & \frac{\lambda_1}{a} & =\frac{3 \lambda_2}{2 a}
\end{array}
\end{aligned}
\)
\(
\begin{aligned}
\lambda_2=\frac{2}{3} \lambda_1 & =\frac{2}{3} \times 660 \mathrm{~nm} \\
& =440 \mathrm{~nm}
\end{aligned}
\)
Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength \(6000 Ã…\). When the slit is illuminated by light of another wavelength, then the angular width decreases by \(30 \%\). Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original apparatus is immersed in a liquid. Find the refractive index of the liquid.
(d) Angular width of central maximum \(=\frac{2 \lambda}{a}\)
\(
\Rightarrow \quad \theta_1=\frac{2 \lambda_1}{a} \dots(i)
\)
\(
\theta_2=\theta_1 \times 0.7=\frac{2 \lambda_2}{a} \dots(ii)
\)
\(
\begin{aligned}
&\text { On dividing Eq. (i) by Eq. (ii), we get }\\
&\frac{1}{0.7}=\frac{\lambda_1}{\lambda_2}=\frac{600}{\lambda_2} \quad\left(\because \lambda_1=60000 Ã…=600 \mathrm{~nm}\right)
\end{aligned}
\)
Wavelength, \(\lambda_2=420 \mathrm{~nm}\)
When immersed in liquid, \(\lambda_2=\lambda_1 / \mu\)
Using Eqs. (i) and (ii), we get
\(
\begin{array}{ll}
\Rightarrow & \frac{2 \lambda_1 / \mu}{a}=\frac{2 \lambda_1}{a} \times 0.7 \\
\Rightarrow & \frac{1}{0.7}=\mu
\end{array}
\)
Refractive index of the liquid, \(\mu=\frac{10}{7}=1.42\)
Note: \(\text { Angular width of central maxima, } \theta=\frac{2 \lambda}{a} \text {. }\)
A diffraction grating has \(1.26 \times 10^4\) rulings uniformly spaced over width, \(w=25.4 \mathrm{~mm}\). It is illuminated at normal incidence by blue light of wavelength 450 nm. At what angles to the central axis do the second order maxima occur.
(b) The grating spacing \(d\) is
\(
\begin{aligned}
d=\frac{w}{N} & =\frac{25.4 \times 10^{-3}}{1.26 \times 10^4} \\
& =2.016 \times 10^{-6} \mathrm{~m}=2016 \mathrm{~nm}
\end{aligned}
\)
The second order maxima corresponds to \(n=2\).
\(
\begin{aligned}
&\text { For } \lambda=450 \mathrm{~nm} \text {, we have }\\
&\begin{aligned}
& & d \sin \theta & =2 \lambda \\
\Rightarrow & & \theta & =\sin ^{-1}\left(\frac{2 \lambda}{d}\right) \\
\Rightarrow & & \theta & =\sin ^{-1}\left(\frac{2 \times 450 \times 10^{-9}}{2.016 \times 10^{-6}}\right) \\
& \therefore & \text { Angle, } \theta & \approx 27^{\circ}
\end{aligned}
\end{aligned}
\)
To observe diffraction, the size of an obstacle
(a) Should be of the same order as the wavelength.
Explanation:
Diffraction occurs when waves bend around obstacles or spread out as they pass through apertures, and this effect is most pronounced when the size of the obstacle or aperture is roughly equal to the wavelength of the wave. When the obstacle is much larger than the wavelength, the bending is minimal, and when the obstacle is much smaller, the wave essentially passes through without significant change.
Why other options are incorrect:
(b) should be much larger than the wavelength: If the obstacle is much larger than the wavelength, the diffraction effect will be minimal.
(c) has no relation to wavelength: Diffraction is a wave phenomenon, so it definitely depends on wavelength.
(d) should be exactly ( \(\lambda / 2\) ): While diffraction can occur when the obstacle size is close to \(\lambda / 2\), it’s not the only size for significant diffraction. The key is that the obstacle size should be on the same order as the wavelength.
Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm. If yellow light is replaced by X-rays, then the observed pattern will reveal
(d) If yellow light in a single-slit diffraction experiment is replaced with X-rays, no diffraction pattern will be observed because the wavelength of X-rays is much smaller than the slit width, making diffraction negligible.
Explanation:
Diffraction occurs when the wavelength of light is comparable to or larger than the size of the obstacle (in this case, the slit width). The wavelength of yellow light is on the order of 500-600 nanometers (0.5-0.6 micrometers), while the wavelength of X-rays is much smaller, ranging from 0.01 to 10 nanometers. Since the slit width ( 0.6 mm or 600 micrometers) is significantly larger than the X-ray wavelength, diffraction will not be noticeable.
