NEET Practice Q & A

Hint

(c) We have, \(\beta=\frac{\lambda D}{d}\)
\(\because \quad\) The arrangement is placed in a liquid of refractive index \(n\).
\(
\therefore \quad \lambda^{\prime}=\frac{\lambda}{n}
\)
Therefore, fringe width, \(\beta^{\prime}=\frac{\lambda D}{d n}\)
\(
\beta^{\prime}=\frac{\beta}{n}
\)

Hint

\(
\begin{array}{ll}
\text { (c) } \theta(\text { in radian }) & =\frac{\lambda}{d} \\
\therefore \quad d =\frac{\lambda}{\theta}=\frac{6 \times 10^{-7}}{\pi / 180^{\circ}}=3.4 \times 10^{-5} \mathrm{~m}
\end{array}
\)

Hint

(b) When phase difference is 0 , intensity is maximum, i.e. \(I_{\text {max }}\).
When \(\quad \phi=\frac{\pi}{2}\) or \(\frac{\phi}{2}=\frac{\pi}{4}\), the intensity will be
\(
I=I_{\max } \cos ^2 \frac{\phi}{2}=\frac{I_{\max }}{2}
\)
\(\therefore\) The desired ratio is \(2: 1\).

Hint

(c) For constructive interference in case of soap film,
\(
2 \mu t=\left(n-\frac{1}{2}\right) \lambda
\)
For minimum thickness \(t, n=1\)
\(
\begin{aligned}
\Rightarrow & 2 \mu t =\frac{\lambda}{2} \\
\therefore & t =\frac{\lambda}{4 \mu}=\frac{750}{4 \times 1.33}=141 \mathrm{~nm}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Fringe width, } \beta=\frac{\lambda D}{d} \Rightarrow \beta \propto \frac{D}{d}\\
&\frac{\beta_2}{\beta_1}=\frac{D_2}{d_2} \times \frac{d_1}{D_1}=\frac{\frac{1}{2} D_1}{2 d_1} \times \frac{d_1}{D_1}=\frac{1}{4} \Rightarrow \beta_2=\frac{\beta_1}{4}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
\Delta y=\frac{5 \lambda D}{d} & -\frac{5 \lambda D}{2 d}=\frac{5 \lambda D}{2 d} \\
& =\frac{5 \times 6.5 \times 10^{-7} \times 1}{2 \times 10^{-3}} \mathrm{~m}=1.63 \mathrm{~mm}
\end{aligned}
\)

Hint

\(
\text { (b) } n_1 \lambda_1=n_2 \lambda_2 \Rightarrow n_2=\frac{n_1 \lambda_1}{\lambda_2}=12 \times \frac{600}{400}=18
\)

Hint

\(
\begin{aligned}
&\text { (c) In single narrow slit, for first minima, }\\
&\begin{aligned}
& d \sin \theta  =\lambda \\
\Rightarrow & \sin \theta  =\frac{\lambda}{d} \\
\therefore & \theta  =\sin ^{-1}\left(\frac{5000 \times 10^{-10}}{0.001 \times 10^{-3}}\right)=30^{\circ}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) We have, } y_1=\frac{10 \lambda_1 D}{d}, y_2=\frac{5 \lambda_2 D}{d}\\
&\therefore \quad \frac{y_1}{y_2}=\frac{2 \lambda_1}{\lambda_2}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { (c) Intensity, }  I=4 I_0 \cos ^2 \frac{\phi}{2} \\
\because  I=I_0 \Rightarrow \cos ^2 \frac{\phi}{2}=\frac{1}{4} \\
\therefore  \phi=\frac{2 \pi}{3}
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (a) We have, } \frac{2(6)-1}{2} \beta=10 \beta^{\prime}\\
&\therefore \quad \frac{\beta}{\beta^{\prime}}=\frac{10}{5.5}=1.8=\mu
\end{aligned}
\)

Hint

(b) Let \(n\)th red fringe coincides with \(m\) th blue fringe, then
\(
\begin{aligned}
& \frac{n \lambda_R D}{d} & =\frac{m \lambda_B D}{d} \\
\therefore & \frac{n}{m} & =\frac{\lambda_B}{\lambda_R}=\frac{480}{600}=\frac{4}{5}
\end{aligned}
\)
i.e. 4th red fringe coincides with 5th blue fringe. Next 8th red fringe will coincide with 10th blue fringe and so on. Thus, \(n=4\)
and \(m=n+1=5\)

Hint

(b) For maximum (brightness) intensity on the screen, path difference should be integer multiple of wavelength, i.e.
\(
\begin{gathered}
d \sin \theta=n \lambda \\
\sin \theta=\frac{n \lambda}{d}=\frac{n(2000)}{7000}=\frac{n}{3.5}
\end{gathered}
\)
As, \(\quad(\sin \theta)_{\max }=1 \Rightarrow n=3.5\)
So, number of maximas \(=0, \pm 1, \pm 2, \pm 3\)
Thus, only seven maximas can be obtained on both sides of the screen.

Hint

(a) When a light beam is tilted, the number of minima observed on the screen remains the same. Therefore, \(\mathrm{n} 1=\mathrm{n} 2\). The fringe pattern shifts, but the total number of fringes (and thus minima) stays constant.

Explanation:
In a double-slit experiment, the number of minima is determined by the wavelength of light and the separation of the slits, not the angle of incidence of the light beam as long as the beam remains parallel to the screen. Tilting the beam simply shifts the entire interference pattern, but the pattern’s width and the number of minima remain unchanged.
Therefore, the relationship between \(n_1\) and \(n_2\) is: \(n_1=n_2\)

Hint

(d) If \(d \sin \theta=(\mu-1) t\), central fringe is obtained at \(O\).
If \(d \sin \theta>(\mu-1) t\), central fringe is obtained above \(O\).
If \(d \sin \theta<(\mu-1) t\), central fringe is obtained below \(O\).

