Wave Optics

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

Using Brewster law

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

To find the fringe separation (also known as fringe width) in Young's double slit experiment, we use the formula:

$ \beta = \dfrac{\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:

$d = 1.5 \, \mathrm{mm} = 1.5 \times 10^{-3} \, \mathrm{m}$

$D = 1 \, \mathrm{m}$

$\lambda = 600 \times 10^{-9} \, \mathrm{m}$

Substituting these values into the formula:

$ \beta = \dfrac{600 \times 10^{-9} \, \mathrm{m} \times 1 \, \mathrm{m}}{1.5 \times 10^{-3} \, \mathrm{m}} $

Calculating the value:

$ \beta = \dfrac{600 \times 10^{-9}}{1.5 \times 10^{-3}} $

$ \beta = 4 \times 10^{-4} \, \mathrm{m} $

Thus, the fringe separation is $4 \times 10^{-4} \, \mathrm{m}$. The correct answer is:

Option D $4 \times 10^{-4} \mathrm{~m}$

The correct answer is Option C: A and C. Here's why:

Interference occurs when two waves superpose, resulting in a new wave with an amplitude that varies depending on the phase difference between the original waves. Let's analyze the given options:

Option A: y = a sin ωt and Option C: y = a sin (ωt - φ)

These two waves have the same frequency (ω) and amplitude (a), but they differ in phase by φ. This phase difference is crucial for interference. When waves with the same frequency and different phases superpose, they create an interference pattern with alternating regions of constructive and destructive interference.

Option B: y = a sin 2ωt and Option D: y = a sin 3ωt

These waves have different frequencies (2ω and 3ω). While superposition occurs, it doesn't lead to a stable interference pattern. The waves will not consistently reinforce or cancel each other out because their frequencies are not synchronized.

In summary, interference patterns are observed when two waves have the same frequency (or a very small difference in frequency, called "beats") and a phase difference. Options A and C satisfy these conditions, leading to interference.

$\begin{aligned} & I_1=\frac{I_0}{2} \\ & I_B=\left(I_1\right) \cos ^2 60^{\circ} \\ & =\frac{I_0}{2}\left(\frac{1}{2}\right)^2=\frac{I_0}{8} \\ & I_C=I_B \cos ^2 45^{\circ} \\ & =\frac{I_0}{8} \times\left(\frac{1}{\sqrt{2}}\right)^2=\frac{I_0}{16} \end{aligned}$

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."

According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.

$\sin {i_c} = {1 \over \mu } \propto \lambda $

${i_c} \propto \lambda $

Yellow, orange, red emerge from air.

In Young's double slit experiment, the angular separation of the fringes (θ) is given by :

$\theta = \frac{m\lambda}{d}$

where m is the fringe order, λ 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 (θ) 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 (λ), 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 (λ) increases, the angular separation of the fringes (θ) 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.

In YDSE,

Separation between fringes is

Related to fringed width $\beta = {{D\lambda } \over d}$

If D is increased $\beta$ increases while

Angular separation is independent of distance between slits and screen (D)

As angular separation $ = {\beta \over D} = {\lambda \over d}$

According to Malu's Law

${I_{out}} = {I_{in}}\,{\cos ^2}\theta $

${I_{out}} = I\,{\cos ^2}(30^\circ )$

${I_{out}} = {{3I} \over 4}$

$\beta = {{\lambda D} \over d}$

Let length of segment of screen = l

$ \Rightarrow l = 8{\beta _1} = {{8{\lambda _1}D} \over d}$ ..... (1)

and $ l = n{\beta _2} = {{n{\lambda _2}D} \over d}$ .... (2)

from (1) and (2)

$8{\lambda _1} = n{\lambda _2}$

8(600 nm) = n(400 nm)

n = 12

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,

$\begin{aligned} & \frac{9 \beta}{2}=4 \times 10^{-3} \\ & \frac{9}{2} \times \frac{D \lambda}{d}=4 \times 10^{-3} \\ & \frac{9}{2} \times \frac{2 \times 6 \times 10^{-7}}{d}=4 \times 10^{-3} \\ & \Rightarrow \quad d=1.35 \times 10^{-3} \mathrm{~m}=1.35 \mathrm{~mm} \end{aligned}$

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 $=\frac{d D}{d t}=4 \mathrm{~m} / \mathrm{s}$

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 $t$, we get

$\begin{aligned} \frac{d \beta}{d t} & =\frac{\lambda}{d} \cdot \frac{d D}{d t}=\frac{5 \times 10^{-7}}{4 \times 10^{-4}} \times 4 \\ & =5 \times 10^{-3} \mathrm{~m} / \mathrm{s}=5 \mathrm{~mm} / \mathrm{s} \end{aligned}$

Distance between position of bright and dark fringe 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 YDSE, fringes are of equal width, i.e., width of bright and dark fringes are same. Hence, both assertion and reason are true and reason is the correct explanation of assertion.

When unpolarised light incoming from sun, enters into atmosphere of earth, then it scattered 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 incoming light incident on reflecting surface of earth at an angle of polarisation, then reflecting light is polarised. Little amount of light incident on polarising direction, hence reflected light is partially polarised. Hence, assertion and reason both are true but reason is not correct explanation of assertion.

The intensity of light after passing through an ideal polaroid is given by Malus's Law, which states that the intensity of the plane-polarized light emerging from a polaroid is half the intensity of the unpolarized light incident on it.