A slit of width \(d\) is illuminated by white light. For red light \((\lambda=6500 Ã…)\), the first minima is obtained at \(\theta=30^{\circ}\). Then, the value of \(d\) will be
\(
\begin{aligned}
&\text { (b) For first minima, } d \sin \theta=n \lambda=\text { (1) } \lambda\\
&\Rightarrow d=\frac{\lambda}{\sin \theta}=\frac{6.5 \times 10^{-7}}{\sin 30^{\circ}}=13 \times 10^{-4} \mathrm{~mm}
\end{aligned}
\)
The light of wavelength \(6328 Ã…\) is incident on a slit of width 0.2 mm perpendicularly, the angular fringe width will be
\(
\text { (b) Angular fringe width, } \theta=\frac{\lambda}{d}=\frac{6328 \times 10^{-10}}{2 \times 10^{-4}} \times \frac{180}{\pi}=0.18^{\circ}
\)
Direction of the first secondary maxima in the Fraunhofer diffraction pattern at a single slit is given by ( \(a\) is the width of the slit)
(d) In single slit, for \(n\)th secondary maxima path difference
\(
a \sin \theta=(2 n+1) \frac{\lambda}{2}
\)
For first secondary maximum, \(n=1\)
\(
\therefore \quad a \sin \theta=\frac{3 \lambda}{2}
\)
A light wave is incident normally over a single slit of width \(24 \times 10^{-5} \mathrm{~cm}\). The angular position of second dark fringe from the central maxima is \(30^{\circ}\). What is the wavelength of light?
(a) For \(n\)th dark fringe, in single slit
\(
a \sin \theta=n \lambda
\)
For second dark fringe, \(a \sin \theta=2 \lambda\)
So, \(24 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}=2 \lambda\)
\(
\therefore \quad \lambda=6 \times 10^{-7} \mathrm{~m}=6000 Ã…
\)
The main difference in the phenomenon of interference and diffraction is that
(b)Â The main difference is that diffraction is due to the interaction of light from the same wavefront, whereas interference is the interaction of two waves derived from the same source.
Explanation:
Diffraction:
When a wave encounters an obstacle or opening, it bends around the edges. This bending, or spreading, of waves is called diffraction. It involves the interaction of waves from different parts of the same wavefront as they pass through a narrow opening or around an obstacle.
Interference:
Interference occurs when two or more waves combine to form a resultant wave. In the context of light, interference patterns arise from the superposition of waves originating from two separate sources (or two separate parts of the same source).
The X-ray cannot be diffracted by means of an ordinary grating due to
(c) Short wavelength.
Explanation: Diffraction occurs when waves pass through an opening or around an obstacle that is comparable in size to the wavelength of the wave. X-rays have very short wavelengths (typically in the range of 0.01 to 10 nanometers). Ordinary diffraction gratings are designed for much longer wavelengths, such as those of visible light. Therefore, X-rays are not diffracted by ordinary gratings because their wavelengths are too short.
Why other options are incorrect:
(a) large wavelength: If X-rays had a large wavelength, they would be able to diffract through ordinary gratings, making this option incorrect.
(b) high speed: While X-rays do have high speed, it is the wavelength that is the key factor for diffraction. The speed is more related to the energy of the X-ray, not its ability to diffract.
(d) All of these: This is incorrect because option (c) is the correct explanation.
Two polaroids are oriented with their planes perpendicular to incident light and transmission axis making an angle of \(30^{\circ}\) with each other. What fraction of incident unpolarised light is transmitted?
(a)
Suppose intensity of unpolarised light \(=I^{\prime}\)
\(\therefore\) Intensity of polarised light
\(
=\frac{I^{\prime}}{2}
\)
Intensity of emergent light is \(I=\frac{I^{\prime}}{2} \cos ^2 30^{\circ}=\frac{I^{\prime}}{2} \times \frac{3}{4}=\frac{3 I^{\prime}}{8}\)
\(\therefore\) Fraction of incident unpolarised light transmitted,
\(
\frac{I}{I^{\prime}}=\frac{3}{8}
\)
\(
\frac{I}{I^{\prime}} \times 100=\frac{300}{8}=37.5 \%
\)
When light of particular wavelength falls on a plane surface at an angle of incidence \(60^{\circ}\), then the reflected light becomes completely plane polarised. Find the refractive index of surface material and the angle of refraction through it.
