Wave Optics

When a plane wave-front (parallel rays) enters a prism, all parts of the wave-front travel at the same speed through the medium. While the wave-front is tilted due to refraction, it remains flat and parallel, thus remaining a plane wave.

According to Huygens' Principle, every point on a wave-front acts as a source of secondary spherical wavelets. When a plane wave encounters a very small opening (pinhole) comparable to the wavelength of light, it undergoes diffraction. The emerging light spreads out in all directions, forming a spherical wave-front.

Hence, the statement I is true.

When light passes through a slit, the degree of spreading (diffraction) depends on the slit width (d) relative to the wavelength ( $\lambda$ ).

Smaller slit widths lead to greater diffraction and more pronounced curvature. As the slit width increases, the light behaves more like a straight beam, meaning the wave-front becomes flatter.

Curvature is the reciprocal of the radius of the wave-front, i.e., $\propto \frac{1}{\mathrm{r}}$. So, a flatter wave has less curvature. Therefore, increasing the slit width decreases the curvature, it does not increase it.

Hence, the statement II is false.

Since Statement I is true and Statement II is false, so the correct option is (C).

The intensity from each slit is equal, $\mathrm{I}_1=\mathrm{I}_2=\mathrm{I}_0$

The slit width is equal to distance between slits $d=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$

Distance of screen from slits $\mathrm{D}=10 \mathrm{~m}$

Wavelength of light $\lambda=6000 \mathop A\limits^o =6 \times 10^{-7} \mathrm{~m}$

The distance from the central maximum to a point directly in front of one slit is :

$ \mathrm{y}=\frac{\mathrm{d}}{2} $

The path difference for a point at height $y$ on the screen is given by the formula :

$ \Delta \mathrm{x}=\frac{\mathrm{dy}}{\mathrm{D}} $

$\Rightarrow $ $\Delta \mathrm{x}=\frac{\mathrm{d}\left(\frac{\mathrm{d}}{2}\right)}{\mathrm{D}}=\frac{\mathrm{d}^2}{2 \mathrm{D}}$

$ \Delta \mathrm{x}=\frac{\left(2 \times 10^{-3}\right)^2}{2 \times 10}=\frac{4 \times 10^{-6}}{20}=2 \times 10^{-7} \mathrm{~m} $

The relation between path difference and phase difference is :

$\phi=\frac{2 \pi}{\lambda} \times \Delta x$

$\Rightarrow $ $ \phi=\frac{2 \pi}{6 \times 10^{-7}} \times\left(2 \times 10^{-7}\right)=\frac{4 \pi}{6}=\frac{2 \pi}{3} $

The formula for resultant intensity when two waves of equal intensity $\mathrm{I}_0$ interfere is :

$ \mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right) $

$\Rightarrow $ $I=4 I_0 \cos ^2\left(\frac{\frac{2 \pi}{3}}{2}\right)=4 I_0 \cos ^2\left(\frac{\pi}{3}\right)$

$\Rightarrow $ $\mathrm{I}=4 \mathrm{I}_0\left(\frac{1}{2}\right)^2=4 \mathrm{I}_0 \times \frac{1}{4}$

$\Rightarrow $ $ \mathrm{I}=\mathrm{I}_0 $

The intensity of light on the screen directly in front of one of the slits is equal to the intensity of a single slit, i.e., $\mathrm{I}_0$.

Hence, the correct option is (B).

When unpolarized light strikes an interface at a specific angle called the Brewster angle ( $\mathrm{i}_{\mathrm{B}}$ or $\mathrm{i}_{\mathrm{p}}$ ), the reflected light is completely linearly polarized. At this specific angle, the reflected ray and the refracted ray are perpendicular $\left(90^{\circ}\right)$ to each other.

• 1. $\mathrm{i}_{\mathrm{p}}=$ Angle of incidence (polarizing angle)

• 2. $\theta_{\mathrm{r}}=$ Angle of reflection

• 3. $\theta_{\mathrm{r}^{\prime}}=$ Angle of refraction

Since the angle of reflection equals the angle of incidence :

$ \theta_{\mathrm{r}}=\mathrm{i}_{\mathrm{p}} $

Since the angle between the reflected and refracted ray is $90^{\circ}$ :

$ \mathrm{i}_{\mathrm{p}}+90^{\circ}+\theta_{\mathrm{r}^{\prime}}=180^{\circ} $

$\Rightarrow $ $\theta_{\mathrm{r}^{\prime}}=90^{\circ}-\mathrm{i}_{\mathrm{p}}$

Applying Snell's Law

$ \mu_1 \sin \left(\mathrm{i}_{\mathrm{p}}\right)=\mu_2 \sin \left(\theta_{\mathrm{r}^{\prime}}\right) $

$\Rightarrow $ $\mu_1 \sin \left(\mathrm{i}_{\mathrm{p}}\right)=\mu_2 \sin \left(90^{\circ}-\mathrm{i}_{\mathrm{p}}\right)$

$\Rightarrow $ $\mu_1 \sin \left(\mathrm{i}_{\mathrm{p}}\right)=\mu_2 \cos \left(\mathrm{i}_{\mathrm{p}}\right)$

$\Rightarrow $ $\tan \left(\mathrm{i}_{\mathrm{p}}\right)=\frac{\mu_2}{\mu_1}$

If the first medium is air ( $\mu_1 \approx 1$ ), the formula simplifies to :

$ \tan \left(\mathrm{i}_{\mathrm{p}}\right)=\mu $

Substituting the values:

$ \tan \left(\mathrm{i}_{\mathrm{p}}\right)=\frac{1.52}{1.00}=1.52 $

$\Rightarrow $ $\mathrm{i}_{\mathrm{p}}=\tan ^{-1}(1.52)=57.7^{\circ}$

As established by the geometry of Brewster's condition :

$ \mathrm{i}_{\mathrm{p}}+\theta_{\mathrm{r}^{\prime}}=90^{\circ} $

$\Rightarrow $ $\theta_{\mathrm{r}^{\prime}}=90^{\circ}-\mathrm{i}_{\mathrm{p}}$

$\Rightarrow $ $\theta_{\mathrm{r}^{\prime}}=90^{\circ}-57.7^{\circ}$

$ \theta_{\mathrm{r}^{\prime}}=32.3^{\circ} $

$\Rightarrow $ The angle of the refracted beam with respect to the normal is $32.3^{\circ}$. Hence, the correct option is (D).

When light travels from one medium to another, its frequency remains constant, but its velocity and wavelength change.