Hint

\(
\text { (a) } I_2=I_1 \cos ^2\left(90^{\circ}-\theta\right)=\frac{I^{\prime}}{2} \cos ^2 \theta \sin ^2 \theta=\frac{I^{\prime}}{8}(\sin 2 \theta)^2
\)

\(
\begin{aligned}
3 & =\frac{32}{8}(\sin 2 \theta)^2 \\
\sin 2 \theta=\frac{\sqrt{3}}{2} & =\sin 60^{\circ} \text { or } \sin 120^{\circ} \\
\theta & =30^{\circ} \text { or } 60^{\circ}
\end{aligned}
\)

Hint

(a) Number of fringes shifted, \(N=\frac{(\mu-1) t}{\lambda}\)
\(\therefore\) Refractive index of the sheet,
\(
\mu=\frac{N \lambda}{t}+1=\frac{30 \times 6000 \times 10^{-10}}{3.6 \times 10^{-5}}+1=1.5
\)

Hint

(c) Fringe width, \(\beta=\frac{D \lambda}{d}\)
i.e. \(\beta \propto \lambda\)
So, wavelength \(\lambda\) and hence fringe width \(\beta[latex] decreases 1.5 or [latex]\frac{3}{2}\) times when immersed in liquid. The distance between central maximum and 10th maxima is 3 cm in vacuum. When immersed in liquid it will reduce to 2 cm. Position of central maximum will not change while 10 th maxima will be obtained at \(y=4 \mathrm{~cm}\).

Hint

\(
\begin{aligned}
&\text { (b) Position of third maxima, }\\
&x_3=\frac{3 \lambda D}{d}=\frac{3\left(5 \times 10^{-5}\right)(200)}{0.02}=1.5 \mathrm{~cm}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (d) } I_1=\frac{I}{4} \text { and } I_2=\frac{9 I}{32} \\
& \therefore \quad \frac{I_1}{I_2}=\frac{8}{9} \\
& \text { or } \sqrt{\frac{I_1}{I_2}}=\frac{2 \sqrt{2}}{3}
\end{aligned}
\)
\(
\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\right)^2=\left(\frac{2 \sqrt{2}+3}{2 \sqrt{2}-3}\right)^2
\)

Hint

\(
\begin{aligned}
&\text { (a) According to the question, }\\
&y=\frac{\omega}{4}=\frac{\lambda D}{4 d} \Rightarrow \Delta x=\frac{y d}{D}=\frac{\lambda}{4}
\end{aligned}
\)
Phase difference, \(\phi=\frac{2 \pi}{\lambda} \cdot \Delta x=\frac{\pi}{2}\)
\(
\therefore \quad \frac{\phi}{2}=\frac{\pi}{4}
\)
Now, intensity, \(I_2=I_1 \cos ^2 \frac{\phi}{2}\)
\(
\frac{I_1}{I_2}=\frac{1}{\cos ^2 \phi / 2}=2
\)

Hint

(c) Intensity in the first case, \(I_1=\left(\sqrt{I_0}+\sqrt{I_0}\right)^2=4 I_0\) where, \(I_0=\) intensity due to each source.
Intensity in second case, \(I_2=I_0+I_0=2 I_0\)
\(
\frac{I_1}{I_2}=\frac{4 I_0}{2 I_0}=\frac{2}{1}
\)

Hint

\(
\begin{aligned}
& \text { (a) Shift }=\frac{(\mu-1) t D}{d}=\omega=\frac{\lambda D}{d} \\
& \therefore \quad t=2 \lambda \text { (as } \mu=1.5)
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Path difference, }\\
&\begin{aligned}
\Delta x=\frac{y d}{D} & =\left(\frac{1.25 \lambda D}{d}\right)\left(\frac{d}{D}\right) \\
& =1.25 \lambda=\frac{5}{4} \lambda
\end{aligned}
\end{aligned}
\)
\(
\begin{array}{ll}
\text { Phase difference, } \phi & =\frac{2 \pi}{\lambda} \cdot \Delta x \\
& =\frac{2 \pi}{\lambda} \times \frac{5}{4} \lambda=\frac{5}{2} \pi \\
\therefore \quad & \frac{\phi}{2}=\frac{5}{4} \pi=225^{\circ} \\
\therefore \quad & \text { Intensity, } I=I_{\max } \cos ^2 \frac{\phi}{2}=\frac{1}{2} I_{\max }
\end{array}
\)

Hint

\(
\text { (c) Intensity, } I=4 I_0 \cos ^2 \frac{\phi}{2}
\)
\(
\begin{aligned}
& I=I_0 \\
& \phi=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \cdot \Delta x=\frac{2 \pi}{\lambda}\left(\frac{y d}{D}\right) \\
& y=\frac{\lambda D}{3 d}
\end{aligned}
\)

Hint

\(
\text { (c) } 2 I_0=4 I_0 \cos ^2 \frac{\phi}{2}
\)
\(
\begin{aligned}
& \phi=\frac{\pi}{2}=\frac{2 \pi}{\lambda} \cdot \Delta x=\frac{2 \pi}{\lambda} \cdot\left(\frac{y d}{D}\right) \\
& y_1=\frac{\lambda D}{4 d}
\end{aligned}
\)
Further, \(I_0=4 I_0 \cos ^2 \frac{\phi}{2}\)
\(
\begin{array}{ll}
\Rightarrow & \phi=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda}\left(\frac{y_2 d}{D}\right) \\
\therefore & y_2=\frac{\lambda D}{3 d}
\end{array}
\)
Minimum separation, \(\Delta y=y_2-y_1=\frac{1}{12}\left(\frac{\lambda D}{d}\right)=\frac{\alpha}{12}\)

