Past NEET Entrance Papers

Hint

(c)

In Young’s double-slit experiment, if a monochromatic source is replaced by white light, which contains multiple wavelengths, the patterns observed on the screen will differ significantly.
First, remember that the pattern formed in the double-slit experiment consists of both bright and dark fringes due to constructive and destructive interference, respectively. The position and intensity of these fringes depend on the wavelength of the light used. For monochromatic light (light of a single wavelength), the interference pattern is stable, with bright and dark fringes evenly spaced. Each fringe is uniformly bright or dark.
When using white light, which is a combination of various wavelengths of light, each color, or each wavelength, forms its own interference pattern with slightly different fringe spacing. This occurs because the separation between fringes \(\Delta y\) is given by the formula:
\(
\Delta y=\frac{\lambda L}{d}
\)
Where:
\(\lambda\) is the wavelength of light
\(L\) is the distance from the slits to the screen
\(d\) is the separation between the slits
Since different wavelengths have different values for \(\lambda\), each color’s fringes will be at slightly different positions. The result is that near the center of the pattern, where there is the least path difference, all wavelengths constructively interfere to form a bright white central fringe. However, moving away from the center, the fringes start to show different colors as the path difference between the light from the two slits increases. This causes a dispersion of colors with different orders of fringes dominated by different colors. The constructive and destructive interference patterns of different wavelengths slightly offset one another.

Thus:
Option a – “Interference pattern will disappear” is incorrect because the interference pattern does not disappear but changes due to the dispersion of colors.
Option b – “There will be a central dark fringe surrounded by a few coloured fringes” is incorrect as the central fringe in the presence of white light is bright, not dark.
Option c – “There will be a central bright white fringe surrounded by a few coloured fringes” is correct. This is because all the wavelengths interfere constructively at the center, creating a bright white fringe, succeeded by colored fringes due to the varying interference conditions for each wavelength.
Option d – “All bright fringes will be of equal width” is incorrect. The width and spacing of the fringes vary by wavelength, so the interference pattern will not have uniform fringe widths.
Therefore, the correct answer is Option c: “There will be a central bright white fringe surrounded by a few coloured fringes.”

Hint

(d) At Brewster’s angle, the reflected light becomes completely polarised perpendicular to the plane of incidence. This happens because the angle between the reflected and refracted rays becomes \(90^{\circ}\)
The refracted light, however, remains partially polarised because it contains components of both polarisation states.

Explanation: Step 1: Understand Brewster’s Law
Brewster’s angle \(\left(\theta_B\right)\) is the angle of incidence at which light reflected from a surface is completely polarized. At this angle, the reflected and refracted rays are perpendicular to each other.

Step 2: Analyze the Reflection and Refraction
At Brewster’s angle, the reflected light is completely polarized in a direction perpendicular to the plane of incidence.
The refracted light is partially polarized because it consists of both components of polarization.

Step 3: Conclusion: The reflected light is completely polarized, while the refracted light is partially polarized.

Hint

(b) In Young’s double-slit experiment, if the wavelength of light used is increased (from violet to red), then the fringe width increases.

Explanation: The formula for fringe width in Young’s double-slit experiment is \(\beta=\frac{\lambda D}{d}\), where \(\lambda\) is the wavelength of light, \(D\) is the distance between the screen and the slits, and \(\)d\(\) is the distance between the slits. Since the fringe width is directly proportional to the wavelength, increasing the wavelength will result in a larger fringe width.

Hint

(b) Polarizing angle or the Brewster’s angle is the angle at which an incident ray gets reflected and refracted through the medium such that the two beams are perpendicular to each other.
By Snell’s Law, \(n_1 \sin i=n_2 \sin r\)
Also \(i+r=90^{\circ}\)
Thus, \(n_1 \sin \left(90^{\circ}-r\right)=n_2 \sin r\)
Hence, \(r=\tan ^{-1}\left(\frac{n_1}{n_2}\right)=\tan ^{-1}\left(\frac{1}{1.73}\right)=30^{\circ}\)

Hint

(d) To find the fringe separation (also known as fringe width) in Young’s double slit experiment, we use the formula:
\(
\beta=\frac{\lambda D}{d}
\)
where:
\(\beta\) is the fringe width (fringe separation).
\(\lambda\) is the wavelength of the light used.
\(D\) is the distance from the slits to the screen.
\(d\) is the distance between the two slits.
Given values are:
\(
\begin{aligned}
& d=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \\
& D=1 \mathrm{~m} \\
& \lambda=600 \times 10^{-9} \mathrm{~m}
\end{aligned}
\)
Substituting these values into the formula:
\(
\beta=\frac{600 \times 10^{-9} \mathrm{~m} \times 1 \mathrm{~m}}{1.5 \times 10^{-3} \mathrm{~m}}
\)
Calculating the value:
\(
\begin{aligned}
& \beta=\frac{600 \times 10^{-9}}{1.5 \times 10^{-3}} \\
& \beta=4 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)

Hint

(c) Interference occurs when waves with the same or very similar frequencies overlap, creating a resulting wave with a varying amplitude depending on the relative phase difference between the waves.

Option A \((y=a \sin \omega t)\) and Option C \((y=a \sin (\omega t-\phi))\) :
These waves both have the same angular frequency ( \(\omega\) ), which is the key requirement for interference. The difference in the phase angle ( \(\phi\) ) only affects the relative position of the waves’ peaks and troughs, not their fundamental frequency.