The absolute refractive index is defined as:

$ \mu=\frac{c}{v}=\frac{\lambda_{\text {vacuum }}}{\lambda_{\text {medium }}} $

Therefore, for any two media, the product of the refractive index and the wavelength is constant:

$ \mu_1 \lambda_1=\mu_2 \lambda_2 $

Wavelength in water, $\lambda_1=540 \mathrm{~nm}$

Refractive index of water, $\mu_1=4 / 3$

Refractive index of the second medium, $\mu_2=3 / 2$

Using the relation $\mu_1 \lambda_1=\mu_2 \lambda_2$,

$ \lambda_2=\frac{\mu_1 \lambda_1}{\mu_2} $

$\Rightarrow $ $ \lambda_2=\frac{\frac{4}{3} \times 540}{3 / 2}=\frac{4 \times 540 \times 2}{3 \times 3}=480 \mathrm{~nm} $

Therefore, the wavelength of the light in the second transparent medium is 480 nm .

Hence, the correct option is (C).

In a single slit diffraction experiment the linear width $(W)$ of the central maximum on a screen at distance $D$ is given by the formula:

$ \mathrm{W}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} $

Where :

• 1. $\lambda$ is the wavelength of the light.

• 2. a is the width of the slit.

From the formula, the width of the central maximum is directly proportional to the wavelength ( $\mathrm{W} \propto \lambda$ ) when the slit width (a) is kept constant.

This means if the wavelength increases, the width of the central maximum also increases. Therefore, Statement A is true and Statement B is false.

From the formula, the width of the central maximum is inversely proportional to the slit width ( $\mathrm{W} \propto \frac{1}{\mathrm{a}}$ ) when the wavelength ( $\lambda$ ) is kept constant.

This means if the slit width decreases, the width of the central maximum increases. Therefore, Statement C is true and Statement D is false.

The peak intensity (brightness) of the central maximum depends on how the superposition of waves on screen occurs. For a given incident intensity, the power passing through the slit is constant for a fixed slit width. When the wavelength decreases, the central maximum becomes narrower (the angular width decreases). Because the exact same amount of energy is now concentrated into a smaller area, the peak brightness of the central maximum increases. Therefore, Statement E is true.

Hence, the true statements are $\mathbf{A}, \mathbf{C}$, and $\mathbf{E}$.

In YDSE, the linear position of the $\mathrm{n}^{\text {th }}$ bright fringe on the screen is given by :

$ \mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{~d}} $

Where :

• $\lambda=$ wavelength of light used.

• $\mathrm{D}=$ distance between the slits and the screen.

• $\mathrm{d}=$ distance between the two slits.

The linear fringe width $(\beta)$, which is the distance between two consecutive bright or dark fringes, is :

$ \beta=y_{n+1}-y_n=\frac{\lambda D}{d} $

The angular separation (or angular fringe width) $\theta_\beta$ is the angle subtended by the fringe width at the slits. For small angles :

$ \theta_\beta \approx \frac{\beta}{D} $

Substituting the value of $\beta$ :

$ \theta_\beta=\frac{\left(\frac{\lambda \mathrm{D}}{\mathrm{~d}}\right)}{\mathrm{D}}=\frac{\lambda}{\mathrm{d}} $

Statement I says that angular separation $\left(\theta_\beta\right)$ increases as the screen is moved away (D increases).

From the formula $\theta_\beta=\lambda / d$, we can see that $\theta_\beta$ depends only on the wavelength ( $\lambda$ ) and the slit separation (d).

$ \theta_\beta \propto \lambda \ldots \text { (i), } \quad \theta_\beta \propto \frac{1}{d} \ldots \text { (ii) } $

It is independent of the distance to the screen (D).

Therefore, Statement I is false.

Statement II says that $\theta$ increases when a source of higher wavelength is used.

From the formula $\theta_\beta \propto \lambda$, i.e., if $\lambda$ increases, $\theta$ must also increase.

Therefore, Statement II is true.

Therefore, Statement I is false and Statement II is true.

Hence, the correct option is (C).

In a Young's Double Slit Experiment (YDSE), the position of the fringes is determined by the path difference between the two waves. When a transparent sheet of thickness t and refractive index n is placed in front of one slit, the light travels through the medium instead of air for that distance.

The optical path through the sheet is nt , while the path in air for the same distance would be t . Thus, an additional path difference $(\Delta \mathrm{x})$ is introduced :

$ \Delta \mathrm{x}_{\text {optical }}=\mathrm{nt}-\mathrm{t}=(\mathrm{n}-1) \mathrm{t} $

The central fringe which originally occurs at the centre where the geometric path difference is zero, shifts to a new position $\Delta y$ where the net path difference is zero.

The geometric path difference for a point at distance $y$ from the center is :

$ \Delta \mathrm{x}_{\mathrm{geo}}=\frac{\mathrm{d} \cdot \mathrm{y}}{\mathrm{D}} $

Where d is the distance between slits and D is the distance to the screen.

So, at the position of central maximum net path difference must be zero, $\Delta \mathrm{x}_{\text {optical }}+\Delta \mathrm{x}_{\text {geo }}=0$

$ \frac{\mathrm{d} \cdot \mathrm{y}}{\mathrm{D}}=(\mathrm{n}-1) \mathrm{t} $

$ \Rightarrow \mathrm{t}=\frac{\mathrm{y} \cdot \mathrm{~d}}{\mathrm{D}(\mathrm{n}-1)} $

Putting the values,

• Distance between slits, $\mathrm{d}=0.1 \mathrm{~cm}$

• Distance to screen $\mathrm{D}=50 \mathrm{~cm}$

• Shift of central fringe $y=0.2 \mathrm{~cm}$

• Refractive index $\mathrm{n}=1.5$

$ \mathrm{t}=\frac{0.2 \times 0.1}{50(1.5-1)} $

$\Rightarrow $ $t=\frac{0.02}{25}$

$\Rightarrow $ $ \mathrm{t}=8 \times 10^{-4} \mathrm{~cm} $

Therefore, the thickness of the sheet to be $8 \times 10^{-4} \mathrm{~cm}$.

Hence, the correct option is (B).