Hint

\(
\begin{aligned}
&\text { (c) Five fringes have to be shifted. }\\
&\begin{aligned}
& \therefore \quad 5=\frac{\left(\mu_1-1\right) t-\left(\mu_2-1\right) t}{\lambda}=\frac{\left(\mu_1-\mu_2\right) t}{\lambda} \\
& \begin{aligned}
\therefore \text { Thickness, } t=\frac{5 \lambda}{\left(\mu_1-\mu_2\right)} & =\frac{5 \times 480 \times 10^{-9}}{1.7-1.4} \\
& =8 \times 10^{-6} \mathrm{~m} \text { or } 8 \mu \mathrm{~m}
\end{aligned}
\end{aligned}
\end{aligned}
\)

Hint

(a) At \(x=0, \Delta x=5 \lambda\)
which corresponds to a maximum.
At \(x=\infty, \Delta x=0\), this also corresponds to a maximum.
\(\therefore\) Other possible maxima between \(x=0\) and \(x=\infty\) are \(\Delta x=\lambda, 2 \lambda, 3 \lambda, 4 \lambda\).

Hint

(a) Location of the point, \(y=\frac{\omega}{4}=\frac{\lambda D}{4 d}\)
\(
\begin{aligned}
\Delta x & =\frac{y d}{D}=\frac{\lambda}{4} \\
\phi & =\frac{2 \pi}{\lambda} \cdot \Delta x=\frac{\pi}{2} \\
I_R & =I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi \\
4 I & =2 I+I_2+0 \\
I_2 & =2 I
\end{aligned}
\)

Hint

(c) Say \(P\) be the point on \(X\)-axis, where constructive interference is obtained.

\(
\begin{aligned}
B P-A P & =2 \lambda \text { or } \lambda \\
\sqrt{9 \lambda^2+x^2}-x & =2 \lambda \\
9 \lambda^2+x^2 & =x^2+4 \lambda^2+4 x \lambda \\
x & =\frac{5 \lambda}{4}
\end{aligned}
\)
Further, \(\sqrt{9 \lambda^2+x^2}-x=\lambda\)
\(
\begin{aligned}
& \therefore 9 \lambda^2+x^2 & =x^2+\lambda^2+2 x \lambda \\
& \therefore x & =4 \lambda
\end{aligned}
\)
Hence, the distance from \(A\) on the \(X\)-axis at which the interference is constructive is \(5 \lambda / 4\).

Hint

(b) Path difference at a point \(Q\) on the circle is

\(
\Delta y=d \cos \theta \dots(i)
\)
For maxima at \(Q\), path difference should be integer multiple of wavelength
\(
\Delta y=n \lambda \dots(ii)
\)
\(
\begin{aligned}
&\text { From Eqs. (i) and (ii), we get }\\
&\begin{array}{ll}
& n \lambda=d \cos \theta \\
\therefore & \theta=\cos ^{-1}\left(\frac{n \lambda}{d}\right) \\
\text { For } & n=4 \\
\text { Angular position, } & \theta=\cos ^{-1}\left(\frac{4 \lambda}{d}\right)
\end{array}
\end{aligned}
\)

Hint

(c) Path difference between the waves reaching at \(O\),
\(
\Delta=\Delta_1+\Delta_2
\)
where, \(\Delta_1=\) initial path difference and \(\Delta_2=\) path difference between the waves after emerging from slits.
\(
\text { Now, } \quad \Delta_1=S S_1-S S_2=\sqrt{\left(D^2+d^2\right)}-D
\)
and \(\quad \Delta_2=S_1 O-S_2 O=\sqrt{\left(D^2+d^2\right)}-D\)
\(
\begin{aligned}
\Delta & =2\left|\sqrt{\left(D^2+d^2\right)}-D\right|=2\left\{\left(D^2+d^2\right)^{1 / 2}-D\right\} \\
& =2\left\{\left(D+\frac{d^2}{2 D}\right)-D\right\} \text { (from binomial expansion) } \\
& =\frac{d^2}{D}
\end{aligned}
\)
For obtaining dark at \(O, \Delta\) must be equals to \((2 n-1) \frac{\lambda}{2}\).
\(
\begin{array}{ll}
\text { i.e. } & \frac{d^2}{D}=(2 n-1) \frac{\lambda}{2} \\
\therefore & d^2=\frac{(2 n-1) \lambda D}{2} \text { or } d=\sqrt{\frac{(2 n-1) \lambda D}{2}}
\end{array}
\)
For minimum distance ( \(n=1\) ),
So,\(d=\sqrt{\frac{\lambda D}{2}}\)

Hint

(b) Path difference between the two rays is given by

\(
\begin{aligned}
\Delta & =C O+P O \\
P R & =d \\
P O & =d \sec \theta \\
C O & =P O \cos 2 \theta=d \sec \theta \cos 2 \theta \\
\Delta & =(d \sec \theta+d \sec \theta \cos 2 \theta)
\end{aligned}
\)
Phase difference between two rays is \(\phi=\pi\) (as one ray is reflected one and another is direct).
Now, for constructive interference, path difference should be even multiple of half-wavelength.
i.e. \(\Delta=\lambda / 2,3 \lambda / 2, \ldots\)
\(
\text { So, } d \sec \theta+d \sec \theta \cos 2 \theta=\frac{\lambda}{2}
\)
\(
\begin{aligned}
d \sec \theta(1+\cos 2 \theta) & =\frac{\lambda}{2} \\
d \sec \theta\left(2 \cos ^2 \theta\right) & =\frac{\lambda}{2} \\
\cos \theta & =\lambda / 4 d
\end{aligned}
\)

