Wave Optics

In a double slit experiment, the bright spots (maxima) from interference are found where:

  $d\,\sin \theta = m\lambda$, where $m$ can be 0, ±1, ±2, and so on.

The dark spots (minima) from single slit diffraction appear at:

  $a\,\sin \theta = \pm\lambda$.

For a bright interference fringe to be inside the central bright area of the single slit (central maximum), it must fall between these two first diffraction minima.

This means we need:

 $|m\lambda/d| \leq \lambda/a$, or $|m| \leq d/a$.

The problem says the distance between the slits ($d$) is 6.1 times the width of each slit ($a$), so $d/a = 6.1$.

The biggest whole number for $m$ is 6, so there are maxima for $m = \pm1, \pm2, ..., \pm6$.

If we count only the bright spots on the sides (not the middle one), that's 6 on one side and 6 on the other, making 12 bright maxima.

If the central bright spot ($m=0$) is counted too, there would be 13, but usually “within the central maximum” means only the fringes between the two minima, not the central line.

So, you will see 12 bright interference fringes inside the central maximum of the single slit envelope.

Answer = 12. (Option C)

The smallest detail a microscope can see (its resolution limit) is given by the formula:

$ d_{\text{min}} \approx \frac{\lambda}{2 \text{N.A.}} $

Here, N.A. stands for numerical aperture. The formula for numerical aperture is:

$ \text{N.A.} = n \sin \alpha $

where n is the refractive index of the oil (the medium), and α is half the angle made by the cone of light rays coming from the specimen.

The resolving power (RP) of the microscope tells us how well it can see small details. It is the inverse of the resolution limit:

$ \text{RP} = \frac{1}{d_{\text{min}}} \propto n \sin \alpha $

Now, the angle subtended by the lens diameter at the focal point is β. From geometry, we see that α (half the cone angle) is related to β by:

$ \sin \alpha = \sin 2\beta $

This means the resolving power becomes:

$ \text{RP} \propto n \sin 2\beta $

So, the microscope will see finer details as the value of $ n \sin 2\beta $ increases. Answer: C.

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 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.

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.

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

To determine the angular spread of the central maximum in a single-slit diffraction pattern, we can use the formula for the angular position of the first minimum (also understood as the boundary of the central maximum) on either side of the center. For a single-slit diffraction pattern, the angle $ \theta $ to the first minimum is given by the condition:

$ a \sin(\theta) = m\lambda $

where :

• $ a $ is the width of the slit,


• $ \lambda $ is the wavelength of the light (or microwaves in this case),


• $ m $ is the order number of the minimum with $ m = \pm1, \pm2, \pm3, ... $ (for the first minimum, $ m = \pm1 $)

However, we are only interested in the angle to the first minimum, so we will only consider $ m = \pm1 $. Since the slit width $ a = 4.0 \mathrm{cm} $ and the wavelength $ \lambda = 2.0 \mathrm{cm} $, we substitute these values into the equation to find $ \theta $:

$ 4.0 \mathrm{cm} \times \sin(\theta) = 1 \times 2.0 \mathrm{cm} $

$ \Rightarrow $ $ 4.0 \sin(\theta) = 2.0 $

$ \Rightarrow $ $ \sin(\theta) = \frac{2.0}{4.0} $

$ \Rightarrow $ $ \sin(\theta) = 0.5 $

The angle whose sine is 0.5 is $ 30^{\circ} $.

This angle of $ 30^{\circ} $ is the angle from the center to the first minimum on one side. The angular spread of the central maximum would cover the range from the first minimum on one side to the first minimum on the other side, totalling twice this angle:

$ \text{Angular spread} = 2 \times \theta = 2 \times 30^{\circ} = 60^{\circ} $

Therefore, the angular spread of the central maxima of the diffraction pattern is $ 60^{\circ} $, which corresponds to Option A.

To find the linear width of the central maximum in a single-slit diffraction pattern, we can use the formula that relates the position of the first minima on either side of the central maximum. The angle, $ \theta $, at which the first minimum occurs is given by:

$ a \sin(\theta) = m\lambda $

Where:

• $ a $ is the width of the slit


• $ \lambda $ is the wavelength of monochromatic light


• $ m $ is the order number of the minimum (for the first minimum, $ m = \pm 1 $)

In our case, we want to find the position of the first minima to determine the width of the central maximum on the screen. So we will use $ m = 1 $ and $ m = -1 $ which correspond to the first minima on either side of the central peak.