(d) Using Brewster’s law, \(\mu=\tan \theta_P=\tan 60^{\circ}=\sqrt{3}\)
At this condition, \(\theta_P+\theta_r=90^{\circ}\)
\(
\therefore \quad \theta_r=90^{\circ}-60^{\circ}
\)
Angle of refraction, \(\theta_r=30^{\circ}\)
Which of the following cannot be polarised?
(d) Ultrasonic waves.
Explanation:
Polarization refers to the direction of the electric field in a transverse wave. Ultrasonic waves are longitudinal waves, meaning the vibrations of the particles travel in the same direction as the wave propagation. Since they don’t have a specific direction of vibration perpendicular to the wave propagation, they cannot be polarized.
Why other options are incorrect:
(a) Radio waves: Radio waves are electromagnetic waves, which are all transverse and can be polarized.
(b) Ultraviolet rays: Ultraviolet rays are also electromagnetic waves and can be polarized.
(c) Infrared rays: Infrared rays are another type of electromagnetic wave and can be polarized.
Light waves can be polarised as they are
(a)Â Light waves can be polarized because they are transverse waves.
Explanation: Polarisation is a property of transverse waves, meaning the oscillations are perpendicular to the direction of wave propagation. Since light waves exhibit this property, they are considered transverse.
An optically active compound
(a)Â An optically active compound rotates the plane of polarized light.
Explanation:
Optical activity: This refers to the ability of a compound to rotate the plane of polarization of polarized light when it passes through it.
Chiral molecules: Optically active compounds are chiral molecules, meaning they have a nonsuperimposable mirror image.
Rotation direction: The direction of rotation can be either clockwise (dextrorotatory) or counterclockwise (levorotatory).
A polaroid is placed at \(45^{\circ}\) to an incoming light of intensity \(I_0\). Now, the intensity of light passing through polaroid after polarisation would be
\(
\text { (b) } I=I_0 \cos ^2 \theta=I_0 \cos ^2 45^{\circ}=\frac{I_0}{2}
\)
When the angle of incidence on a material is \(60^{\circ}\), then the reflected light is completely polarised. The velocity (in \(\mathrm{ms}^{-1}\) ) of the refracted ray inside the material is
\(
\begin{aligned}
&\text { (c) From Brewster’s law, } \mu=\tan i_p \Rightarrow \frac{c}{v}=\tan 60^{\circ}=\sqrt{3}\\
&\Rightarrow \quad v=\frac{c}{\sqrt{3}}=\frac{3 \times 10^8}{\sqrt{3}}=\sqrt{3} \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
\)
The angle of polarisation for any medium is \(60^{\circ}\), what will be critical angle for this
(d) By using \(\mu=\tan \theta_P \Rightarrow \mu=\tan 60^{\circ}=\sqrt{3}\)
Also, \(\quad C=\sin ^{-1}\left(\frac{1}{\mu}\right) \Rightarrow C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
A light has amplitude \(A\) and angle between analyser and polariser is \(60^{\circ}\). Light reflected by analyser has amplitude
\(
\text { (d) The amplitude will be } A \cos 60^{\circ}=A / 2 \text {. }
\)
Light is incident on a glass surface at polarising angle of \(57.5^{\circ}\). Then, the angle between the incident ray and the refracted ray is
\(
\text { (d) Required angle }=2 i_p+90^{\circ}=2 \times 57.5^{\circ}+90^{\circ}=205^{\circ}
\)
When unpolarised light beam is incident from air onto glass ( \(n=1.5\) ) at the polarising angle,
(a) According to Brewster’s law, when a beam of ordinary light (i.e. unpolarised) is reflected from a transparent medium (like glass), the reflected light is completely plane polarised at certain angle of incidence called the angle of polarisation.
The Brewster angle for the glass-air interface is \(54.74^{\circ}\). If a ray of light going from air to glass strikes at an angle of incidence \(45^{\circ}\), then the angle of refraction is \(\left(\right.\) take, \(\left.\tan 54.74^{\circ}=\sqrt{2}\right)\)
(b)
\(
\begin{aligned}
& \text { From Brewster’s law, } \mu=\tan i_p=\tan 54.74^{\circ}=\sqrt{2} \\
& \qquad \sqrt{2}=\frac{\sin 45^{\circ}}{\sin r} \Rightarrow \sin r=\frac{1}{2} \Rightarrow r=30^{\circ}
\end{aligned}
\)
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