Option \(\mathrm{B}(y=a \sin 2 \omega t)\) and Option \(\mathrm{D}(y=a \sin 3 \omega t)\) :
These waves have different angular frequencies (2\\(lomega\\) and 3\\(lomega\\) respectively) compared to Option A and C. Waves with different frequencies cannot interfere constructively or destructively, as their wavelengths are different and they do not align in a meaningful way.

Why other options are incorrect:
B and D: As explained above, options B and D have different angular frequencies from options A and C , making interference impossible.

C: While option C has the same angular frequency as option A , it is not a valid combination for interference because the question asks for the superposition of two waves, and option C represents only one wave.

Hint

(b) Here, polaroid \(A\) polarises light.
Intensity of polarised light from \(A=\frac{I_0}{2}\)
According to law of Malus, Intensity of light emerging from \(B\),
\(
I_B=\left(\frac{I_0}{2}\right)\left(\cos 45^{\circ}\right)^2=\frac{I_0}{2}\left(\frac{1}{\sqrt{2}}\right)^2=\frac{I_0}{4}
\)

Hint

(b) In Young’s double slit experiment, the angular separation of the fringes \((\theta)\) is given by :
\(
\theta=\frac{m \lambda}{d}
\)
where \(m\) is the fringe order, \(\lambda\) is the wavelength of the light, and \(d\) is the distance between the slits.

Now, let’s analyze both statements:
Statement I: If the screen is moved away from the plane of the slits, the angular separation of the fringes remains constant.
This statement is true because the angular separation of the fringes \((\theta)\) does not depend on the distance between the screen and the slits. It only depends on the fringe order ( m ), the wavelength of the light \((\lambda)\), and the distance between the slits (d).
Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.
This statement is false because if the wavelength \((\lambda)\) increases, the angular separation of the fringes \((\theta)\) also increases according to the formula :
\(
\theta=\frac{m \lambda}{d}
\)

So, the correct answer is :
Option b: Statement I is true but Statement II is false.

Hint

(a) Angle of incidence is critical angle for green.
Now \(\sin \theta_C=\frac{1}{\mu} \quad \mu \propto \frac{1}{\lambda}\)
So \(\sin \theta_C \propto \lambda\)
Wavelengths for colours above green have critical angle more than the given angle, so they will be transmitted.
So yellow, orange and red have more wavelength than green, so they will came out in air.

Hint

(d) Number of fringes observed initially: \({n}_1=8\)
Initial wavelength of light: \(\lambda_1=600 \mathrm{~nm}\)
Final wavelength of light: \(\lambda_2=400 \mathrm{~nm}\)
Step 1: Set up the proportion
The number of fringes is inversely proportional to the wavelength:
\(\frac{n_1}{n_2}=\frac{\lambda_2}{\lambda_1}\)

Step 2: Substitute the given values
Substitute \(n_1=8, \lambda_1=600 \mathrm{~nm}\), and \(\lambda_2=400 \mathrm{~nm}\) into the equation:
\(\frac{8}{n_2}=\frac{400}{600}\)

Step 3:Solve for \(n_2\)
\(n_2=\frac{8 \times 600}{400}\)
\(n_2=\frac{4800}{400}\)
\(n_2=12\)

Solution: The number of fringes observed with the new wavelength is 12 .

Hint

(a)

Explanation:
To find the angle between the direction of polarization of the light and that of the polarizer. we can use Malus’s Law, which states that the intensity of polarized light after passing through a polarizer is given by:
\(
I=I_0 \cos ^2(\theta)
\)
where:
\(I_0\) is the initial intensity (10 lumen).
\(I\) is the transmitted intensity ( 2.5 lumen).
\(\theta\) is the angle between the light’s polarization direction and the polarizer’s axis.
We can rearrange this equation to find \(\theta\) :
Substitute the known values into Malus’s Law:
\(
2.5=10 \cos ^2(\theta)
\)
Find the angle \(\theta\) :
\(
\theta=\cos ^{-1}(0.5)=60^{\circ}
\)

Hint

(c) Frequency of light: \(f=500 \mathrm{THz}=500 \times 10^{12} \mathrm{~Hz}\)
Distance between slits: \(d=0.2 \mathrm{~mm}=0.2 \times 10^{-3} \mathrm{~m}\)
Distance to the screen: \({D}=1 \mathrm{~m}\)
Speed of light: \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Wavelength of light: \(\lambda=\frac{\boldsymbol{c}}{{f}}\)
Fringe width: \(\beta=\frac{\lambda D}{d}\)
\(
\begin{aligned}
& \lambda=\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{500 \times 10^{12} \mathrm{~Hz}} \\
& \lambda=6 \times 10^{-7} \mathrm{~m}
\end{aligned}
\)
Calculate the fringe width
Use the formula \(\beta=\frac{\lambda D}{d}\).
Substitute the values:
\(\beta=\frac{\left(6 \times 10^{-7} \mathrm{~m}\right)(1 \mathrm{~m})}{0.2 \times 10^{-3} \mathrm{~m}}\)
\(\beta=3 \times 10^{-3} \mathrm{~m}\)
Calculate the width of 10 fringes
Multiply the fringe width by 10.
\(10 \beta=10 \times 3 \times 10^{-3} \mathrm{~m}\)
\(10 \beta=3 \times 10^{-2} \mathrm{~m}=30 mm\)