Given the width of the slit $ a = 0.01 $ mm, which we need to convert to meters for consistency with the wavelength:

$ a = 0.01 \times 10^{-3} \text{m} $

And the wavelength $ \lambda = 6000 \mathring{A} $, also converting to meters:

$ \lambda = 6000 \times 10^{-10} \text{m} $

We're using a convex lens of focal length $ f = 20 $ cm to project the pattern onto a screen. We'll need to convert the focal length to meters as well:

$ f = 20 \times 10^{-2} \text{m} $

We can calculate the angle $ \theta $ needed for the first minima using the approximation of small angles, where $ \sin(\theta) \approx \theta $:

$ a \theta = m\lambda $

Now, solve for $ \theta $ for the first minimum (m = 1):

$ \theta = \frac{\lambda}{a} $

Plugging in the values:

$ \theta = \frac{6000 \times 10^{-10}}{0.01 \times 10^{-3}} $

$ \theta = \frac{6000 \times 10^{-10}}{10^{-5}} $

$ \theta = 6000 \times 10^{-5} $

Now, linear width of the central maximum on the screen (from -m to m, or -1 to +1) will be twice the distance from the center to the first minimum, which can be found using the focal length of the lens $ f $ and the angle $ \theta $:

$ y = 2f\theta $

Plugging the focal length and calculated theta:

$ y = 2 \times 20 \times 10^{-2} \times 6000 \times 10^{-5} $

$ y = 40 \times 10^{-2} \times 6000 \times 10^{-5} $

$ y = 240 \times 10^{-3} \text{m} $

$ y = 24 \times 10^{-2} \text{m} $

$ y = 24 \text{mm} $

This means the linear width of the central maximum is 24 mm. Hence, the correct answer is Option B.

By Brewster's law

At complete reflection refracted ray and reflected ray are perpendicular.

Intensity of emergent light

$=\frac{\mathrm{I}_0}{2} \cos ^2 45^{\circ}=\frac{\mathrm{I}_0}{4}$

Width of $1^{\text {st }}$ secondary maxima $=\frac{\lambda}{a} \cdot D$

Here

$\begin{aligned} & a=0.2 \times 10^{-3} \mathrm{~m} \\ & \lambda=400 \times 10^{-9} \mathrm{~m} \\ & D=100 \times 10^{-2} \end{aligned}$

Width of $1^{\text {st }}$ secondary maxima

$\begin{aligned} & =\frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 100 \times 10^{-2} \\ & =2 \mathrm{~mm} \end{aligned}$

$\begin{aligned} & \Delta x=\frac{7 \lambda}{4} \\ & \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4}=\frac{7 \pi}{2} \\ & \mathrm{I}=\mathrm{I}_{\max } \cos ^2\left(\frac{\phi}{2}\right) \\ & \frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)=\cos ^2\left(\frac{7 \pi}{2 \times 2}\right)=\cos ^2\left(\frac{7 \pi}{4}\right) \\ & =\cos ^2\left(2 \pi-\frac{\pi}{4}\right) \\ & =\cos ^2 \frac{\pi}{4} \\ & =\frac{1}{2} \\ & \end{aligned}$

Let $\mathrm{I}_0$ be intensity of unpolarised light incident on first polaroid.

$\mathrm{I}_1=$ Intensity of light transmitted from $1^{\text {st }}$ polaroid $=\frac{\mathrm{I}_0}{2}$

$\theta$ be the angle between $1^{\mathrm{st}}$ and $2^{\text {nd }}$ polaroid

$\phi$ be the angle between $2^{\text {nd }}$ and $3^{\text {rd }}$ polaroid

$\theta+\phi=90^{\circ}$ (as $1^{\text {st }}$ and $3^{\text {rd }}$ polaroid are crossed)

$\phi=90^{\circ}-\theta$

$\mathrm{I}_2=$ Intensity from $2^{\text {nd }}$ polaroid

$\mathrm{I}_2=\mathrm{I}_1 \cos ^2 \theta=\frac{\mathrm{I}_0}{2} \cos ^2 \theta$

$\mathrm{I}_3=$ Intensity from $3^{\text {rd }}$ polaroid

$\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \phi$

$\mathrm{I}_3=\mathrm{I}_1 \cos ^2 \theta \cos ^2 \phi$

$\mathrm{I}_3=\frac{\mathrm{I}_0}{2} \cos ^2 \theta \cos ^2 \phi$

$\phi=90-\theta$

$\begin{aligned} & I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta \\ & I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2 \\ & I_3=\frac{I_0}{8} \sin ^2 2 \theta \end{aligned}$

$\mathrm{I}_3$ will be maximum when $\sin 2 \theta=1$

$\begin{aligned} & 2 \theta=90^{\circ} \\ & \theta=45^{\circ} \end{aligned}$

In a Young's double slit experiment, the intensity at a point on the screen is given by the formula :

$I = 4I_0\cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right)$

where $I_0$ is the intensity at the center of the pattern, $d$ is the distance between the slits, $\theta$ is the angle between the line joining the point to the center of the pattern and the line passing through the center of the pattern and the slits, and $\lambda$ is the wavelength of the light used.

