Wave Optics

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} $

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 $4 \, \mu\mathrm{m}$.

In Young's double slit experiment, we have two slits separated by a distance of 2 mm. The interference pattern is projected onto a screen positioned 2 meters away from the slits. We are using light with a wavelength of 400 nm.

To find the fringe width ($\beta$), we use the formula:

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

Where:

$\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$ is the wavelength of light.

$D = 2 \text{ m}$ is the distance from the slits to the screen.

$d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$ is the distance between the slits.

Plugging in the values:

$ \beta = \frac{400 \times 10^{-9} \times 2}{2 \times 10^{-3}} = 4 \times 10^{-4} \text{ m} = 0.4 \times 10^{-3} \text{ m} $

Thus, the fringe width $\beta$ is $0.4 \times 10^{-3} \text{ m}$.

In Young's Double Slit Experiment, we have the following:

Path difference, $ x = \frac{\lambda}{6} $

To find the phase difference $\phi$, we use the formula:

$ \phi = \frac{2\pi}{\lambda} \times x = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} $

Solving this gives us:

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

The intensity $ I $ at a point in terms of maximum intensity $ I_0 $ is given by:

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

Substituting $\phi = \frac{\pi}{3}$ into the equation, we get:

$ I = I_0 \cos^2 \left(\frac{\pi}{6}\right) $

We know that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$, so substituting that value in gives:

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

Thus, the ratio $\frac{I}{I_0}$ is:

$ \frac{I}{I_0} = \frac{3}{4} $

In Young's double slit experiment using monochromatic light with a wavelength of 6000 Å, the initial fringe width ($\beta_1$) is 3 mm. We are tasked with determining the new fringe width when the distance between the screen and the slits is increased by 50%, and the distance between the slits is decreased by 10%.

Given:

Wavelength, $\lambda = 6000 \, \mathring{A}$

Initial fringe width, $\beta_1 = 3 \, \text{mm}$

The formula for fringe width ($\beta$) in Young's experiment is:

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

where $D$ is the distance between the screen and the slits, and $d$ is the distance between the slits. From the given information:

$ 3 = \frac{D \times 6000}{d} \tag{1} $

Now, recalculating for the changed conditions:

The new distance between the screen and slits, $D'$, when increased by 50%:

$ D' = D + \frac{D \times 50}{100} = 1.5D $

The new distance between the slits, $d'$, when decreased by 10%:

$ d' = d - \frac{d \times 10}{100} = 0.9d $

Now, the new fringe width, $\beta'$, is given by:

$ \beta' = \frac{D' \lambda}{d'} = \frac{1.5 \times D \times 6000}{0.9 \times d} $

Substituting Eq. (1) into the equation:

$ \beta' = \frac{1.5}{0.9} \times 3 $

Calculating the above expression:

$ \beta' = \frac{45}{9} = 5 \, \text{mm} $

Thus, the new fringe width is 5 mm.

Given, wavelength, $\lambda=5000 \mathop A\limits^o=5 \times 10^{-7} \mathrm{~m}$

Slit separation, $d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$

Distance between slit and screen,

$D=20 \mathrm{~cm}=0.2 \mathrm{~m}$

Thickness of transparent plate $=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$ Fringe shift, $\Delta x=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}$

Fringe shift is given by

$\begin{aligned} & \Delta x=\frac{D}{d}(\mu-1) t \\ & \Rightarrow \quad 6 \times 10^{-3}=\frac{0.2}{3 \times 10^{-3}}(\mu-1) \times 10^{-3} \\ & \Rightarrow \frac{6 \times 3 \times 10^{-3}}{0.2}=\mu-1 \Rightarrow 9 \times 10^{-2}=\mu-1 \\ & \Rightarrow \quad \mu=1.09 \\ \end{aligned}$

Hence, refractive index of transparent plate is 1.09.

