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