When two waves of intensities $I_1$ and $I_2$ interfere with a phase difference $\phi$, the resultant intensity $I_R$ is given by:

$ \mathrm{I}_{\mathrm{R}}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi $

Since the slits are identical, the intensities due to individual slits are equal ( $\mathrm{I}_1=\mathrm{I}_2=\mathrm{I}_0$ ). The formula simplifies to:

$ \mathrm{I}_{\mathrm{R}}=\mathrm{I}_0+\mathrm{I}_0+2 \sqrt{\mathrm{I}_0^2} \cos \phi $

$\Rightarrow $ $\mathrm{I}_{\mathrm{R}}=2 \mathrm{I}_0(1+\cos \phi)$

Using the trigonometric identity $1+\cos \phi=2 \cos ^2\left(\frac{\phi}{2}\right)$

$ \mathrm{I}_{\mathrm{R}}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right) $

It is given that the resultant intensity at point P is equal to the intensity of an individual slit:

$ \mathrm{I}_{\mathrm{R}}=\mathrm{I}_0 $

Substituting this into equation:

$ \mathrm{I}_0=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right) $

$\Rightarrow $ $\cos ^2\left(\frac{\phi}{2}\right)=\frac{1}{4}$

$\Rightarrow $ $\cos \left(\frac{\phi}{2}\right)=\frac{1}{2}$

$\Rightarrow $ $ \frac{\phi}{2}=\frac{\pi}{3} \Rightarrow \phi=\frac{2 \pi}{3} $

The phase difference is related to the path difference by the formula:

$ \phi=\frac{2 \pi}{\lambda}(\Delta x) $

$\Rightarrow $ $ \Delta \mathrm{x}=\frac{\phi \cdot \lambda}{2 \pi} $

The wavelength of light is given as $\lambda=6000 Å$ and $\phi=\frac{2 \pi}{3}$ :

$ \Delta \mathrm{x}=\frac{\left(\frac{2 \pi}{3}\right) \cdot 6000 Å}{2 \pi} $

$\Rightarrow $ $\Delta x=\frac{6000}{3} Å$

$\Rightarrow $ $ \Delta \mathrm{x}=2000 Å $

Therefore, the path difference between interfering waves is $2000 Å$.

Thus, the correct option is (C).

The relation between phase difference $(\phi)$ and path difference $(\Delta x)$ is:

$ \phi=\frac{2 \pi}{\lambda} \Delta x $

For two waves of equal intensity $\mathrm{I}_0$, the resultant intensity I at a point is:

$ \mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right) $

At point A the path difference is $\Delta \mathrm{x}_{\mathrm{A}}=\frac{\lambda}{3}$.

$ \phi_{\mathrm{A}}=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3} $

So, the resultant intensity at A is,

$ I_A=4 I_0 \cos ^2\left(\frac{2 \pi / 3}{2}\right)=4 I_0 \cos ^2\left(\frac{\pi}{3}\right)=4 I_0\left(\frac{1}{2}\right)^2=I_0 $

At point B the path difference is $\Delta \mathrm{x}_{\mathrm{B}}=\frac{\lambda}{6}$.

$ \phi_{\mathrm{B}}=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3} $

So, the resultant intensity at B is,

$ I_B=4 I_0 \cos ^2\left(\frac{\pi / 3}{2}\right)=4 I_0 \cos ^2\left(\frac{\pi}{6}\right)=4 I_0\left(\frac{\sqrt{3}}{2}\right)^2=3 I_0 $

So, the ratio of intensities at A and B is;

$ \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=\frac{\mathrm{I}_0}{3 \mathrm{I}_0}=\frac{1}{3} $

Thus, the correct option is (C).

In Young’s double slit experiment, intensity at any point on the screen is

$ I = I_0 \cos^2 \left(\frac{\phi}{2}\right) $

where $I_0$ is the maximum intensity and $\phi$ is the phase difference.

Now we need the intensity at the point on the screen exactly opposite to one of the slits.

Let the slits be $S_1$ and $S_2$, separated by distance $d$.

At the point directly opposite to slit $S_1$:

• distance from $S_1$ to that point $= D$

• distance from $S_2$ to that point $= \sqrt{D^2 + d^2}$

So, path difference is

$ \Delta = \sqrt{D^2+d^2}-D $

Given:

$ D = 10d $

Therefore,

$ \Delta = \sqrt{(10d)^2+d^2}-10d = \sqrt{101d^2}-10d = d(\sqrt{101}-10) $

Since $d = 5\lambda$,

$ \Delta = 5\lambda(\sqrt{101}-10) $

Now,

$ \sqrt{101} \approx 10.05 $

So,

$ \Delta \approx 5\lambda(10.05-10)=5\lambda(0.05)=0.25\lambda=\frac{\lambda}{4} $

Hence phase difference is

$ \phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{4} = \frac{\pi}{2} $

Now intensity becomes

$ I = I_0 \cos^2\left(\frac{\phi}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} $

So the required intensity is

$ \boxed{\frac{I_0}{2}} $

Therefore, the correct option is B.

In Young’s double slit experiment, fringe width is given by

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

where

$\lambda$ = wavelength in the medium,

$D$ = distance between slit and screen,

$d$ = distance between the slits.

When the experiment is performed in a medium of refractive index $\mu$, the wavelength becomes

$ \lambda' = \frac{\lambda}{\mu} $

So the new fringe width is

$ \beta' = \frac{\lambda' D}{d} = \frac{1}{\mu}\frac{\lambda D}{d} $

Thus,

$ \beta' = \frac{\beta}{\mu} $

Given:

$ \beta = 2.4\ \mu m,\qquad \mu = 1.2 $

Therefore,

$ \beta' = \frac{2.4}{1.2} = 2\ \mu m $

So, the correct answer is:

$ \boxed{2\ \mu m} $

Option B

Let the intensity of the unpolarized incident light be $I_0$.

We use Malus’ law:

If plane polarized light passes through a polarizer making angle $\theta$ with the direction of polarization, then transmitted intensity is

$ I = I_{\text{in}} \cos^2 \theta $

Also, when unpolarized light passes through a polarizer, its intensity becomes

$ \frac{I_0}{2} $

1) Without the third polarizer

The two polarizers have transmission axes at $30^\circ$ and $90^\circ$.

So, the angle between their axes is

$ 90^\circ - 30^\circ = 60^\circ $

After passing through the first polarizer, intensity becomes

$ I_1 = \frac{I_0}{2} $

Now this light passes through the second polarizer, so

$ I_{\text{without}} = I_1 \cos^2 60^\circ $

Since

$ \cos 60^\circ = \frac{1}{2} $

therefore,

$ I_{\text{without}} = \frac{I_0}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{4} = \frac{I_0}{8} $

2) With the third polarizer inserted between them

Now the polarizers are at $30^\circ$, $60^\circ$, and $90^\circ$.