Hint

(c)

\(
\begin{aligned}
&\text { According to Malu’s Law }\\
&\begin{aligned}
& I_{o u t}=I_{i n} \cos ^2 \theta \\
& I_{o u t}=I \cos ^2\left(30^{\circ}\right) \\
& I_{o u t}=\frac{3 I}{4}
\end{aligned}
\end{aligned}
\)

Hint

(d) Separation between fringes is related to fringed width \(\beta=\frac{D \lambda}{d}\)
If \(D\) is increased \(\beta\) increases while
Angular separation is independent of distance between slits and screen (D)
As angular separation \(=\frac{\beta}{D}=\frac{\lambda}{d}\)

Hint

(b) Given that the slit distance is made half and the screen distance made double than the original value, then,
Fringe width, \(\beta=\frac{\lambda D}{d}\)
Now, \(d^{\prime}=\frac{d}{2}[latex] and [latex]D^{\prime}=2 D\)
So, \(\beta^{\prime}=\frac{\lambda(2 D)}{d / 2}=\frac{4 \lambda D}{d}\)
\(
\beta=4 \beta
\)

Hint

(b) To find the Brewster’s angle \(i_b\) for an interface, we can use the relationship between the refractive index \(\mu\) and the Brewster’s angle. The Brewster’s angle is defined by the formula:
\(
\mu=\tan \left(i_b\right)
\)
where \(\mu\) is the refractive index of the medium.
1. Understanding Brewster’s Angle:
Brewster’s angle \(i_b\) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.
2. Using the Formula:
The relationship between the refractive index \(\mu\) and Brewster’s angle is given by:
\(
\mu=\tan \left(i_b\right)
\)
3. Finding the Angle:
Rearranging the formula gives us:
\(
i_b=\tan ^{-1}(\mu)
\)
4. Considering the Range of \(\mu\) :
The refractive index \(\mu\) can range from 1 (for vacuum) to infinity (for a very dense medium).
5. Calculating for \(\mu=1\) :
If \(\mu=1\) :
\(
i_b=\tan ^{-1}(1)=45^{\circ}
\)
6. Calculating for \(\mu \rightarrow \infty\) :
If \(\mu[latex] approaches infinity:
[latex]
i_b=\tan ^{-1}(\infty)=90^{\circ}
\)
7. Conclusion:
Therefore, the Brewster’s angle \(i_b\) for an interface can vary from \(45^{\circ}\) to \(90^{\circ}\). The correct option for the question is that the Brewster’s angle should be less than \(90^{\circ}\).

Hint

(a) The phase difference \((\varphi)\) between the two waves can be related to the path difference using the formula:
\(
\phi=\frac{2 \pi}{\lambda} \Delta x
\)
where \(\lambda\) is the wavelength of the light.
Since we have established that the path difference \((\Delta x)\) at the central maximum is zero:
\(
\Delta x=0 \Rightarrow \phi=\frac{2 \pi}{\lambda} \times 0=0
\)

Hint

(d) Limit of Resolution for a Telescope is
\(
\theta=1.22 \frac{\lambda}{d}
\)
Given, \(\lambda=600 \times 10^{-9} ; \mathrm{d}=2 \mathrm{~m}\)
By substituting the values, we get
\(
\theta=3.66 \times 10^{-7} \mathrm{rad}
\)

Hint

(a) Angular fringe width (in air) \(\theta_{\text {air }}=\frac{\beta}{D}\)
Angular Fringe width (in water)
\(
\begin{aligned}
& \theta_{\mathrm{w}}=\frac{\beta}{\mu D}=\frac{\theta_{\text {air }}}{\mu} \\
& =\frac{0.2^{\circ}}{\frac{4}{3}}=0.15^{\circ}
\end{aligned}
\)

Hint

(c) The path difference \((\Delta \mathrm{x})\) for the \(n\)th minimum is given by the formula:
\(
\Delta x=(2 n-1) \frac{\lambda}{2}
\)
where \(n\) is the order of the minimum \((n=1,2,3, \ldots)\).
\(
\begin{aligned}
&\text { Here, we need to find the path difference for the } 5 \text { th minimum, so we set } n=5 \text { : }\\
&\Delta x=(2 \times 5-1) \frac{\lambda}{2}
\end{aligned}
\)
\(
\Delta x=(10-1) \frac{\lambda}{2}=9 \frac{\lambda}{2}
\)