The phase difference between the waves from the two slits at a point on the screen is given by :

$\Delta \phi = \frac{2\pi d\sin\theta}{\lambda}$

For a phase difference of $\pi/3$ between the waves from the two slits at point P, we have :

$\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{3}$

For a phase difference of $\pi/2$ between the waves from the two slits at point Q, we have :

$\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{2}$

Solving for $\sin\theta$ in both cases, we get:

$\sin\theta_P = \frac{\lambda}{6d},\quad \sin\theta_Q = \frac{\lambda}{4d}$

Substituting these values in the formula for intensity, we get :

$\frac{I_P}{I_Q} = \frac{4\cos^2\left(\frac{\pi}{6}\right)}{4\cos^2\left(\frac{\pi}{4}\right)} = \frac{3}{2}$

Therefore, the ratio of intensities at points P and Q is 3 : 2

In the 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,
• $D$ is the distance between the screen and the double-slit,
• $d$ is the separation between the two slits.

If we are considering a change in wavelength but the fringe width is changing and the setup of the experiment (the values of $D$ and $d$) stays the same, we can see that the fringe width is directly proportional to the wavelength. This is because $D$ and $d$ are constants in this case, so we can write :

$ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} $

Here, the given wavelengths are $\lambda_1 = 400 \, \text{nm}$ and $\lambda_2 = 600 \, \text{nm}$, and the given fringe width for the light of wavelength $400 \, \text{nm}$ is $\beta_1 = 2 \, \text{mm}$. We are asked to find the fringe width $\beta_2$ for the light of wavelength $600 \, \text{nm}$.

Substituting the given values into the proportionality equation, we get :

$ \frac{2 \, \text{mm}}{\beta_2} = \frac{400 \, \text{nm}}{600 \, \text{nm}} $

Solving this equation for $\beta_2$ gives:

$ \beta_2 = 2 \, \text{mm} \times \frac{600 \, \text{nm}}{400 \, \text{nm}} = 3 \, \text{mm} $

Case I :

Case II :

$i+r=90^{\circ}$ as transmitted is $\perp$ to reflected.

$\tan i=\frac{\mu_{a}}{\mu_{g}} \Rightarrow i=\tan ^{-1} \frac{\mu_{a}}{\mu_{g}}=\frac{\pi}{2}-\theta_{B}$

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

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

$\therefore$ ${{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{2({I_1} + {I_2})} \over {4 \times \sqrt {{I_1}{I_2}} }}$

$ = {1 \over 2} \times {{\left( {{{{I_1}} \over {{I_2}}} + 1} \right)} \over {\sqrt {{{{I_1}} \over {{I_2}}}} }}$

$ = {1 \over 2} \times {{\left( {{1 \over 4} + 1} \right)} \over {\left( {{1 \over 2}} \right)}}$

$ = {5 \over 4} = {{2 \times 2 + 1} \over {1 + 3}}$

$\therefore$ ${\alpha \over \beta } = {2 \over 1} = 2$

$B = 12 \times {10^{ - 3}}$

$\beta ' = {\beta \over \mu } = {{12 \times {{10}^{ - 3}}} \over {{4 \over 3}}}$

$ = 9 \times {10^{ - 3}}$ m = 9 mm

Let,

Width of first slit = w₁ and width of second slit = w₂

Given, ${{{w_1}} \over {{w_2}}} = {9 \over {16}}$

Given,

Intensity of light $(I) \propto w$

$\therefore$ ${{{I_1}} \over {{I_2}}} = {{{w_1}} \over {{w_2}}} = {9 \over {16}}$

We know,

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

$ = {\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}$

$ = {\left( {{{{3 \over 4} + 1} \over {{3 \over 4} - 1}}} \right)^2}$

$ = {\left( {{7 \over 1}} \right)^2}$

$ = {{49} \over 1}$

Fringe width = ${{\lambda D} \over d}$

$\Rightarrow$ Fringe width $\propto$ ${\lambda \over d}$

$\Rightarrow$ New fringe width = 0.5 mm $ \times {{1.2} \over 2} = 0.3$ mm

For the intensity to remain same the position must be of a maxima so path difference must be n$\lambda$ so

(1.5 $-$ 1) x$\lambda$ = n$\lambda$

x = 2n (n = 0, 1, 2 ....)

So, value of x will be

x = 0, 2, 4, 6 ....