After first polarizer:

$ I_1 = \frac{I_0}{2} $

From first to third polarizer:

Angle difference is

$ 60^\circ - 30^\circ = 30^\circ $

So intensity after the third polarizer is

$ I_2 = I_1 \cos^2 30^\circ $

Since

$ \cos 30^\circ = \frac{\sqrt{3}}{2} $

we get

$ I_2 = \frac{I_0}{2} \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3I_0}{8} $

From third to second polarizer:

Angle difference is again

$ 90^\circ - 60^\circ = 30^\circ $

So final intensity is

$ I_{\text{with}} = I_2 \cos^2 30^\circ $

$ I_{\text{with}} = \frac{3I_0}{8} \cdot \frac{3}{4} = \frac{9I_0}{32} $

3) Required ratio

$ \frac{I_{\text{with}}}{I_{\text{without}}} = \frac{\frac{9I_0}{32}}{\frac{I_0}{8}} = \frac{9}{32} \cdot 8 = \frac{9}{4} $

So, the required ratio is

$ \boxed{\frac{9}{4}} $

Hence, the correct option is Option C.

(P) Colorful sky in north polar region (Aurora Borealis)

The Aurora Borealis occurs when high-speed charged particles (electrons and protons) from the solar wind collide with atoms and molecules in Earth's upper atmosphere. These collisions transfer energy to the atmospheric gases, exciting the electrons within oxygen and nitrogen atoms to higher energy levels. When these excited electrons return to their ground states, they emit photons of specific characteristic wavelengths (such as green and red from oxygen, or blue and violet from nitrogen).

So, this phenomenon is emission of radiation from oxygen and nitrogen atoms excited by charged particles.

$ \text { P → } 5 $

(Q) Partially polarized sun light

Direct sunlight entering the Earth's atmosphere is initially unpolarized. However, when this light strikes air molecules and small particles suspended in the atmosphere, it undergoes Rayleigh scattering. The scattered light viewed at an angle perpendicular ( $90^{\circ}$ ) to the incident beam becomes completely plane-polarized, while light scattered at other intermediate angles is partially polarized.

So, it is due to scattering of light by molecules in the atmosphere.

$ Q \rightarrow 4 $

(C) Rainbow

A rainbow is formed when sunlight enters a spherical raindrop suspended in the air. The process involves:

Refraction and Dispersion: Sunlight bends and splits into its constituent colors as it enters the drop due to different wavelengths having slightly different refractive indices.

Reflection: The separated colors travel through the droplet and reflect off the back inside surface of the water drop.

Refraction: The light exits the droplet, bending once more before reaching the observer's eye.

So, during the process of rainbow formation the light undergoes dispersion and reflection.

$ \mathrm{R} \rightarrow 1 $

(D) Dark and bright fringes

The regular alternation of dark and bright bands (fringes) is the pattern produced by the phenomena of wave interference and diffraction. When light encounters a small aperture, obstacle, or slit system, it bends around edges (diffraction) and superimposes, generating constructive (bright) and destructive (dark) regions.

Therefore, the dark and bright fringes are formed by diffraction.

$ S \rightarrow 3 $

$\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 1, \mathrm{~S} \rightarrow 3$.

Thus, the correct option is (A).

When white light is used in Young's double slit experiment and one slit is covered by a red filter and the other by a green filter, several effects occur:

Different Fringe Widths: Each color will produce fringes of different widths due to their varying wavelengths. This means that red and green will not overlap perfectly.

Color Overlap: As the fringe patterns develop, some regions will have overlap due to the differing fringe spacings of red and green light. This overlap will create areas where the colors mix, potentially leading to an appearance of intermediate colors.

Overall, because each color produces its own pattern and these do not align, a clear interference pattern as seen with monochromatic light (a single color) will not be formed. Instead, there will be a complex pattern where the fringes do not distinctly appear as solid bands of red or green due to the overlapping and different fringe widths.

To find the amplitude of the resultant wave for the given electric field components, we use the formula for the resultant amplitude of two combining waves. These components are:

$ E_1 = E_0 \sin \omega t $

$ E_2 = E_0 \sin \left(\omega t + \frac{\pi}{3}\right) $

The general formula for the amplitude $ E $ of two superimposed waves $ E_1 $ and $ E_2 $ with a phase difference is:

$ E = \sqrt{E_1^2 + E_2^2 + 2E_1E_2 \cos\phi} $

where $ \phi $ is the phase difference between the two waves. In this scenario, $ \phi = \frac{\pi}{3} $.

Substituting the given expressions and phase difference into the formula, we get:

$ E = \sqrt{(E_0)^2 + (E_0)^2 + 2(E_0)(E_0)\cos\left(\frac{\pi}{3}\right)} $

Given that $\cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$, this simplifies to:

$ E = \sqrt{2E_0^2 + E_0^2} = \sqrt{3}E_0 $

Thus, the amplitude of the resultant wave is $\sqrt{3}E_0 \approx 1.73E_0$.

Through $P_2 I_1=I_0 \sin ^2\left(\frac{\pi}{2}-\theta\right)$

$I_1=I_0 \cos ^2 \theta$

Through $P_3 I_{\text {net }}=\left(I_0 \cos ^2 \theta\right) \sin ^2 \theta$

$I_{\text {net }}=\frac{I_0}{4}[\sin (2 \theta)]^2 \text { for max } I_{\text {net }} \theta=45^{\circ}$

So angle between $\mathrm{P}_2$ and $\mathrm{P}_3=\frac{\pi}{4}$

Correct Ans. (1)

In Young's double slit experiment, the fringe width $\beta$ is given by:

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

where:

$\lambda$ is the wavelength of light,

$D$ is the distance from the slits to the screen,

$d$ is the separation between the slits.

Here's how the change affects the fringe width:

Initial Situation:

When $d = 0.2 \text{ mm}$, the fringe width is:

$\beta_{\text{initial}} = \frac{\lambda D}{0.2 \text{ mm}}$

After Increasing Slit Separation:

When $d$ is increased to $0.4 \text{ mm}$:

$\beta_{\text{new}} = \frac{\lambda D}{0.4 \text{ mm}}$

Comparing the Two Fringe Widths:

Notice that:

$\beta_{\text{new}} = \frac{1}{2} \times \frac{\lambda D}{0.2 \text{ mm}} = \frac{1}{2} \beta_{\text{initial}}$

This means the fringe width is halved, which is a reduction of $50\%$.

Thus, the fringe width decreases by $50\%$ when the slit separation is increased from $0.2 \text{ mm}$ to $0.4 \text{ mm}$.

The correct answer is Option B: $50\%$.

To find the ratio of the intensities of the maximum to the minimum in an interference pattern formed by two monochromatic light beams with intensity ratios of 1:9, you use the formula:

$ \frac{I_{\max }}{I_{\min }} = \frac{\left(\sqrt{I_1} + \sqrt{I_2}\right)^2}{\left(\sqrt{I_1} - \sqrt{I_2}\right)^2} $

Given that the intensity ratio is 1:9, we have $I_1 = 1$ and $I_2 = 9$.