Hint

(a) Given, distance between source and observer,
\(
D=2 \mathrm{~m}
\)
Distance of 5th dark fringe from centre
\(
=4 \times 10^{-3} \mathrm{~m}
\)
Wavelength, \(\lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}\)
Distance of 5th dark fringe from centre
\(
=\beta+\beta+\beta+\beta+\frac{\beta}{2}=\frac{9 \beta}{2}
\)
where, \(\beta\) = fringe width.
According to question, \(\frac{9 \beta}{2}=4 \times 10^{-3}\)
\(
\begin{gathered}
\frac{9}{2} \times \frac{D \lambda}{d}=4 \times 10^{-3} \\
\frac{9}{2} \times \frac{2 \times 6 \times 10^{-7}}{d}=4 \times 10^{-3} \\
\Rightarrow \quad \begin{aligned}
d & =1.35 \times 10^{-3} \mathrm{~m} \\
& =1.35 \mathrm{~mm}
\end{aligned}
\end{gathered}
\)

Hint

(a) Given, wavelength of light,
\(
\lambda=500 \mathrm{~nm}=5 \times 10^{-7} \mathrm{~m}
\)
Distance between slit and screen,
\(
D=1.8 \mathrm{~m}
\)
Distance between two slits,
\(
d=0.4 \mathrm{~mm}=4 \times 10^{-4} \mathrm{~m}
\)
Velocity of screen \(=\frac{d D}{d t}=4 \mathrm{~ms}^{-1}\)
Speed of first maxima \(=\frac{d \beta}{d t}=\) ?
In Young’s double slit experiment, fringe width,
\(
\beta=\frac{D \lambda}{d}
\)
Differentiating with respect to \(t[latex], we get
[latex]
\begin{aligned}
\frac{d \beta}{d t}=\frac{\lambda}{d} \cdot \frac{d D}{d t} & =\frac{5 \times 10^{-7}}{4 \times 10^{-4}} \times 4 \\
& =5 \times 10^{-3} \mathrm{~ms}^{-1} \\
& =5 \mathrm{mms}^{-1}
\end{aligned}
\)

Hint

(a) Distance between position of bright and dark fringes is called fringe width in Young’s double slit experiment, which is given by \(\beta=\frac{D \lambda}{d}\).
where, \(\beta=\) fringe width,
\(D=\) distance between source and screen,
\(d=\) distance between two slits
and \(\lambda=\) wavelength of monochromatic light wave.
Therefore, in Young’s double slit experiment, fringes are of equal width, i.e. width of bright and dark fringes are same. Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Hint

(b) When unpolarised light coming from sun, enters the atmosphere of earth, then it scatters due to presence of atmospheric particles. The scattered light seen in a direction perpendicular to the direction of incidence is found to be plane polarised.
When the light from the sun incident on reflecting surface of earth at an angle of polarisation, then reflecting light is polarised. Little amount of light is incident on polarising direction, hence reflected light is partially polarised.
Hence, Assertion and Reason both are correct but Reason is not the correct explanation of Assertion.

Hint

(d) Huygens’ principle use the phenomenon of light as reflection, refraction and diffraction but does not use point of spectra origin.

Hint

(c) In Young’s double slit experiment, position of bright fringes on the screen are given by
\(
x=\frac{n D \lambda}{d}
\)
Hence, distance of \(n\)th maxima will be \(x=n \lambda \frac{D}{d} \propto \lambda\)
\(
\lambda_{\text {blue }}=4360 Å, \lambda_{\text {green }}=5480 Å
\)
where, \(x=\) distance of 4th maxima from the central one.
\(
\begin{array}{ll}
\text { As, } & \lambda_{\text {blue }}<\lambda_{\text {green }} \\
\therefore & x_{\text {blue }}<x_{\text {green }}
\end{array}
\)

Hint

(c) The situation is shown in the figure

The path difference between the waves reaching the point \(P\) is
\(
\begin{aligned}
\Delta p & =B P-A P \\
& =\left(d^2+b^2\right)^{1 / 2}-d \\
& =d\left(1+\frac{b^2}{d^2}\right)^{1 / 2}-d \\
& \approx d\left[1+\frac{1}{2} \cdot \frac{b^2}{d^2}\right]-d=\frac{b^2}{2 d}
\end{aligned}
\)
For a dark band, \(\Delta p=\frac{b^2}{2 d}=(2 n-1) \frac{\lambda}{2}\) where, \(n=1,2,3\) or \(\lambda=\frac{1}{(2 n-1)} \cdot \frac{b^2}{d}\)
Hence, the missing wavelengths are \(\frac{b^2}{d}, \frac{b^2}{3 d}, \frac{b^2}{5 d}, \ldots\).

Hint

(b) The condition of interference maxima is \(d \sin \theta=n \lambda, \quad \sin \theta=\frac{n \lambda}{d}\)
Given, \(d=2 \lambda\) and \(\sin \theta=\frac{n \lambda}{2 \lambda}=\frac{n}{2}\)
The magnitude of \(\sin \theta\) lies between 0 and 1.
When \(n=0, \sin \theta=0 \Rightarrow \theta=0^{\circ}\)
When \(n=1, \sin \theta=1 / 2 \Rightarrow \theta=30^{\circ}\)
When \(n=2, \sin \theta=1 \Rightarrow \theta=90^{\circ}\)
Thus, there is central maximum at \(\theta=0^{\circ}\), on either side of it maxima lie at \(\theta=30^{\circ}\) and \(\theta=90^{\circ}\), so maximum number of possible interference maxima is five.