Hint

(b) The angular width is written as;
\(
\theta=\frac{2 \lambda}{d}
\)
Here we have \(\boldsymbol{\theta}\) is the angular width, \(\boldsymbol{\lambda}\) is the wavelength and \(\boldsymbol{d}\) is the distance.
Given: Wavelength, \(\lambda=6000 Å\)
As we know,
\(
\theta=\frac{2 \lambda}{d} \dots(1)
\)

and When the same slit is illuminated by another monochromatic light, the angular width decreases.
\(
\begin{aligned}
& \theta^{\prime}=\frac{2 \lambda^{\prime}}{d} \\
& \theta\left(1-\frac{30}{100}\right)=\frac{2 \lambda^{\prime}}{d} \dots(2)
\end{aligned}
\)
Now, on dividing the equation (1) and (2) we have;
\(
\begin{aligned}
& \frac{\theta}{\theta\left(1-\frac{30}{100}\right)}=\frac{\frac{2}{d}}{\frac{2^{\prime}}{d}} \\
& \frac{10}{7}=\frac{\lambda}{\lambda^{\prime}} \\
& \Rightarrow \lambda^{\prime}=\frac{7 \times 6000}{10} \\
& \Rightarrow \lambda^{\prime}=4200 Å
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) Angular width }=\frac{\lambda}{d} \\
& 0.20^{\circ}=\frac{\lambda}{2 \mathrm{~mm}} \text { and } 0.21^{\circ}=\frac{\lambda}{d}
\end{aligned}
\)
\(
\begin{aligned}
& \text { On dividing we get, } \frac{0.20}{0.21}=\frac{d}{2 \mathrm{~mm}} \\
& \therefore \quad d=1.9 \mathrm{~mm}
\end{aligned}
\)

Hint

(c) For telescope, angular magnification \(=\frac{f_0}{f_e}\) Angular resolution \(=\frac{D}{1.22 \lambda}\) should be large.
So, objective lens should have large focal length \(\left(f_0\right)\) and large diameter \(D\) for large angular magnification and high angular resolution.

Hint

(b) When reflected light and refracted light are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.
Also, \(\tan i=\mu\) (i=Brewster angle)

Hint

(c) Position of \(8^{\text {th }}\) bright fringe in medium,
\(
x=\frac{8 \lambda_m D}{d}
\)
Position of \(5^{\text {th }}\) dark fringe in air,
\(
x^{\prime}=\frac{\left(5-\frac{1}{2}\right) \lambda_{\mathrm{air}} D}{d}=\frac{4.5 \lambda_{\mathrm{air}} D}{d}
\)
Given \(x=x^{\prime} \Rightarrow \frac{8 \lambda_m D}{d}=\frac{4.5 \lambda_{\mathrm{air}} D}{d}\)
\(
\mu_m=\frac{\lambda_{\mathrm{air}}}{\lambda_m}=\frac{8}{4.5}=1.78
\)

Hint

(b) The resolving power of an optical microscope,
\(
\mathrm{RP}=\frac{2 \mu \sin \theta}{\lambda}
\)
For wavelength \(\lambda_1=4000 Å\), resolving power will be
\(
\mathrm{RP}_1=\frac{2 \mu \sin \theta}{4000} \dots(i)
\)
For wavelength \(\lambda_2=6000 Å\), resolving power will be
\(
\mathrm{RP}_2=\frac{2 \mu \sin \theta}{6000} \dots(ii)
\)
On dividing eqn. (i) by eqn. (ii)
\(
\frac{\mathrm{RP}_1}{\mathrm{RP}_2}=\frac{6000}{4000}=\frac{3}{2}
\)

Hint

(b) The intensity of transmitted light through \(P_1\),
\(
I_1=\frac{I_0}{2}
\)
The intensity of transmitted light through \(P_3\),
\(
I_2=I_1 \cos ^2 45^{\circ}=\frac{I_0}{2}\left(\frac{1}{\sqrt{2}}\right)^2=\frac{I_0}{2} \cdot \frac{1}{2}=\frac{I_0}{4}
\)
Angle between polaroids \(P_3\) and \(P_2\)
\(
=\left(90^{\circ}-45^{\circ}\right)=45^{\circ}
\)
\(\therefore \quad\) Intensity of transmitted light through \(P_2\),
\(
I_3=I_2 \cos ^2 45^{\circ}=\frac{I_0}{4}\left(\frac{1}{\sqrt{2}}\right)^2=\frac{I_0}{8}
\)

Hint

(b) Here, \(d=5 \lambda, D=10 d, y=\frac{d}{2}\)
Resultant Intensity at \(y=\frac{d}{2}, I_y=\) ?
The path difference between two waves at \(y=\frac{d}{2}\)
\(
\Delta x=d \tan \theta=d \times \frac{y}{D}=\frac{d \times \frac{d}{2}}{10 d}=\frac{d}{20}=\frac{5 \lambda}{20}=\frac{\lambda}{4}
\)
Corresponding phase difference, \(\phi=\frac{2 \pi}{\lambda} \Delta x=\frac{\pi}{2}\)
Now, maximum intensity in Young’s double slit experiment,
\(
\begin{aligned}
& I_{\max }=I_1+I_2+2 I_1 I_2 \text { or } I_0=4 I \quad\left(\because I_1=I_2=I\right) \\
\therefore & I=\frac{I_0}{4}
\end{aligned}
\)
Required intensity,
\(
I_y=I_1+I_2+2 I_1 I_2 \cos \frac{\pi}{2}=2 I=\frac{I_0}{2}
\)

Hint

(d) Here, \(a=0.02 \mathrm{~cm}=2 \times 10^{-4} \mathrm{~m}\)
\(
\begin{aligned}
& \lambda=5 \times 10^{-5} \mathrm{~cm}=5 \times 10^{-7} \mathrm{~m} \\
& D=60 \mathrm{~cm}=0.6 \mathrm{~m}
\end{aligned}
\)
Position of first minima on the diffraction pattern,
\(
y=\frac{D \lambda}{a}=\frac{0.6 \times 5 \times 10^{-7}}{2 \times 10^{-4}}=15 \times 10^{-4} \mathrm{~m}=0.15 \mathrm{~cm}
\)