Calculate $\sqrt{I_1}$ and $\sqrt{I_2}$:

$ \sqrt{I_1} = \sqrt{1} = 1 $

$ \sqrt{I_2} = \sqrt{9} = 3 $

Substitute these values into the formula:

$ \frac{I_{\max }}{I_{\min }} = \frac{(1 + 3)^2}{(1 - 3)^2} = \frac{4^2}{2^2} = \frac{16}{4} = 4 $

Thus, the ratio of the intensities of the maximum to the minimum is 4:1.

In Young's double-slit interference experiment, the width of one slit is half that of the other. The intensity, $ \mathrm{I} $, is proportional to the slit width. This gives us:

$ \begin{align*} \mathrm{I}_1 &= \mathrm{I}_0, \\ \mathrm{I}_2 &= 2 \mathrm{I}_0. \end{align*} $

The maximum intensity, $ \mathrm{I}_{\max} $, occurs when the amplitudes add constructively:

$ \mathrm{I}_{\max} = \left(\sqrt{\mathrm{I}_1} + \sqrt{\mathrm{I}_2}\right)^2. $

The minimum intensity, $ \mathrm{I}_{\min} $, occurs when the amplitudes interfere destructively:

$ \mathrm{I}_{\min} = \left(\sqrt{\mathrm{I}_1} - \sqrt{\mathrm{I}_2}\right)^2. $

Now, using $ \mathrm{I}_1 = \mathrm{I}_0 $ and $ \mathrm{I}_2 = 2 \mathrm{I}_0 $, we find:

$ \frac{\mathrm{I}_{\max}}{\mathrm{I}_{\min}} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}. $

To find the wavelength of the refracted light in a medium with a refractive index of 2, we can use the relationship between frequency, wavelength, and the speed of light. The frequency ($f$) of the light remains the same when it enters the medium.

The formula relating frequency, wavelength, and velocity is:

$ v = f \lambda $

Where:

$ v $ is the speed of light in the medium,

$ f $ is the frequency of the light,

$ \lambda $ is the wavelength of the light.

The wavelength of light in the medium ($\lambda_{\text{medium}}$) can be found using the refractive index ($\mu$) and the wavelength in a vacuum ($\lambda_{\text{vacuum}}$) as follows:

$ \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{\mu} $

Given:

Frequency of the light, $ f = 5 \times 10^{14} \, \text{Hz} $

Speed of light in a vacuum, $ v_{\text{vacuum}} = 3 \times 10^{8} \, \text{m/s} $

Refractive index of the medium, $ \mu = 2 $

First, calculate $\lambda_{\text{vacuum}}$:

$ \lambda_{\text{vacuum}} = \frac{v_{\text{vacuum}}}{f} = \frac{3 \times 10^{8}}{5 \times 10^{14}} $

Now calculate $\lambda_{\text{medium}}$:

$ \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{2} = \frac{3 \times 10^{8}}{2 \times 5 \times 10^{14}} $

$ \lambda_{\text{medium}} = 0.3 \times 10^{-6} \, \text{m} $

Converting to nanometers:

$ 0.3 \times 10^{-6} \, \text{m} = 300 \, \text{nm} $

Therefore, the wavelength of the refracted light in the medium is 300 nm.

The light wave is propagating with plane wave fronts described by the equation $x + y + z = \text{constant}$. The wave propagates in a direction perpendicular to these wave fronts. This direction is symmetric with respect to the $x$, $y$, and $z$ axes.

Since the wave is symmetric about these axes, the angle it makes with each axis is the same. Let these angles be $\alpha$, $\beta$, and $\gamma$, which are the angles made by the light with the $x$, $y$, and $z$ axes, respectively.

This implies:

$ \cos \alpha = \cos \beta = \cos \gamma $

According to the sum of the squares of the direction cosines:

$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $

Substituting the equality of direction cosines, we get:

$ 3 \cos^2 \alpha = 1 $

$ \cos^2 \alpha = \frac{1}{3} $

Thus, the angle $\alpha$ is given by:

$ \alpha = \cos^{-1} \left(\frac{1}{\sqrt{3}}\right) $

Let the wave be incident from the denser medium onto the rarer one in each case. Since $n_3>n_2>n_1$ the critical‐angle sines are

– at the $n_2\kern1pt|\kern1ptn_1$ interface:

$ \sin\theta_{1C}=\frac{n_1}{n_2}, $

– at the $n_3\kern1pt|\kern1ptn_1$ interface:

$ \sin\theta_{2C}=\frac{n_1}{n_3}. $

We’re given

$ \sin\theta_{1C}-\sin\theta_{2C}=\frac12 \quad\text{and}\quad \frac{n_2}{n_3}=\frac25. $

Hence

$ \frac{n_1}{n_2}-\frac{n_1}{n_3} =\;n_1\Bigl(\frac1{n_2}-\frac1{n_3}\Bigr) =\frac12, $

and with $n_2=\tfrac25\,n_3$ this becomes

$ n_1\Bigl(\frac1{n_2}-\frac1{n_3}\Bigr) =n_1\Bigl(\frac{5}{2n_3}-\frac1{n_3}\Bigr) =n_1\frac{3}{2n_3} =\frac12 \;\Longrightarrow\; n_3=3\,n_1,\quad n_2=\frac{2}{5}n_3=\frac{6}{5}n_1. $

Therefore

$ \sin\theta_{1C} =\frac{n_1}{n_2} =\frac{n_1}{\tfrac{6}{5}n_1} =\frac56, $

so

$ \theta_{1C}=\sin^{-1}\!\Bigl(\frac56\Bigr). $

That corresponds to Option C.

after passing through third poleriser, Intensity of both the waves must be $\frac{I_0}{4}$

now, at a point where path diff is $\frac{\lambda}{3}$, phase difference

$\begin{aligned} & \Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}\right)=\frac{2 \pi}{3} \\ & \therefore \mathrm{I}_{\text {res }}=\sqrt{\left(\frac{\mathrm{I}_0}{4}\right)^2+\left(\frac{\mathrm{I}_0}{4}\right)^2+2\left(\frac{\mathrm{I}_0}{4}\right)^2} \cos \frac{2 \pi}{3} \\ & =\frac{\mathrm{I}_0}{4} \end{aligned}$

In Young's double-slit experiment, the fringe width ($\beta$) can be calculated using the formula:

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

where:

$\lambda'$ is the wavelength of light in the medium,

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

$d$ is the separation between the slits.

