Wave Optics

• 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} $

$ \begin{aligned} &\text { According to Brewster's law }\\ &\begin{aligned} & & \mu_{g o} & =\frac{\mu_g}{\mu_w}=\tan i_p=\sqrt{3} \\ \therefore & & \mu_g & =\sqrt{3} \cdot \sqrt{2}=\sqrt{6} \end{aligned} \end{aligned} $

For 5th bright fringe

$ x^5=n \frac{D \lambda}{d}=\frac{5 \times 1 \times \lambda}{1 \times 10^{-3}} $

and 3rd dark fringe

$ x^3=\frac{(2 n-1) D \lambda}{2 d}=\frac{5 \times 1 \times \lambda}{2 \times 1 \times 10^{-3}} $

Now, $x^5-x^3=1.63 \mathrm{~mm}$

$ \begin{aligned} \frac{5 \lambda}{10^{-3}}-\frac{5 \lambda}{2 \times 10^{-3}} & =1.63 \times 10^{-3} \\ \frac{5 \lambda}{2} & =1.63 \times 10^{-3} \\ \lambda & =\frac{3.26}{5} \times 10^{-6}=6.5 \times 10^{-7} \mathrm{~m} \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.

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

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 $\sin 2 \theta=1$

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

We know that the resultant intensity in Young's double slit experiment is given as

$ I_{\mathrm{res}}=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}} \cos \phi $

where $ \phi= $ phase difference,

$ \begin{aligned} I_{1} & =I_{2}=I_{0} \\\\ \phi & =\frac{2 \pi}{\lambda} \Delta x \end{aligned} $

For path difference is, $ \Delta x=\lambda $

$ \begin{aligned} \phi_{1} & =\frac{2 \pi}{\lambda} \times \lambda=2 \pi \\\\ I_{\text {res }} & =I_{0}+I_{0}+2 \sqrt{I_{0} I_{0}} \cos (2 \pi) \\\\ I_{\text {res }} & =I=4 I_{0} \\\\ I_{0} & =\frac{I}{4} \end{aligned} $

For path difference, $ \Delta x=\frac{\lambda}{3} $

$ \begin{aligned} \phi_{2} & =\frac{2 \pi}{3} \\\\ \cos \frac{2 \pi}{3} & =\frac{-1}{2} \\\\ I^{\prime} & =I_{0}+I_{0}+2 \sqrt{I_{0} I_{0}} \cos \left(\frac{2 \pi}{3}\right) \\\\ I^{\prime} & =2 I_{0}-I_{0} \end{aligned} $

where $ \rho $ is resistivity

(iii) $ \vdots $ Subsitute the value of $ I_{0} $ )

$ I^{\prime}=\frac{I}{4} $

Given, $d=36 \mathrm{~m}$

$ \lambda=540 \mathrm{~nm} $

From Rayleigh criterion, $\theta=1.22 \frac{\lambda}{D}$

$ \begin{aligned} & \quad=\frac{1.22 \times 540 \times 10^{-9}}{3.6} \\\\ & =183 \times 10^{-9} \mathrm{rad} \\\\ & =1.83 \times 10^{-7} \mathrm{rad} \end{aligned} $

In YDSE, the intensity of light at a point on the screen is

$ I=I_0 \cos ^2\left(\frac{\pi \Delta x}{\lambda}\right) $

Given, $\lambda=6000 \times 10^{-10} \mathrm{~m}$

Path difference,

$ \Delta x_1=2000 \times 10^{-10} \mathrm{~m} $

Path difference,

$ \Delta x_2=1000 \times 10^{-10} \mathrm{~m} $

For $I_1$,

$ \begin{aligned} & I_1=I_0 \cos ^2\left(\frac{\pi \times 2000 \times 10^{-10}}{6000 \times 10^{-10}}\right) \\\\ & I_1=I_0 \cos ^2\left(\frac{\pi}{3}\right) \\\\ & I_1=\frac{I_0}{4} \end{aligned} $

For $I_2$,

$ \begin{aligned} & I_2=I_0 \cos ^2\left(\frac{\pi \times 1000 \times 10^{-10}}{6000 \times 10^{-10}}\right) \\\\ & I_2=I_0 \cos ^2\left(\frac{\pi}{6}\right) \\\\ & I_2=\frac{3 I_0}{4} \end{aligned} $

Therefore, the ratio $I_1: I_2$ is

$ \begin{aligned} & I_1: I_2=\frac{I_0}{4}: \frac{3 I_0}{4} \\\\ & I_1: I_2=1: 3 \end{aligned} $

So, the ratio of the intensities is $1: 3$.