Hint

(b) For first minimum, the path difference between extreme waves,
\(
\begin{aligned}
& \quad a \sin \theta=\lambda \\
& \text { Here } \theta=30^{\circ} \Rightarrow \sin \theta=\frac{1}{2} \\
& \therefore \quad a=2 \lambda \dots(i)
\end{aligned}
\)
For first secondary maximum, the path difference between extreme waves
\(a \sin \theta^{\prime}=\frac{3}{2} \lambda \quad\) or \(\quad(2 \lambda) \sin \theta^{\prime}=\frac{3}{2} \lambda \quad[\) Using eqn \((\mathrm{i})]\)
or \(\sin \theta^{\prime}=\frac{3}{4} \quad \therefore \quad \theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)\)

Hint

(c) For double slit experiment,
\(
\begin{aligned}
& d=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}, D=1 \mathrm{~m}, \lambda=500 \times 10^{-9} \mathrm{~m} \\
& \text { Fringe width } \beta=\frac{D \lambda}{d}
\end{aligned}
\)
Width of central maxima in a single slit \(=\frac{2 \lambda D}{a}\) As per question, width of central maxima of single slit pattern \(=\) width of 10 maxima of double slit pattern
\(
\begin{aligned}
\frac{2 \lambda D}{a} & =10\left(\frac{\lambda D}{d}\right) \text { or } a=\frac{2 d}{10}=\frac{2 \times 10^{-3}}{10} \\
& =0.2 \times 10^{-3} \mathrm{~m}=0.2 \mathrm{~mm}
\end{aligned}
\)

Hint

(a) The situation is shown in the figure.

In figure \(A\) and \(B\) represent the edges of the slit \(A B\) of width \(a\) and \(C\) represents the midpoint of the slit.
For the first minimum at \(P\),
\(
a \sin \theta=\lambda \dots(i)
\)
where \(\lambda\) is the wavelength of light.
The path difference between the wavelets from \(A\) to \(C\) is
\(
\Delta x=\frac{a}{2} \sin \theta=\frac{1}{2}(a \sin \theta)=\frac{\lambda}{2} \quad(\text { using }(\mathrm{i}))
\)
The corresponding phase difference \(\Delta \phi\) is
\(
\Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi
\)

Hint

(c) As, intensity \(I \propto\) width of slit \(W\)
Also, intensity \(I \propto\) square of amplitude \(A\)
\(
\begin{aligned}
& \therefore \quad \frac{I_1}{I_2}=\frac{W_1}{W_2}=\frac{A_1^2}{A_2^2} \\
& \text { But } \frac{W_1}{W_2}=\frac{1}{25} \text { (given) }
\end{aligned}
\)
\(
\therefore \quad \frac{A_1^2}{A_2^2}=\frac{1}{25} \quad \text { or } \quad \frac{A_1}{A_2}=\sqrt{\frac{1}{25}}=\frac{1}{5}
\)
\(
\frac{I_{\max }}{I_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}=\frac{\left(\frac{A_1}{A_2}+1\right)^2}{\left(\frac{A_1}{A_2}-1\right)^2}
\)
\(
=\frac{\left(\frac{1}{5}+1\right)^2}{\left(\frac{1}{5}-1\right)^2}=\frac{\left(\frac{6}{5}\right)^2}{\left(-\frac{4}{5}\right)^2}=\frac{36}{16}=\frac{9}{4}
\)

Hint

\(
\begin{aligned}
&\text { (d) Linear width of central maxima Y }\\
&=\mathrm{D}(2 \theta)=2 \mathrm{D} \theta=\frac{2 \mathrm{D} \lambda}{\mathrm{a}} \quad \therefore \theta=\frac{\lambda}{a}
\end{aligned}
\)

The central maxima lies between the first minima on both sides.
The angular width ‘ \(d\) ‘ central maxima \(=2 \theta=\frac{2 \lambda}{b}\)
Linear width of central maxima \(=2 \mathrm{D} \theta=\frac{2 D \lambda}{b}\)
\(b\) – width of slits and D – distance between slit and screen.

Hint

(c) Intensity at any point on the screen is
\(
I=4 I_0 \cos ^2 \frac{\phi}{2}
\)
where \(I_0\) is the intensity of either wave and \(\phi\) is the phase difference between two waves.
Phase difference, \(\phi=\frac{2 \pi}{\lambda} \times\) Path difference
When path difference is \(\lambda\), then
\(
\phi=\frac{2 \pi}{\lambda} \times \lambda=2 \pi
\)
\(
I=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right)=4 I_0 \cos ^2(\pi)=4 I_0=K \dots(i)
\)
When path difference is \(\frac{\lambda}{4}\), then
\(
\begin{aligned}
& \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2} \\
& I=4 I_0 \cos ^2\left(\frac{\pi}{4}\right)=2 I_0=\frac{K}{2} \text { [Using (i)] }
\end{aligned}
\)