The wavelength in the medium ($\lambda'$) is given by:

$ \lambda' = \frac{\lambda_0}{n} $

where:

$\lambda_0 = 690 \, \text{nm}$ is the wavelength of light in air,

$n = 1.44$ is the refractive index of the liquid.

Substituting the values,

$ \lambda' = \frac{690 \, \text{nm}}{1.44} \approx 479.17 \, \text{nm} $

Now, convert $\lambda'$ to meters:

$ \lambda' = 479.17 \, \text{nm} = 479.17 \times 10^{-9} \, \text{m} $

Next, use the values for $D$ and $d$:

$D = 0.72 \, \text{m}$

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

Calculating the fringe width:

$ \beta = \frac{479.17 \times 10^{-9} \, \text{m} \cdot 0.72 \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} $

$ \beta \approx \frac{344.2 \times 10^{-9} \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} $

$ \beta \approx 0.229 \times 10^{-3} \, \text{m} = 0.229 \, \text{mm} $

The closest answer choice is:

Option D: 0.23 mm

$ d \sin \theta = m \lambda $

For two wavelengths, $\lambda_1 = 600\,\text{nm}$ and $\lambda_2 = 480\,\text{nm}$, the bright fringes coincide when the conditions

$ m \lambda_2 = n \lambda_1 $

are simultaneously satisfied for integers $m$ and $n$, representing the bright fringe order for each wavelength.

Rearranging the equation gives:

$ \frac{m}{n} = \frac{\lambda_1}{\lambda_2} = \frac{600}{480} = \frac{5}{4} $

Thus, the smallest positive integers that satisfy this ratio are:

$ m = 5 \quad \text{and} \quad n = 4. $

This means the first coincidence of the bright fringes occurs at the 5th bright fringe for the 480 nm light.

Let the amplitude from the slit of width $d$ be proportional to $E$ and from the slit of width $xd$ be proportional to $xE$ (with $x>1$, as is evident from the given options).

For two coherent waves with amplitudes $E_1$ and $E_2$, the resultant intensity when added in phase (maximum) and out of phase (minimum) is given by

$ I_{\max} \propto (E_1+E_2)^2 \quad \text{and} \quad I_{\min} \propto (E_2-E_1)^2. $

Here, setting

$E_1 = E$ (from the narrower slit of width $d$) and

$E_2 = xE$ (from the wider slit of width $xd$),

we have

$ I_{\max} \propto (E+xE)^2 = E^2 (1+x)^2, $

and since $x > 1$ the subtraction in the destructive case gives

$ I_{\min} \propto (xE-E)^2 = E^2 (x-1)^2. $

The ratio of maximum to minimum intensity is therefore

$ \frac{I_{\max}}{I_{\min}} = \frac{(1+x)^2}{(x-1)^2}. $

We are given that

$ \frac{(1+x)^2}{(x-1)^2} = \frac{9}{4}. $

Taking the square root of both sides (noting that all quantities are positive) leads to

$ \frac{1+x}{x-1} = \frac{3}{2}. $

To solve for $x$, cross-multiply:

$ 2(1+x) = 3(x-1). $

Expanding both sides:

$ 2 + 2x = 3x - 3. $

Rearrange the equation to isolate $x$:

$ 2 + 2x + 3 = 3x \quad \Longrightarrow \quad 5 + 2x = 3x, $

which implies

$ x = 5. $

Thus, the value of $x$ is

$ \boxed{5}. $

For transmitted green light to be maxima, reflected green should be minima.

$\begin{aligned} & \Delta \mathrm{P}=2 \mu_0 \mathrm{t}=\mathrm{n} \lambda \\ & \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu_0} \therefore \mathrm{t}_{\min }=\frac{\lambda}{2 \mu_0}=\frac{550}{2 \times 2}=137.5 \end{aligned}$

In Young's double-slit experiment, the width of the fringes, denoted as $\beta$, can be calculated using the formula:

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

In this formula:

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

$D$ is the distance from the slits to the screen.

$d$ is the distance between the slits.

According to this relationship, the fringe width $\beta$ is directly proportional to the wavelength $\lambda$. This means that as the wavelength increases, the fringe width also increases.

When comparing red and blue light:

Red light has a longer wavelength ($\lambda_R$) compared to blue light ($\lambda_b$).

Therefore, the fringe width $\beta_R$ for red light is greater than the fringe width $\beta_b$ for blue light, as $\lambda_R > \lambda_b$.

This shows that with red light, the fringes are actually farther apart compared to the fringes produced by blue light. Thus, the assertion that red light fringes are closer than those produced by blue light is false, while the reason about fringe width being proportional to the wavelength is true.

We know, in YDSE,

$\beta$ (fringe width) $ = {{\lambda D} \over d}$

In denser medium, $\lambda \downarrow \Rightarrow \beta \downarrow $

$\Rightarrow$ fringes come closer.

Also, $\mu = {c \over v} \Rightarrow v = {c \over \mu }$

$ \Rightarrow \mu \uparrow \Rightarrow v \downarrow \Rightarrow $ speed of light reduces.

frequency remains same,

$ \Rightarrow \mu = {{{\lambda _{vac.}}f} \over {{\lambda _{med.}}f}} \Rightarrow {\lambda _{med}} = {{{\lambda _{vac.}}} \over \mu }$ (as $v = \lambda f$)

Hence, option C is correct.

Here, $a=0.3 \mathrm{~cm}$

$ \begin{aligned} & =3 \times 10^{-3} \mathrm{~m} \\ \lambda & =6000 \mathop {\rm{A}}\limits^{\rm{o}}=6 \times 10^{-7} \mathrm{~m} \end{aligned} $

∵ Fresnel's distance

$ Z_F=\frac{a^2}{\lambda}=\frac{\left(3 \times 10^{-3}\right)^2}{6 \times 10^{-7}}=15 \mathrm{~m} $

• A) Steam engine: A steam engine operates by converting thermal energy (from steam) into mechanical work. This process is fundamentally governed by the Laws of thermodynamics, which describe how heat and work are related. Thus, A matches with II.

• B) Electron microscope: An electron microscope uses a beam of electrons instead of light to image a sample. The resolving power of an electron microscope is much higher than a light microscope because electrons, when accelerated, have a much shorter de Broglie wavelength than visible light. This relies on the wave nature of electrons, a concept from quantum mechanics (de Broglie hypothesis). Thus, B matches with III.

• C) Non-reflecting coatings: These coatings are thin films applied to optical surfaces to reduce reflection. They work by manipulating the phase of light waves. Light reflected from the top surface of the coating and light reflected from the bottom surface of the coating (at the coating-glass interface) interfere destructively, thereby reducing the overall reflected light. This phenomenon is Interference of light. Thus, C matches with IV.