The original intensity of the unpolarized light beam is $2a^2$. When this beam passes through the polaroid, the intensity of the emergent plane-polarized light is:

$ I = \frac{1}{2} \times 2a^2 = a^2 $

Thus, the intensity of the emergent plane-polarized light is:

Option B

$a^2$

Here, wavelength ($\lambda$) = 5400 $\mathop A\limits^o=5.4\times10^{-7}$ m, $a=0.80 \mathrm{~mm}=8 \times 10^{-4} \mathrm{~m}, D=1.4 \mathrm{~m}$

$\therefore$ Distance between first two dark bands on each side of central maximum is the width of central maximum

$\begin{aligned} & =2 x=\frac{2 \lambda D}{d} \\ & =\frac{2 \times 5.4 \times 10^{-7} \times 1.4}{8 \times 10^{-4}} \\ & =1.89 \times 10^{-3} \mathrm{~m}=1.89 \mathrm{~mm} \end{aligned}$

To determine the correct answer, let's analyze both the Assertion and the Reason separately.

Assertion:

"If a glass slab is placed in front of one of the slits, then fringe width will decrease."

Analysis:

In a Young's double-slit experiment, the fringe width ($ \beta $) is given by the formula:

$ \beta = \frac{\lambda D}{d} $

Where:

$ \lambda $ is the wavelength of the light used.

$ D $ is the distance between the slits and the screen.

$ d $ is the distance between the two slits.

When a glass slab of thickness $ t $ and refractive index $ \mu $ is placed in front of one of the slits, it introduces an additional optical path difference ($ \Delta $) given by:

$ \Delta = (\mu - 1) t $

Effects of the Glass Slab:

Shift of Fringe Pattern: The additional path difference causes the entire interference pattern to shift laterally but does not alter the fringe width.

Fringe Width Remains Unchanged: The fringe width $ \beta $ depends only on $ \lambda $, $ D $, and $ d $, none of which are affected by the introduction of the glass slab.

Conclusion on Assertion:

The assertion claims that the fringe width will decrease, which is incorrect.

The fringe width remains constant; only the position of the fringes shifts.

Reason:

"Glass slab will produce an additional path difference."

Analysis:

The glass slab increases the optical path length for the light passing through it.

This introduces an additional path difference ($ \Delta $) between the two interfering waves.

The calculation for the additional path difference is:

$ \Delta = (\mu - 1) t $

Conclusion on Reason:

The reason correctly states that the glass slab produces an additional path difference.

This is a true statement.

Final Conclusion:

Assertion: Incorrect

The fringe width does not decrease when a glass slab is placed in front of one slit.

Reason: Correct

The glass slab does produce an additional path difference, causing a shift in the interference pattern.

Relation Between Assertion and Reason:

The reason is correct, but it does not explain the (incorrect) assertion.

Answer: Option D

Assertion is incorrect and Reason is correct.

Fringe width $(W)=\frac{D \lambda}{d}$

Now, distance between the coherent sources is halved i.e., $d^{\prime}=d / 2$ and distance of coherent source and screen is doubled i.e., $D^{\prime}=2 D$

$\begin{aligned} & \therefore W^{\prime}=\frac{D^{\prime} \lambda}{d^{\prime}}=\frac{2 D \lambda}{(d / 2)}=\frac{4 D \lambda}{d} \Rightarrow \frac{W^{\prime}}{W}=4 \\ & \Rightarrow W^{\prime}=4 W \text { (becomes four times) } \end{aligned}$

The specific rotation < of a solution is given

$\alpha=\frac{\theta}{l \times c}$

Where $\theta$ is the optical rotation in degrees, $l$ is the length of the tube in decimeter and $c$ the strength of the solution in $\mathrm{gm} / \mathrm{cc}$,

$\therefore \quad c=\frac{\theta}{l \times \alpha}$

Here $\theta=13^{\circ}, l=20 \mathrm{~cm}=2 \cdot 0$ decimeter and $\alpha=65^{\circ}$

$\therefore c=\frac{13^{\circ}}{2 \times 65^{\circ}}=0.1 \mathrm{~g} / \mathrm{cc}$

$\begin{aligned} & \text { At } S_3, \Delta x=S_1 S_3-S_2 S_3=0 \\ & \therefore \quad \phi=\frac{2 \pi}{\lambda} \square \Delta x=0 \\ & \quad I_3=I_0+I_0+2 \sqrt{I_0 \times I_0} \cos 0^{\circ} \\ & \Rightarrow \quad I_3=4 I_0 \end{aligned}$

The path difference at $S_4$ is

$\begin{aligned} \Delta x^{\prime} & =S_1 S_4-S_2 S_4 \\ & =\frac{d y}{D}=\frac{d}{D} \times \frac{\lambda D}{2 d}=\frac{\lambda}{2} \quad\left(\because y=\frac{\lambda D}{2 d}\right) \\ \therefore \quad \phi^{\prime} & =\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi \\ \Rightarrow I_4 & =I_0+I_0+2 I_0 \cos \pi=0 \Rightarrow \frac{I_3}{I_4}=\frac{4 I_0}{0}=\infty \end{aligned}$

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

$\begin{aligned} x & =n \lambda \frac{D}{d} \propto \lambda \\ \lambda_{\text {blue }} & =4360 \mathop A\limits^o \\ \lambda_{\text {green }} & =5480 \mathop A\limits^o \end{aligned}$

$x=$ distance of 4th maxima from the central are

$\begin{array}{ll} \text { As } & \lambda_{\text {blue }}< \lambda_{\text {green }} \\ \therefore & x_{\text {blue }}< x_{\text {green }} \end{array}$

According to Newton's corpuscular theory of light, the light should travel faster in denser media than in rarer media, it is contrary to present theory of light which explains that light travels faster in air than in water.