Hint

(d) Here, \(\lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}\)
\(
a=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, D=2 \mathrm{~m}
\)
Distance between the first dark fringes on either side of the central bright fringe is also the width of central maximum.
\(
\begin{aligned}
& \text { Width of central maximum }=\frac{2 \lambda D}{a} \\
& \qquad=\frac{2 \times 600 \times 10^{-9} \mathrm{~m} \times 2 \mathrm{~m}}{10^{-3} \mathrm{~m}}
\end{aligned}
\)
\(
=24 \times 10^{-4} \mathrm{~m}=2.4 \times 10^{-3} \mathrm{~m}=2.4 \mathrm{~mm}
\)

Hint

(d) Let \(n_1\) bright fringe of \(\lambda_1\) coincides with \(n_2\) bright fringe of \(\lambda_2\). Then
\(
\begin{aligned}
& \frac{n_1 \lambda_1 D}{d}=\frac{n_2 \lambda_2 D}{d} \text { or } n_1 \lambda_1=n_2 \lambda_2 \\
& \frac{n_1}{n_2}=\frac{\lambda_2}{\lambda_1}=\frac{10000}{12000}=\frac{5}{6}
\end{aligned}
\)
Let \(x\) be given distance.
\(
\therefore \quad x=\frac{n_1 \lambda_1 D}{d}
\)
Here, \(n_1=5, D=2 \mathrm{~m}, d=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
\(
\begin{aligned}
& \lambda_1=12000 \AA=12000 \times 10^{-10} \mathrm{~m}=12 \times 10^{-7} \mathrm{~m} \\
& x=\frac{5 \times 12 \times 10^{-7} \mathrm{~m} \times 2 \mathrm{~m}}{2 \times 10^{-3} \mathrm{~m}}=6 \times 10^{-3} \mathrm{~m}=6 \mathrm{~mm}
\end{aligned}
\)

Hint

(b) Fringe width, \(\beta=\frac{\lambda D}{d}\)
where \(D\) is the distance between slits and screen and \(d\) is the distance between the slits.
When \(D\) is doubled and \(d\) is reduced to half, then fringe width becomes
\(
\beta^{\prime}=\frac{\lambda(2 D)}{(d / 2)}=\frac{4 \lambda D}{d}=4 \beta
\)

Hint

(a) De-Broglie Wavelength: The wavelength associated with a body in motion is known as De – Broglie wavelength. The De – Broglie wavelength of a body is given as
\(
\lambda=\frac{h}{m v} \dots(1)
\)
Angular width \((\omega)[latex] : The angular width of the central maximum is given as
[latex]
\omega=2 \lambda / d \dots(2)
\)
Angular width is directly proportional to the wavelength.
From Eq (1) it is clear that If the speed of the electron is increased the De -Broglie Wavelength associated with it will decrease.
From Eq (2) we get that the angular width will decrease if wavelength will decrease. So, overall, an increase in speed will decrease the angular width.

Hint

(c) For the second minimum,
Path difference \(=2 \lambda\)
Therefore, corresponding value of phase difference is
\(
\Delta \phi=\frac{2 \pi}{\lambda} \times \text { Path difference }=\frac{2 \pi}{\lambda} \times 2 \lambda=4 \pi
\)

Hint

(d) The resultant intensity of two periodic waves at a point is given by
\(
\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cdot \cos \phi
\)
Resultant intensity is maximum if
\(
\cos \phi=-1
\)
i.e. \(I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2}\)
Resultant intensity is minimum if
\(
\begin{aligned}
& \cos \phi=+1 \\
& \text { i.e, } I_{\min }=I_1+I_2-2 \sqrt{I_1 I_2}
\end{aligned}
\)
Therefore, the sum of the maximum and minimum intensities is \(I_{\text {max }}+I_{\text {min }}\)
\(
\begin{aligned}
& =\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2}+\mathrm{I}_1+\mathrm{I}_2-2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \\
& =2\left(\mathrm{I}_1+\mathrm{I}_2\right)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } \mu=\frac{\text { velocity of light in vacuum }(c)}{\text { velocity of light in medium }(v)} \\
& v=v \lambda=2 \times 10^{14} \times 5000 \times 10^{-10}=10^8 \mathrm{~m} / \mathrm{s} \\
& \mu=\frac{c}{v_{\text {med }}}=\frac{3 \times 10^8}{10^8}=3
\end{aligned}
\)

Hint

(c) R.P. \(=1 / \Delta \theta\)
The angular resolution, \(\Delta \theta=\frac{1.22 \lambda}{D}\) \(=\frac{1.22 \times 5000 \times 10^{-8}}{0.1}=6.1 \times 10^{-4} \simeq 10^{-4} \mathrm{rad}\)

Hint

\(
\begin{aligned}
&\text { (c) Resolution of telescope }\\
&d \theta=1.22 \frac{\lambda}{D}=1.22 \times \frac{5000 \times 10^{-8}}{10}
\end{aligned}
\)

\(
x=d \theta \times d=\frac{1.22 \times 5000 \times 10^{-8} \times 10^5}{10}=6.1 \times 10^{-1} \mathrm{~cm} \simeq 5 \mathrm{~mm}
\)

Hint

(c) Given \(d=\) diameter of lens \(=2 \mathrm{~mm}=2 \times 10^{-1} \mathrm{~cm}\), \(\lambda=5000 Å=5000 \times 10^{-8} \mathrm{~cm}\)
\(
\begin{aligned}
&\text { Resolving power of eye lens }\\
&=\frac{d}{\lambda}=\frac{2 \times 10^{-1}}{5000 \times 10^{-8}}=\frac{1}{d \theta}
\end{aligned}
\)