• D) Tokamak: A Tokamak is a device that uses a strong magnetic field to confine plasma within a toroidal chamber for nuclear fusion research. The charged particles in the plasma are held in place by the magnetic field, preventing them from interacting with the reactor walls and maintaining the extremely high temperatures required for fusion. This principle is Magnetic confinement of plasma. Thus, D matches with I.

The highest intensity when two light waves mix is given by:

$ I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2 $

Case 1: Both waves have the same intensity

If both intensities are equal, let them be $ I_1 = I_2 = I_0 $.

$ I = \left(2 \sqrt{I_0}\right)^2 = 4 I_0 $

This means the maximum intensity is 4 times the intensity of one wave.

So, we can write: $ I_0=\frac{I}{4} $

Case 2: One wave is four times stronger

If the first wave is four times stronger, $ I_1 = 4I_0 $ and $ I_2 = I_0 $.

Maximum intensity becomes:

$ I' = (\sqrt{4I_0} + \sqrt{I_0})^2 $

$ I' = (2\sqrt{I_0} + \sqrt{I_0})^2 = (3\sqrt{I_0})^2 = 9I_0 $

Now, substitute $ I_0 = \frac{I}{4} $:

$ I' = 9 \times \frac{I}{4} = \frac{9I}{4} $

Step 1: Find the position equations for the fringes.

The position of the $n$th bright fringe is given by: $y_n = \frac{n D \lambda}{d}$

So, for the 5th bright fringe: $y_5 = \frac{5 D \lambda}{d}$

The position of the $n$th dark fringe is: $y_n = \frac{(2n - 1) D \lambda}{2d}$

For the 7th dark fringe: $y_7 = \frac{(2 \times 7 - 1) D \lambda}{2d} = \frac{13 D \lambda}{2d}$

Step 2: Use the given distance between 5th bright and 7th dark fringes.

Distance between the 7th dark and 5th bright fringe: $y_7 - y_5 = \frac{13 D \lambda}{2 d} - \frac{5 D \lambda}{d}$

First, make the denominators the same: $\frac{13 D \lambda}{2 d} - \frac{10 D \lambda}{2 d} = \frac{3 D \lambda}{2 d}$

The distance given in the question is $3~\mathrm{mm}$: $\frac{3 D \lambda}{2 d} = 3 \times 10^{-3}~\mathrm{m}$

Step 3: Find the value of $\frac{D \lambda}{d}$.

Divide both sides by $3$: $\frac{D \lambda}{2 d} = 1 \times 10^{-3}$ Now multiply both sides by $2$: $\frac{D \lambda}{d} = 2 \times 10^{-3}~\mathrm{m}$ So, $\frac{D \lambda}{d} = 2~\mathrm{mm} \quad \ldots (i)$

Step 4: Find the positions needed for the second distance.

Position of the 7th bright fringe: $y_7' = \frac{7 D \lambda}{d}$

Position of the 5th dark fringe: $y_5' = \frac{(2 \times 5 - 1) D \lambda}{2 d} = \frac{9 D \lambda}{2 d}$

Step 5: Calculate the distance between 5th dark and 7th bright fringes.

$y_7' - y_5' = \frac{7 D \lambda}{d} - \frac{9 D \lambda}{2 d}$ Make denominators the same: $\frac{14 D \lambda}{2 d} - \frac{9 D \lambda}{2 d} = \frac{5 D \lambda}{2 d}$

Step 6: Find the final answer using the value from earlier.

From (i), $\frac{D \lambda}{d} = 2~\mathrm{mm}$, so: $\frac{5 D \lambda}{2 d} = \frac{5}{2} \times 2~\mathrm{mm} = 5~\mathrm{mm}$

Final Answer: Distance between 5th dark and 7th bright fringes is 5 mm.

$ \begin{aligned} &\text { According to Fresnel distance }\\ &\begin{aligned} Z_F & =\frac{a^2}{\lambda} \\ \lambda & =\frac{a^2}{Z_F}=\frac{\left(5 \times 10^{-3}\right)^2}{50} \\ & =0.5 \times 10^{-6} \mathrm{~m} \\ & =5000 \times 10^{-10} \mathrm{~m} \\ & =5000 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} \end{aligned} $

Given, $\Delta x=\frac{\lambda}{3}$

$ \begin{aligned} & \theta=\pi \cdot \frac{\Delta x}{\lambda} \\ & \theta=\frac{\pi \cdot \lambda}{3 \cdot \lambda}=\frac{\pi}{3} \end{aligned} $

So, $I^{\prime}=I_{\text {max }} \cos ^2 \theta$

$ \begin{aligned} & I^{\prime}=I_{\max } \cdot \cos ^2\left(\frac{\pi}{3}\right) \\ & I^{\prime}=\frac{I_{\max }}{4} \quad \therefore I_{\max }=I \\ & I^{\prime}=\frac{I}{4} \end{aligned} $

According to Rayleigh’s law of scattering, the intensity of scattered light $ I $ is inversely proportional to the fourth power of the wavelength of light, i.e.

$ I \propto \frac{1}{\lambda^4} $

This can be expressed as scattering $\propto \lambda^n$.

So,

$ n = -4 $

✅ Correct Option: (B) −4

$ \begin{aligned} & \text { } \beta_1=\frac{D_1 \lambda_1}{d_1}=0.36 \times 10^{-3} \\ & \Rightarrow \quad \frac{1 \times \lambda_1}{2 \times 10^{-3}}=0.36 \times 10^{-3} \\ & \Rightarrow \quad \lambda_1=0.72 \times 10^{-6} \mathrm{~m} \end{aligned} $

$ \begin{aligned} d_2 & =(2-0.5) \mathrm{mm}=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \\ D_2 & =(100+50) \mathrm{cm}=1.5 \mathrm{~m} \\ \lambda_2 & =\lambda_1=0.72 \times 10^{-6} \mathrm{~m} \\ \therefore \beta & =\frac{D_2 \lambda_2}{d_2}=\frac{1.5 \times 0.72 \times 10^{-6}}{1.5 \times 10^{-3}} \\ & =0.72 \times 10^{-3} \mathrm{~m}=0.72 \mathrm{~mm} \end{aligned} $

$I_0 \rightarrow$ Intensity of unpolarised light.

Intensity of polarised light emerging from first polariod $=\frac{I_0}{2}$

According to law of Malus, intensity of light emerging from second polaroid.