Hint

(b) In diffraction, fringe width, \(\beta=\frac{\lambda D}{d}\)
When \(\frac{d}{2}=\beta\), then \(\frac{d}{2}=\frac{\lambda D}{d} \Rightarrow D=\frac{d^2}{2 \lambda}\)

Hint

(a) To find the point of coincidence of bright fringes, we can equate the distance of bright fringes from the central maximum, made by both the wavelengths of light.
Given, distance between the screen and slit,
\(
D=120 \mathrm{~cm}=120 \times 10^{-2} \mathrm{~m}
\)
Slit width,
\(
\begin{aligned}
d & =2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} \\
\lambda_1 & =6500 Å \text { and } \lambda_2=5200 Å
\end{aligned}
\)
Let \(n\)th fringe due to \(\lambda_2=5200 Å\) coincide with \((n-1)\) th bright fringe due to \(\lambda_1=6500 Å\).
\(
\begin{array}{rlrl}
& \therefore & \frac{D n \lambda_2}{d} & =\frac{D(n-1) \lambda_1}{d} \\
\text { or } & n \times 5200 & =(n-1) 6500 \\
\text { or } & 4 n & =5 n-5
\end{array}
\)
\(
\begin{aligned}
&\Rightarrow \quad n=5\\
&\text { So, the least distance required, } y_n=\frac{n \lambda_2 D}{d}\\
&\begin{aligned}
& =\frac{5 \times 5200 \times 10^{-10} \times 120 \times 10^{-2}}{2 \times 10^{-3}} \\
& =0.16 \times 10^{-2} \mathrm{~m}=0.16 \mathrm{~cm}
\end{aligned}
\end{aligned}
\)

Hint

(c) The correct condition for Fraunhofer diffraction due to a single slit is: Source is at infinity and the incident beam is parallel.

Explanation: Fraunhofer diffraction requires a plane wavefront incident on the slit, which is achieved when the light source is infinitely far away and the beam is parallel.

Hint

(a) Most of the light is diffracted between the two first order minima.
These minima occur at angles given by \(b \sin \theta= \pm \lambda\).
\(\Rightarrow \quad \sin \theta= \pm \frac{\lambda}{b}= \pm \frac{4.5 \times 10^{-7}}{0.2 \times 10^{-3}}= \pm 2.25 \times 10^{-3}\)
\(\Rightarrow \quad \theta \simeq 2.25 \times 10^{-3} \mathrm{rad} \quad\) (angle is very small)
The angular divergence, \(2 \theta=4.5 \times 10^{-3} \mathrm{rad}\)

Hint

(b) The correct condition for obtaining secondary maxima in the diffraction pattern due to a single slit is given by option (b) \(a \sin \theta=(2 n-1) \lambda / 2\).

Explanation:
In single-slit diffraction, secondary maxima are observed as less intense bright fringes located between the dark fringes (minima). The condition for these secondary maxima to occur is that the path difference between wavelets from different parts of the slit is an odd multiple of half the wavelength.

For secondary maxima, we need to consider the odd multiples of the wavelength. The correct condition for secondary maxima can be derived from the equation for minima, which occurs at:
\(
a \sin \theta=\left(n+\frac{1}{2}\right) \lambda
\)
where \(n\) is a positive integer \((1,2,3, \ldots)\)

Hint

\(
\begin{aligned}
&\text { (b) Fringe width, } \beta=\frac{D \lambda}{d}\\
&\Rightarrow \quad d=\frac{D \lambda}{\beta}=\frac{6000 \times 10^{-10} \times 40 \times 10^{-2}}{0.012 \times 10^{-2}}=0.2 \mathrm{~cm}
\end{aligned}
\)

Hint

(d) Given, \(\theta_P=57^{\circ}\)
According to Brewster’s law, we have
\(
\theta_P+\theta_r=90^{\circ} \Rightarrow \theta_r=90^{\circ}-\theta_P=90^{\circ}-57^{\circ}
\)
Angle of refraction, \(\theta_r=33^{\circ}\)

Hint

(c) To observe diffraction, the size of the obstacle should be of the order of wavelength, i.e. \(\lambda \approx d\).

Hint

(a) According to the question,
\(
\lambda=6250 \AA=6250 \times 10^{-10} \times 10^2 \mathrm{~cm}=6250 \times 10^{-8} \mathrm{~cm}
\)
Width of the slit, \(a=2 \times 10^{-2} \mathrm{~cm}\)
The angular width of central maximum, \(\beta=2 \theta=2 \frac{\lambda}{a}\)
Thus, linear width is
\(
\begin{aligned}
L & =\beta \times D=\frac{2 \lambda}{a} \times D=\frac{2 \times 6250 \times 10^{-8}}{2 \times 10^{-2}} \times 50 \\
& =6250 \times 10^{-6} \times 50=3125 \times 10^{-4} \\
& =312.5 \times 10^{-3} \mathrm{~cm}
\end{aligned}
\)

Hint

(a) Given, wavelength, \(\lambda=590 \mathrm{~nm}=590 \times 10^{-9} \mathrm{~m}\)
Diameter of lens \(=10 \times 10^{-2} \mathrm{~m}\)
Radius of lens \(=5 \times 10^{-2} \mathrm{~m}\)
Focal length of lens \(=20 \times 10^{-2} \mathrm{~m}\)
Now, the diameter of image disc
\(
\begin{aligned}
D=2 R=2\left(1.22 \frac{\lambda D}{b}\right) & =\left(\frac{1.22 \times 590 \times 10^{-9} \times 20 \times 10^{-2}}{5 \times 10^{-2}}\right) \times 2 \\
& \simeq 5.76 \times 10^{-4} \mathrm{~cm}
\end{aligned}
\)

Hint

(c) According to law of Malus, when a beam of completely plane polarised light is incident on an analyser, the resultant intensity of light \(I\) transmitted from an analyser varies directly as the square of cosine angle between plane of analyser and polariser.
Let initial intensity is \(I\) and final intensity is \(I^{\prime}\), then
\(
I^{\prime}=I \cos ^2 60^{\circ}=I\left(\frac{1}{2}\right)^2=\frac{I}{4}
\)
Now, % difference between initial and final intensities,
\(
\begin{aligned}
& =\frac{I-(I / 4)}{I} \times 100=\frac{3 I}{4} \times \frac{1}{I} \times 100 \\
& =\frac{3}{4} \times 100=75 \%
\end{aligned}
\)