Given :

$\begin{aligned} \lambda_1 &= 3750 \, \mathring{A} \\ \lambda_2 &= 7500 \, \mathring{A} \\ D &= 4 \, \mathrm{m} \\ d &= 3 \, \mathrm{mm} = 3 \times 10^{-3} \, \mathrm{m} \end{aligned}$

Since,

$ \frac{\lambda_1}{\lambda_2} = \frac{3750}{7500} = \frac{1}{2} $

We know that,

$ \Rightarrow n_1 \lambda_1 = n_2 \lambda_2 \Rightarrow \frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2} = \frac{1}{2} $

Therefore, the required minimum distance is:

$\begin{aligned} x &= \frac{n_1 \lambda_1 D}{d} \\ &= \frac{2 \times 3750 \times 10^{-10} \times 4}{3 \times 10^{-3}} \\ &= \frac{3 \times 10^4 \times 10^{-10}}{3 \times 10^{-3}} \\ &= 10^{-3} \, \mathrm{m} = 1 \, \mathrm{mm} \end{aligned}$

For the given condition,

$\frac{n_1 \lambda_1}{x_1}=\frac{n_2 \lambda_2}{x_2} \quad \text{.... (i)}$

Here, $n_1=5$ (for fifth order)

$\begin{aligned} & \lambda_1=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m} \\ & x_1=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m} \\ & n_2=3(\text { for third order }) \\ & \lambda_2=400 \mathrm{~nm}=4 \times 10^{-7} \mathrm{~m} \\ & x_2=? \end{aligned}$

Putting these values in Eq. (i), we get

$\frac{5 \times 6 \times 10^{-7}}{6 \times 10^{-3}}=\frac{3 \times 4 \times 10^{-7}}{x_2}$

$\begin{aligned} & \Rightarrow \quad 5 x_2=12 \times 10^{-3} \\ & \Rightarrow \quad x_2=2.4 \times 10^{-3} \mathrm{~m} \\ & \qquad =2.4 \mathrm{~mm} \\ & \end{aligned}$

As we know that, wavefront is defined as the locus of all points in same phase.

As we know that,

for diffraction, $d\sin\theta=n\lambda$

where, d is slit width, $\theta$ is angular width and $\lambda$ is wavelength.

Now, if $\lambda$ decreases then $\sin\theta$ decreases and pattern shrinks. So, images moves towards O.

Initially, separation between slits $=d$

Distance of slit from screen $=D$

Initial and final fringe widths be $\beta$ and $\beta^{\prime}$.

Finally, separation between slits, $d^{\prime}=d / 2$

Distance from screen $D^{\prime}=2 D$

As,

$\begin{aligned} \beta & =\frac{\lambda D}{d} \\ \therefore \beta^{\prime} & =\frac{\lambda D^{\prime}}{d^{\prime}}=\frac{\lambda 2 D}{d / 2}=\frac{4 \lambda D}{d} \end{aligned}$

$\Rightarrow \quad \beta^{\prime}=4 \beta$

Angular width for Ist minima, $\theta_1=30^\circ=\pi/6$ rad

Wavelength, $\lambda=500$ nm = $500\times10^{-9}$ m

Let first secondary maximum occur at angle $\theta_2$.

$\because$ $n\lambda = a\sin {\theta _1}$

where, a is slit width.

$\therefore$ $a = {{n\lambda } \over {\sin 30\Upsilon }} = 2\lambda $ [$\because$ $n = 1$]

and for maxima

$(2n + 1){\lambda \over 2} = a\sin {\theta _2}$

$ \Rightarrow (2 + 1){\lambda \over 2} = a\sin {\theta _2}$

$ \Rightarrow \sin {\theta _2} = 3\lambda /2a$ ..... (i)

Substituting the value $a=2\pi$ in Eq. (i), we get

$\begin{array}{ll}\therefore & \sin \theta_2=\frac{3 \lambda}{2 \times 2 \lambda}=\frac{3}{4} \\ \Rightarrow & \theta_2=\sin ^{-1}(3 / 4)\end{array}$