When two light waves superimpose, the resultant intensity is influenced by both their individual intensities and the phase difference between them. In this case, the path difference is 12.5% of the wavelength, which translates into a phase difference:

$ \frac{12.5}{100} \times 360^{\circ} = 45^{\circ} $

The formula for the resultant intensity when two waves interfere is given by:

$ I_r = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\phi) $

Given:

$ I_1 = I $

$ I_2 = 2I $

The expression for the resultant intensity becomes:

$ I_r = I + 2I + 2 \sqrt{2I \times I} \cos(45^{\circ}) $

Calculating further:

$ I_r = 3I + 2I\sqrt{2} \times \frac{\sqrt{2}}{2} $

$ I_r = 3I + 2I = 5I $

Therefore, the resultant intensity at the point where the two light waves with a path difference of 12.5% of the wavelength superimpose is $5I$.

Given:

Intensity of unpolarized light, $ I $

Angle between the polarizer and analyser, $ \theta = 45^{\circ} $

Step-by-step Breakdown

Initial Intensity after Polariser:

When unpolarized light passes through a polarizer, its intensity is reduced by half. Therefore, the intensity of polarized light emerging from the polarizer is:

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

Intensity after Analyser using Malus's Law:

According to Malus's law, the intensity $ I_{\text{out}} $ of the light after passing through the analyser is given by:

$ I_{\text{out}} = I_0 \cos^2 \theta $

Substituting the given values:

$ I_{\text{out}} = \frac{I}{2} \cos^2 45^{\circ} $

Knowing that $\cos 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{1}{2}^{\frac{1}{2}} $, this becomes:

$ I_{\text{out}} = \frac{I}{2} \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I}{2} \times \frac{1}{2} $

$ I_{\text{out}} = \frac{I}{4} $

Thus, this derivation shows that the intensity of the light after it has passed through both the polariser and the analyser is $\frac{I}{4}$.

Given the following parameters:

Width of slit, $ x $

Wavelength, $ \lambda = 6500 \, \text{Å} $

First minima angle, $ \theta = 30^{\circ} $

To find the slit width using the diffraction principle, we apply:

$ a \sin \theta = n \lambda $

For the first minima ($ n = 1 $):

$ a \sin 30^{\circ} = 1 \times (6500 \times 10^{-10}) $

Calculating this step-by-step:

$ a \times 0.5 = 6500 \times 10^{-10} $

$ a \times 0.5 = 65 \times 10^{-8} $

Solving for $ a $:

$ a = \frac{65 \times 10^{-8}}{0.5} $

$ a = 1.3 \times 10^{-6} $

Converting to micrometers:

$ a = 1.3 \, \mu \text{m} $

Therefore, the width of the slit $ x $ is $ 1.3 \, \mu \text{m} $.

For diffraction to occur significantly, the slit width $ a $ should be comparable to the wavelength $ \lambda $ of the incident light. This means that the condition $ a \leq \lambda $ or equivalently $ \frac{a}{\lambda} \leq 1 $ must be met.

To determine the minimum separation at which two objects can be seen as distinct under a microscope, we use the formula:

$ d = \frac{\lambda}{2 \cdot \text{NA}} $

where $\text{NA}$, the numerical aperture, is defined as:

$ \text{NA} = n \sin \theta $

Here, $n$ represents the refractive index of the medium.

Case 1: Microscope in Air

In air, the refractive index ($n$) is 1. The given minimum separation to see objects distinctly is $d = 6 \, \mu\mathrm{m}$. The relationship between minimum separation $d$ and refractive index $n$ is:

$ d \propto \frac{1}{n} $

Case 2: Microscope in a Medium with Refractive Index 1.5

In this scenario, the refractive index $n$ is 1.5. The separation $d'$ can be calculated as:

$ d' = \frac{d}{1.5} $

Substitute the given values:

$ d' = \frac{6 \, \mu\mathrm{m}}{1.5} = 4 \, \mu\mathrm{m} $

Thus, when the microscope is placed in a medium with a refractive index of 1.5, the minimum separation at which two objects can be seen as distinct is $4 \, \mu\mathrm{m}$.