Hint

(a) A hyperbola.
Explanation:
In Young’s double-slit experiment, a point \(P\) on the screen will experience constructive interference (bright fringe) when the path difference between the light waves traveling from the two slits to \(P\) is a multiple of the wavelength. This condition can be mathematically represented as:
\(
d \sin (\theta)=n \lambda
\)
where \(d\) is the distance between the slits, \(\theta\) is the angle between the line connecting \(P\) to the center of the slits and the normal to the slits, \(n\) is an integer representing the order of the fringe ( 0 for the central maximum, 1 for the first bright fringe, etc.), and \(\lambda\) is the wavelength of the light.
If we rearrange this equation to solve for \(d \sin (\theta)\) :
\(
d \sin (\theta)=n \lambda
\)
The equation of a hyperbola in standard form is:
\(
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\)
If we make the substitutions \(x=d \sin (\theta)\) and \(y=n \lambda\), we can see that the equation for a hyperbola describes the locus of points \(P\) with a constant path difference.

Hint

(a) Phase.
Explanation: A wavefront is defined as the locus of all points in a medium that are vibrating in the same phase. This means that all particles on a wavefront are at the same point in their oscillation cycle (e.g., all at crests or all at troughs).

Hint

\(
\text { (c) For net intensity, } I^{\prime}=4 I_0 \cos ^2 \frac{\phi}{2}
\)
\(
\text { Case I: } K=4 I_0 \cos ^2\left(\frac{2 \pi}{2}\right) \quad\left(\because \phi=\frac{2 \pi}{\lambda} \times \lambda\right)
\)
\(
K=4 I_0 \cos ^2(\pi) \Rightarrow K=4 I_0 \dots(i)
\)
\(
\text { Case II: } \quad K^{\prime}=4 I_0 \cos ^2\left(\frac{\pi / 2}{2}\right) \quad\left(\because \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}\right)
\)
\(
K^{\prime}=4 I_0 \cos ^2(\pi / 4) \Rightarrow K^{\prime}=2 I_0 \dots(ii)
\)
\(
\begin{aligned}
&\text { Comparing Eqs. (i) and (ii), we get }\\
&K^{\prime}=K / 2
\end{aligned}
\)

Hint

(d) We know that,
The maximum intensity, \(I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2 \dots(i)\)
The minimum intensity, \(I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \dots(ii)\)
So, the ratio of the maximum and minimum intensities is
\(
\begin{aligned}
& \frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\frac{(\sqrt{4 I}+\sqrt{I})^2}{(\sqrt{4 I}-\sqrt{I})^2} \\
\Rightarrow \quad & \frac{I_{\max }}{I_{\min }}=\left(\frac{3 \sqrt{I}}{\sqrt{I}}\right)^2=\frac{9}{1}=9: 1
\end{aligned}
\)
Hence, the maximum and minimum intensities in the resulting pattern are \(9 I\) and \(I\).

Hint

(a) The diagram below shows a thin soap film and soap bubbles.

The rays 1 and 2 produce the phenomenon of interference when they meet. The path difference between the rays are proportional to thickness of film, i.e. path difference \((\Delta x) \propto t\). As, thickness \(t\) varies, so \(\Delta x\) varies. Maximas and minimas appear for different wavelengths in incident white light. When white light is used, the path difference (phase difference) will satisfy the condition of maxima for certain wavelengths and condition of minima for other wavelengths. This gives coloured appearance of the film.

Hint

(c) \(\frac{I_{\max }}{I_{\min }}=\left(\frac{a_{\max }}{a_{\min }}\right)^2=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2\)
According to given question, \(\frac{I_{\max }}{I_{\min }}=\frac{9}{1}\)
\(
\begin{aligned}
9 & =\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}\right)^2 \Rightarrow(3)^2=\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}\right)^2 \\
3 & =\frac{a_1 / a_2+1}{a_1 / a_2-1} \Rightarrow 3\left(\frac{a_1}{a_2}\right)-3=\frac{a_1}{a_2}+1 \\
2\left(\frac{a_1}{a_2}\right) & =4 \Rightarrow \frac{a_1}{a_2}=\frac{4}{2}=\frac{2}{1} \Rightarrow a_1: a_2=2: 1
\end{aligned}
\)

Hint

(c) According to given question,
\(
I_1: I_2=3: 1, a_1: a_2=?
\)
The intensity of light due to an illuminated slit is directly proportional to the width of the slit, if \(w_1\) and \(w_2\) be the width of two silts and \(I_1\) and \(I_2\) are the intensities of light on the screen due to the respective slits, then
\(
\frac{w_1}{w_2}=\frac{I_1}{I_2}=\left(\frac{a_1}{a_2}\right)^2
\)
where, \(a_1\) and \(a_2\) are the amplitudes of the corresponding light wave.
Here, \(\frac{w_1}{w_2}=\frac{3}{1}\)
\(
\Rightarrow \quad\left(\frac{a_1}{a_2}\right)^2=\frac{3}{1} \quad \text { or } a_1: a_2=\sqrt{3}: 1
\)

Hint

(d) Suppose, \(i_p\) is the Brewster’s angle, when unpolarised light is incident at an polarising angle, the reflected component along \(O B\) and refracted component along \(O C\) in the following figure.