$ \begin{aligned} & I^{\prime}=\frac{I_0}{2} \cos ^2 \theta \Rightarrow 0.375 I_0=\frac{I_0}{2} \cos ^2 \theta \\ & \Rightarrow \cos ^2 \theta=0.75=\frac{3}{4} \\ & \quad \cos \theta=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}=\cos 30^{\circ} \\ & \quad \therefore \quad \theta=30^{\circ} \end{aligned} $

Position of first minima is given as

$ \begin{aligned} y_1 & =\frac{m \lambda D}{a} \\ & =\frac{\lambda D}{a} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because m=1]\\ & =\frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}} \\ & =0.25 \times 10^{-3} \mathrm{~m} \\ & =0.25 \mathrm{~mm} \end{aligned} $

The first minima occur symmetrically on either side of the central maximum. Therefore, separation between the first $\operatorname{minima}=2 y_1$

$ \begin{aligned} & =2 \times 0.25 \mathrm{~mm} \\ & =0.5 \mathrm{~mm} \end{aligned} $

Let’s recall the formula for fringe width in Young’s double-slit experiment:

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

where

$\lambda$ = wavelength of light,

$D$ = distance between slits and screen,

$d$ = distance between the two slits.

Given:

• Initially, slit separation = $d$

• Finally, slit separation = $3d$

Find:

Ratio of initial and final fringe widths, i.e.

$ \frac{\beta_{\text{initial}}}{\beta_{\text{final}}} $

Calculation:

$ \beta_{\text{initial}} = \frac{\lambda D}{d} $

$ \beta_{\text{final}} = \frac{\lambda D}{3d} $

Thus,

$ \frac{\beta_{\text{initial}}}{\beta_{\text{final}}} = \frac{\lambda D / d}{\lambda D / 3d} = 3 $

So the ratio of initial : final fringe width is 3 : 1.

✅ Correct Option: D) 3 : 1

Given, $d=0.2 \mathrm{~cm}=0.002 \mathrm{~m}$

$ \lambda=5000 \mathop {\rm{A}}\limits^{\rm{o}}=5000 \times 10^{-10} \mathrm{~m} $

The distance between two consecutive dark fringes is given by

$ \begin{aligned} & \beta=\frac{\lambda D}{d}=\frac{\left(5000 \times 10^{-10}\right) \mathrm{m} \times 1 \mathrm{~m}}{0.002 \mathrm{~m}} \\ & \beta=\frac{5 \times 10^{-7}}{2 \times 10^{-3}} \mathrm{~m} \\ & \beta=2.5 \times 10^{-4} \mathrm{~m}=0.25 \mathrm{~mm} \end{aligned} $

$ \begin{aligned} &\text { Fringe width, } \beta=\frac{D \lambda}{d}\\ &\begin{aligned} & \frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2} \cdot \frac{d_2}{d_1} \\ & =\left(\frac{\lambda_1}{\lambda_1+20 \% \text { of } \lambda_1}\right) \cdot \frac{\left(d_1-25 \% \text { of } d_1\right)}{d_1} \\ & =\left(\frac{\lambda_1}{\lambda_1+\frac{\lambda_1}{5}}\right)\left(\frac{d_1-\frac{d_1}{4}}{d_1}\right)=\frac{5}{6} \times \frac{3}{4} \\ & \Rightarrow \quad \frac{\beta_1}{\beta_2}=\frac{5}{8} \\ & \Rightarrow \beta_2=\frac{8}{5} \beta_1=\frac{8}{5} \times 0.3=0.48 \mathrm{~mm} \end{aligned} \end{aligned} $

When unpolarised light of intensity $\left(I_0\right)$ passes through the first polariser, the intensity reduces to half $\left(\frac{I_0}{2}\right)$.

According to law of Malus, intensity after second polariser.

$ I_2=\frac{I_0}{2} \cos ^2 30^{\circ}=\frac{I_0}{2} \times \frac{3}{4}=\frac{3 I_0}{8} $

The angle between the second and third polariser is also $30^{\circ}$. The intensity after the third polariser

$ \begin{aligned} & I_3=I_2 \cos ^2 30^{\circ} \\ & =\frac{3 I_0}{8} \times \frac{3}{4}=\frac{9 I_0}{32} \\ \therefore \quad & \frac{I_2}{I_3}=\frac{\frac{3 I_0}{8}}{\frac{9 I_0}{32}}=\frac{4}{3} \\ \therefore \quad & I_2: I_3=4: 3 \end{aligned} $

For $n$th bright fringe of red light,

$ d \sin \theta_r=n \lambda_r $

For $(n+1)$ th bright fringe of blue light

$ d \sin \theta_b=(n+1) \lambda_b $

Since, the fringes coincide,

$ \begin{aligned} & \text { i.e. } \theta_r=\theta_b \\ & \Rightarrow \quad d \sin \theta_r=d \sin \theta_b \\ & \Rightarrow \quad n \lambda_r=(n+1) \lambda_b \\ & \Rightarrow \quad n \times 7.5 \times 10^{-5}=(n+1) 5 \times 10^{-5} \\ & \Rightarrow \quad 7.5 n=5 n+5 \\ & \Rightarrow \quad 2.5 n=5 \Rightarrow n=2 \end{aligned} $

The correct answer is Option A: Both Statement I and Statement II are true.

Explanation:

Statement I discusses the dispersion of white light through a prism, which is a phenomenon where white light splits into its component colors when passed through a prism. This happens because different colors (or wavelengths) of light bend (refract) by different amounts upon passing through a prism. Red light bends the least while violet bends the most, with yellow falling somewhere in between. This is why the statement, "When the white light passed through a prism, the red light bends lesser than yellow and violet," is true.

Statement II addresses the underlying cause of the dispersion of light, which is that the refractive index of a medium varies with wavelength (or color) of light. This property of the medium is known as dispersion. A refractive index determines how much light bends when entering a medium. Since the refractive index varies with the wavelength, different colors of light bend by different amounts when they pass through a dispersive medium like a glass prism. This is why violet light bends more than red light - because the refractive index for violet light is higher than that for red light in glass. Therefore, the statement, "The refractive indices are different for different wavelengths in dispersive medium," is also true.

Thus, both statements I and II accurately describe the principles behind the phenomenon of light dispersion through a prism, making Option A the correct choice.

$\begin{aligned} & I_{\max }=\left(\sqrt{4 I_0}+\sqrt{I_0}\right)^2=\left(3 \sqrt{I_0}\right)^2=9 I_0 \\ & I_{\min }=\left(\sqrt{4 I_0}-\sqrt{I_0}\right)^2=I_0 \\ & \frac{I_{\max }}{I_{\min }}=9: 1 \end{aligned}$