Wave Optics

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.