Now, \(\quad \angle B O Y+\angle Y O C=90^{\circ}\)
\(
\left(90^{\circ}-i_p\right)+\left(90^{\circ}-r\right)=90^{\circ}
\)
Here, \(r\) is the angle of refraction.
\(
\Rightarrow \quad 90^{\circ}-i_p=r
\)
According to Snell’s law, \(\mu=\frac{\sin i}{\sin r}\)
When \(i=i_p, r=90^{\circ}-i_p\), then
\(
\begin{aligned}
&\begin{aligned}
& \mu=\frac{\sin i_p}{\sin \left(90^{\circ}-i_p\right)}=\frac{\sin i_p}{\cos i_p}=\tan i_p \\
\therefore \quad & i_p=\tan ^{-1}(\mu)
\end{aligned}\\
&\text { where, } i_p \text { is polarised angle. }
\end{aligned}
\)

Hint

(d) By Malus’s law, \(\quad I=I_0 \cos ^2 \theta\)
(where, \(I=\) intensity of emergent polarised light, \(I_0=\) intensity passed through polariser)
Given, \(\theta=60^{\circ}, I=\) ?
\(
\Rightarrow \quad I=I_0 \times \cos ^2 60^{\circ}=I_0 \times\left(\frac{1}{2}\right)^2=\frac{I_0}{4}
\)

Hint

(c) The distance of dark fringe from centre is given by
\(
x_n=(2 n+1) \frac{\lambda D}{2 d}
\)
As, \(n=5\) for 5 th dark fringe.
As, \(n=5\) for 5 th dark fringe.
So, \(x_5=\frac{11}{2} \frac{\lambda D}{d}=\frac{11}{2} \beta \left(\because \frac{\lambda D}{d}=\beta\right)\)
\(
=11 / 2 \times 0.002=0.011 \mathrm{~cm}=1.1 \times 10^{-2} \mathrm{~cm}
\)

Hint

\(
\begin{aligned}
& \text { (c) Fringe width, } \beta=\frac{D \lambda}{d} \\
& \text { Given, } d=0.4 \mathrm{~mm}, D=200 \mathrm{~cm}=2000 \mathrm{~mm} \\
& \qquad \begin{array}{l}
\beta=2 \mathrm{~mm} \\
\therefore \quad 2=\frac{2000 \times \lambda}{0.4} \Rightarrow \lambda=400 \mathrm{~nm}
\end{array}
\end{aligned}
\)

Hint

\(
\text { (b) Here, the condition for the formation of rings is } \lambda \approx \frac{d^2}{4 D} \text {. }
\)

Hint

(c) Wave theory of light explains the phenomenon of interference, diffraction and velocity of light in a denser medium or rarer medium, but this theory fails to explain the scattering of light.

Hint

(a) Fringe width \(\beta \propto \lambda\). As given that \(\lambda_1\) produces photoelectric effect, so \(\lambda_1<\lambda_2\).
Hence,\(\beta_1<\beta_2\)

Hint

(d) When one slit in a Young’s double-slit experiment is closed, the interference pattern disappears, and a single-slit diffraction pattern is observed. Therefore, the correct answer is (d) the diffraction pattern is observed.

Explanation:
Interference pattern:
In a typical Young’s double-slit experiment, light waves from two slits interfere, creating a series of bright and dark fringes on a screen.

Single-slit diffraction:
When only one slit is open, the light diffracts through that single slit, creating a central bright maximum and dimmer secondary maxima on either side.

Hint

(c) Given, \(\mu=1.54\)
We know that, \(\theta=\tan ^{-1}(\mu) \Rightarrow \theta=\tan ^{-1}(1.54) \Rightarrow \theta=57^{\circ}\)
On rotating the polaroid, the reflected light is fully polarised. Hence, the intensity gradually reduced to zero and then again increases.

Hint

(c) In Young’s double-slit experiment, dark fringes occur at positions where the phase difference between the two waves is an odd multiple of \(\pi\). The formula for destructive interference (dark fringes) is given as:
\(
\Delta \phi=(2 n-1) \pi
\)
where \(\boldsymbol{\Delta} \boldsymbol{\phi}\) is the phase difference and \(\boldsymbol{n}\) is the order of the dark fringe, with \(n=1,2,3, \ldots\).
For the third dark fringe, \(\boldsymbol{n}=\mathbf{3}\).
\(
\begin{aligned}
& \Delta \phi=(2 \times 3-1) \pi \\
& \Delta \phi=(6-1) \pi \\
& \Delta \phi=5 \pi
\end{aligned}
\)
The phase difference between the two waves at the location of the third dark fringe is \(5 \pi\).

Hint

(a) Fringe width, \(\beta=\frac{D \lambda}{d}\) According to question, \(\frac{\beta_2}{\beta_1}=\frac{\lambda_2 D_2 d_1}{\lambda_1 D_1 d_2}\) Here, \(D_2=2 D_1\) and \(d_2=\frac{d_1}{2}\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{\beta_2}{\beta_1}=\frac{\lambda_2\left(2 D_1\right) \times d_1}{\lambda_1\left(D_1\right) \times\left(\frac{d_1}{2}\right)}=4 \frac{\lambda_2}{\lambda_1}=\frac{4 \times 4 \times 10^{-7}}{6.4 \times 10^{-7}} \\
& \Rightarrow \quad \beta_2=\frac{4 \times 4 \times 10^{-7} \times 1 \times 10^{-4}}{6.4 \times 10^{-7}}=2.5 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)

Hint

(c) Let us first discuss the terms diffraction and scattering of light. The phenomenon in which the bending of light waves around the edges or corners of an obstacle or aperture happens is known as diffraction of light. When light passes through a medium in which the size of the particles present in it is of the order of wavelength of light, then light striking these particles deviates in different directions. This phenomena is known as scattering of light.
The size of the particles in the clouds is very large as compared to the wavelength of the incident light from the sun. Due to which diffraction/scattering of light hardly happens. Therefore, all the rays reflected are seen white. Hence, clouds normally appear to be whitish in colour.
Therefore, the assertion is true but reason is false.