Let \(S\) be the minimum distance between two points so that it may be resolved.
\(
\therefore \quad S=r d \theta \text {. Here } r=50 \mathrm{~m}=5000 \mathrm{~cm}
\)
\(
S=5000 \times \frac{5000 \times 10^{-8}}{2 \times 10^{-1}}=1.25 \mathrm{~cm}
\)

Hint

(b) Bragg’s Law is given by the equation:
\(
2 d \sin \theta=n \lambda
\)
where:
\({d}\) = interplanar distance
\(\theta=\) angle of diffraction
\({n}=\) order of diffraction (an integer)
\(\lambda=\) wavelength of the incident light
We are given:
Interplanar distance \(d=2.8 \times 10^{-8} \mathrm{~m}\)
For the maximum wavelength \(\left(\lambda_{\max }\right)\), we consider the case when \(\sin \theta\) is at its maximum value, which is 1 . Therefore, we can rewrite Bragg’s Law as:
\(
2 d=n \lambda_{\max }
\)
If we take \({n}=\mathbf{1}\) (first order diffraction), we can simplify this to:
\(
\lambda_{\max }=2 d
\)
Substituting the value of \(d\) :
\(
\lambda_{\max }=2 \times\left(2.8 \times 10^{-8}\right)=5.6 \times 10^{-8} \mathrm{~m}
\)

Hint

(a) The colors appearing on a thin soap film or soap bubble are due to the phenomenon of (a) interference.

Explanation: When light reflects off the front and back surfaces of a thin film like a soap bubble, the reflected waves can interfere with each other, creating constructive interference for certain wavelengths (colors) and destructive interference for others. This results in the vibrant colors we see.
Why other options are incorrect:
(b) Dispersion: Dispersion refers to the separation of light into its constituent colors based on their wavelengths. While light does undergo dispersion when it passes through a prism, it’s not the primary cause of the colors in a soap bubble.
(c) Refraction: Refraction is the bending of light as it passes from one medium to another. While light does refract within the soap film, it’s the interference of the reflected light that creates the colorful patterns.
(d) Diffraction: Diffraction is the bending of light waves around an obstacle or through an opening. While diffraction can play a role in some optical phenomena, it’s not the dominant mechanism behind the colors in soap bubbles. Thin-film interference is the key.

Hint

(a)

\(
\begin{aligned}
& n \lambda=2 d \sin \theta, \quad \theta=60^{\circ}, n=2 \\
& d=\frac{2 \times 1 \times 2 \times 10^{-10}}{2 \times \sqrt{3}}=1.15 Å
\end{aligned}
\)

Hint

(d) Frequency \(=n\); Wavelength \(=\lambda\); Velocity of light in air \(=v\) and refractive index of glass slab \(=\mu\). Frequency of light remains the same, when it changes the medium. Refractive index is the ratio of wavelengths in vacuum and in the given medium. Similarly refractive index is also the ratio of velocities in vacuum and in the given medium.

Hint

(d) Refractive index of water \(\left(\mu_2\right)=1.33\).
\(
\frac{v_2}{v_1}=\frac{\mu_1}{\mu_2}=\frac{1}{1.33}
\)
Therefore \(v_2=\frac{v_1}{1.33}=\frac{3 \times 10^8}{1.33}=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)

Hint

\(
\text { (b) Time }=\frac{\text { distance }}{\text { velocity }}=\frac{t}{v}=\frac{t}{c / \mu}=\frac{\mu t}{c}
\)

Hint

(a) Wavelength \((\lambda)=5000 Å\) and velocity \((v)=1.5 \times 10^4 \mathrm{~m} / \mathrm{s}\)
Wavelength of the approaching star,
\(\lambda^{\prime}=\lambda \frac{c-v}{c}\)
or \(\frac{\lambda^{\prime}}{\lambda}=1-\frac{v}{c}\)
or, \(\frac{v}{c}=1-\frac{\lambda^{\prime}}{\lambda}=\frac{\lambda-\lambda^{\prime}}{\lambda}=\frac{\Delta \lambda}{\lambda}\)
Therefore \(\Delta \lambda=\lambda \times \frac{v}{c}=5000 Å \times \frac{1.5 \times 10^6}{3 \times 10^8}=25 Å\)
(here \(\Delta \lambda\) is the change in the wavelength)

Hint

(d) Separations between the slits \(d_1=16 \mathrm{~cm}\) and \(d_2=9 \mathrm{~cm}\)
Actual distance of separation
\(
d=\sqrt{d_1 d_2}=\sqrt{16 \times 9}=12 \mathrm{~cm}
\)

Hint

(d) Ray optics is valid when characteristic dimensions are (d) much larger than the wavelength of light.

Explanation: Ray optics is an approximation that works well when objects are significantly larger than the wavelength of light, allowing us to treat light as rays traveling in straight lines.
Why other options are incorrect:
(a) much smaller than the wavelength of light: When objects are much smaller than the wavelength of light, wave effects like diffraction become dominant, making ray optics invalid.
(b) of the same order as the wavelength of light: If the dimensions are similar to the wavelength of light, diffraction effects become significant, and ray optics is not a good approximation.
(c) of the order of one millimetre: One millimetre is about 1000 times the wavelength of visible light. While this might seem large to us, it’s not large enough for ray optics to be fully valid. For ray optics to be accurate, objects need to be significantly larger than a millimetre.