Hint

\(
\begin{aligned}
&\text { (b) By Young’s double slit interference experiment, } \beta=\frac{\lambda D}{d}\\
&\begin{array}{ll}
\text { Given, } & \beta_1=1.2 \mathrm{~mm}, \frac{d_2}{d_1}=1 \frac{1}{2}=1.5 \\
\text { So, } & \beta \propto \frac{1}{d} \Rightarrow \frac{\beta_1}{\beta_2}=\frac{1 / d_1}{1 / d_2} \\
\Rightarrow & \frac{\beta_1}{\beta_2}=\frac{d_2}{d_1}=1.5 \Rightarrow \frac{1.2}{\beta_2}=1.5 \\
\Rightarrow & \beta_2=\frac{1.2}{1.5}=\frac{4}{5}=0.8 \mathrm{~mm}
\end{array}
\end{aligned}
\)

Hint

(d) Let \(\lambda\) be the wavelength of monochromatic light incident on slit \(S\), then the angular distance between two consecutive fringes, i.e. the angular fringe width is
\(
\theta=\lambda / d \dots(i)
\)
where, \(d=\) distance between coherent sources.
Given, \(\frac{\Delta \theta}{\theta}=\frac{10}{100}\)
So, from Eq. (i),
\(
\begin{aligned}
& \frac{\Delta \lambda}{\lambda}=\frac{\Delta \theta}{\theta}=\frac{10}{100}=0.1 \\
\Rightarrow \quad \Delta \lambda & =0.1 \lambda=0.1 \times 5890 Å=589 Å(\text { increases })
\end{aligned}
\)

Hint

(a) \(\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{2 I}+\sqrt{I})^2}{(\sqrt{2 I}-\sqrt{I})^2}=\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2 \text { (given) }\)
\(
\begin{aligned}
\frac{I_{\max }}{I_{\min }} & =\left[\frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}\right]^2 \\
& =\left(\frac{2+1+2 \sqrt{2}}{2-1}\right)^2=\frac{9+8+12 \sqrt{2}}{1}=34
\end{aligned}
\)

Hint

(b) The bandwidth is inversely proportional to the distance between the slits \(d\).
\(
\text { As, } \beta=\frac{\lambda D}{d} \Rightarrow \beta \propto \frac{1}{d}
\)

Hint

(b) If a Young’s double slit set up is shifted from air to water, the fringe width decreases.

Explanation: When submerged in water, the wavelength of light decreases compared to air, which results in a smaller fringe width.
Key points:
Fringe width formula: The fringe width is directly proportional to the wavelength of light.
Refractive index: Water has a higher refractive index than air, meaning light travels slower in water, resulting in a shorter wavelength.
Therefore, options (a), (c), and (d) are incorrect: because the fringe width does not become infinite, increase, or remain unchanged when the experiment is moved to water.

Hint

(b) The path difference \(S_1 P-S_2 P\) is a whole number of wavelength, for the interference at an arbitrary point \(P\).
So, \(S_1 P-S_2 P=n \lambda \text { (for maximum intensity) }\)

Hint

\(
\begin{aligned}
&\text { (c) In Young’s double slit experiment, we have }\\
&\begin{aligned}
\beta & =\frac{\lambda D}{d} \\
\Delta \beta & =\frac{\lambda \Delta D}{d} \\
\lambda & =\frac{\Delta \beta d}{\Delta D} \\
\lambda & =\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}}=6 \times 10^{-7} \mathrm{~m}=600 \mathrm{~nm}
\end{aligned}
\end{aligned}
\)

Hint

(c) The direction in which the first minima occurs is \(\theta\) (say), then \(e \sin \theta=\lambda\)
or \(\quad e \theta=\lambda\) or \(\theta=\frac{\lambda}{e}(\because \theta \simeq \sin \theta\), when \(\theta\) is small \()\) Width of the central maximum \(=2 b \theta+e=\frac{2 b \lambda}{e}+e\)

Hint

(d) Grating with \(10^6\) lines \(/ \mathrm{cm}\) is suitable for constructing spectrometer for visible and ultraviolet region because this corresponds to the wavelength, equivalent to the wavelengths of visible and ultraviolet region.

Hint

(a) Width of central maximum,
\(
\beta=\frac{2 D \lambda}{d}
\)
As, we increase \(d\) to \(2 d, \beta\) becomes \(\frac{\beta}{2}\).
So, it becomes narrower and fainter.

Hint

(c) When rays of monochromatic light of wavelength \(\lambda\) are incident on a diffraction grating in which slit separation is \(d\), then for angle of diffraction \(\theta\), the following relation holds true.
\(
d \sin \theta=n \lambda
\)
where, \(n\) is called the spectrum order.
Given, \(n=1, \lambda=6500 Å=6500 \times 10^{-10} \mathrm{~m}\)
and \(\quad \sin 30^{\circ}=\frac{1}{2}\)
\(
\begin{array}{ll}
\Rightarrow & d=\frac{n \lambda}{\sin 30^{\circ}}=\frac{6500 \times 10^{-10}}{(1 / 2)} \\
\Rightarrow & d=1.3 \times 10^{-6} \mathrm{~m}
\end{array}
\)

Hint

(b) For polarising angle, \(\mu=\tan i_p\)
\(
\begin{aligned}
\mu & =\frac{1}{\sin C} \\
\tan i_p & =\frac{1}{\sin C}
\end{aligned}
\)
where, \(C=\) critical angle.
\(
\begin{array}{rlrl}
& \therefore \cot i_p =\sin \left(\sin ^{-1}\left(\frac{3}{5}\right)\right) \\
\Rightarrow & \cot i_p =\frac{3}{5} \text { or } \tan i=\frac{5}{3} \\
& \text { or } i_p =\tan ^{-1}\left(\frac{5}{3}\right)
\end{array}
\)