Hint

\(
\begin{aligned}
& \text { (c) } v_g=\frac{c}{\mu}=\frac{3 \times 10^8}{\frac{3}{2}}=2 \times 10^8 \mathrm{~m} / \mathrm{s} \\
& t=\frac{x}{v_g}=\frac{4 \times 10^{-3}}{2 \times 10^8}=2 \times 10^{-11} \mathrm{~s}
\end{aligned}
\)

Hint

(d) In vacuum, \(\lambda\) increases very slightly compared to that in air. As \(\beta \propto \lambda\), therefore, the width of interference fringe increases slightly.

Hint

\(
\text { (c)  } \lambda^{\prime} \text { of refracted light is smaller, because } \lambda^{\prime}=\frac{\lambda}{\mu}
\)

Hint

(a) As \(\beta=\frac{\lambda D}{d}\) and \(\lambda_b<\lambda_y\),
\(\therefore \quad\) Fringe width \(\beta\) will decrease.

Hint

(a)

\(
\lambda_g=\frac{\lambda_a}{\mu}=\frac{5460}{1.5}=3640 Å
\)

Hint

(a)

\(
\frac{I_1}{I_2}=\frac{a^2}{b^2}=\frac{4}{1} \quad \therefore \frac{a}{b}=\frac{2}{1}
\)

Hint

\(
\begin{aligned}
&\text { (d) For dark fringe, } x=(2 n-1) \frac{\lambda D}{2 d}\\
&\begin{gathered}
\lambda=\frac{2 x d}{(2 n-1) D}=\frac{2 \times 10^{-3} \times 0.9 \times 10^{-3}}{(2 \times 2-1) \times 1} \\
\lambda=0.6 \times 10^{-6} \mathrm{~m}=6 \times 10^{-5} \mathrm{~cm}
\end{gathered}
\end{aligned}
\)

Hint

(a)

\(
\beta^{\prime}=\frac{\beta}{\mu}=\frac{0.4}{4 / 3}=0.3 \mathrm{~mm}
\)

Hint

(c) Distance of \(n^{\text {th }}\) maxima \(x=n \lambda \frac{D}{d} \propto \lambda\)
As \(\lambda_b<\lambda_g\)
\(\therefore \quad x\) (blue) \(<x\) (green).

Hint

(c) Interference is a wave phenomenon shown by both the light waves and sound waves.

Hint

(d) Sound waves can not be polarised as they are longitudinal. Light waves can be polarised as they are transverse.

Hint

(d)  Huygen’s construction of wavefront does not apply to origin of spectra which is explained by quantum theory.

Hint

\(
\begin{aligned}
&\text { (c) For first minimum, a } \sin \theta=n \lambda=1 \lambda\\
&\begin{aligned}
& \sin \theta=\frac{\lambda}{a}=\frac{5000 \times 10^{-10}}{0.001 \times 10^{-3}}=0.5 \\
& \theta=30^{\circ}
\end{aligned}
\end{aligned}
\)

Hint

(b) Angular limit of resolution of eye, \(\theta=\frac{\lambda}{d}\), where, d is diameter of eye lens.
Also, if \(Y\) is the minimum separation between two objects at distance \(D\) from eye then,
\(
\begin{array}{r}
\theta=\frac{Y}{D} \\
\Rightarrow \frac{Y}{D}=\frac{\lambda}{d} \Rightarrow Y=\frac{\lambda D}{d} \dots(1)
\end{array}
\)
Here, wavelength \(\lambda=5000 Å=5 \times 10^{-7} \mathrm{~m}\)
\(
D=50 \mathrm{~m}
\)
Diameter of eye lens \(=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\) From eq. (1), minimum separation is
\(
Y=\frac{5 \times 10^{-7} \times 50}{2 \times 10^{-3}}=12.5 \times 10^{-3} \mathrm{~m}=12.5 \mathrm{~cm}
\)

Hint

(d)

\(
\begin{aligned}
& I_1=I_0 \cos ^2\left(\frac{45}{2}\right) \\
& I_2=I_1 \cos ^2\left(90-\frac{45}{2}\right) \\
& =I_0 \cos ^2\left(\frac{45}{2}\right) \sin ^2\left(\frac{45}{2}\right) \\
& =\frac{I_0}{4}\left(4 \cos ^2\left(\frac{45}{2}\right) \sin ^2\left(\frac{45}{2}\right)\right) \\
& =\frac{I_0}{4} \sin ^2 45^{\circ}=\frac{I_0}{8}
\end{aligned}
\)

Hint

(b) Using Brewster law

\(
\begin{aligned}
&\begin{aligned}
& \mu=\tan \theta_p \\
& \Rightarrow 1.73=\tan \theta_p \\
& \Rightarrow \sqrt{3}=\tan \theta_p \\
& \Rightarrow \theta_p=60^{\circ}
\end{aligned}\\
&\text { At this polarising angle, reflected light is perfectly polarized and transmitted light is partially polarised. }
\end{aligned}
\)