Wave Optics

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.

In Young's double slit experiment within a medium with a refractive index of 1.33, the maximum intensity is represented as $ I_0 $.

To find the intensity at a point on the screen where the path difference between the light from the slits is $ \frac{\lambda}{4} $, follow these steps:

Path Difference: Given as $ \Delta x = \frac{\lambda}{4} $.

Phase Difference Calculation:

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

Substituting the given path difference:

$ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} $

Intensity Calculation: The intensity at any point on the screen with a phase difference $ \phi $ is given by:

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

Substituting the phase difference:

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

Simplifying further:

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

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

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

Thus, the intensity at the specified point on the screen is $\frac{I_0}{2}$.

In the Young's Double Slit Experiment (YDSE), for monochromatic light shining on a screen 1.80 m away from two slits separated by 2.08 mm, the separation of the first and second order bright fringes is 0.553 mm. To find the wavelength of light used, we use the following calculations:

The position of the first bright fringe is given by:

$ y_1 = \frac{\lambda D}{d} $

The position of the second bright fringe is:

$ y_2 = \frac{2\lambda D}{d} $

The separation between the first and second order bright fringes is:

$ y_2 - y_1 = \frac{2\lambda D}{d} - \frac{\lambda D}{d} = \frac{\lambda D}{d} $

From this, we can derive the formula for the wavelength $\lambda$ as:

$ \lambda = \frac{(y_2 - y_1) d}{D} $

Substituting the given values into this equation:

$ \lambda = \frac{0.553 \times 10^{-3} \times 2.08 \times 10^{-3}}{1.8} $

Solving this gives:

$ \lambda = 639 \times 10^{-9} \text{ m} $

Therefore, the wavelength of the light used is:

$ \lambda = 639 \text{ nm} $

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.

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

Given,

Wavelength $d=6000\mathop {\rm{A}}\limits^{\rm{o}}$

Decrease in angular width $=30 \%$

Let original angular width $=\beta$

New width $=\beta-30 \% \beta=0.7 \beta$

Fringe width is given by

$ \begin{aligned} & \beta=\frac{\lambda D}{d} \\ & \beta \propto \lambda\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

$ \beta_1 \propto \lambda_1 $

From Eqs. (i) and (ii)

$ \begin{aligned} & \frac{\beta_1}{\beta}=\frac{\lambda_1}{\lambda} \\ & \lambda_1=\frac{\beta_1 \lambda}{\beta} \Rightarrow \frac{\beta \times 0.7 \times 6000}{\beta} \\ & \lambda_1=4200 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

Hence, the correct option is (b).

The point source emits the spherical wavefront as the light diverge from a point source in all direction. Hence, incident wave front will be spherical in nature.

When the rays got reflected by concave mirror the reflected wavefront will be planet.

$ \text { Hence, option (b) is the correct option. } $

$ \begin{aligned}Given,\,\, d & =1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m} \\ \lambda & =6.5 \times 10^{-7} \mathrm{~m} \\ D & =1 \mathrm{~m} \end{aligned} $

Position of third dark fringes,

$ \begin{aligned} x_3 & =(2 n-1) \frac{D \lambda}{2 d} \\ & =(2 \times 3-1) \frac{D \lambda}{2 d}=\frac{5 D \lambda}{2 d} \end{aligned} $

Position of fifth bright fringe,

$ \begin{aligned} & x_5=n \frac{D \lambda}{d}=5 \frac{D \lambda}{d} \\ \therefore \quad & x_5-x_3=\frac{5 D \lambda}{d}-\frac{5 D \lambda}{2 d} \\ & =\frac{5 D \lambda}{2 d} \\ & =\frac{5 \times 1 \times 6.5 \times 10^{-7}}{2 \times 1 \times 10^{-3}} \\ & =1.625 \times 10^{-3} \mathrm{~m} \\ & =1.625 \mathrm{~mm} \end{aligned} $

Given, $d=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$

$ \begin{aligned} & D=1 \mathrm{~m} \\ & \lambda=500 \mathrm{~nm}=5 \times 10^{-7} \mathrm{~m} \end{aligned} $

Fringe width,

$ \begin{aligned} \beta & =\frac{D \lambda}{d}=\frac{1 \times 5 \times 10^{-7}}{10^{-3}} \\ & =5 \times 10^{-4} \mathrm{~m}=0.5 \times 10^{-3} \mathrm{~m} \\ & =0.5 \mathrm{~mm} \end{aligned} $

Given:

Slit width $ a = 2 \, \mathrm{mm} = 2 \times 10^{-3} \, \mathrm{m} $

Wavelength of light $ \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \mathrm{m} $

To find the Fresnel distance $ f_z $, use the formula:

$ f_z = \frac{a^2}{\lambda} $

Substitute the values into the formula:

$ f_z = \frac{(2 \times 10^{-3} \, \mathrm{m})^2}{4000 \times 10^{-10} \, \mathrm{m}} $

Calculate:

$ f_z = \frac{4 \times 10^{-6} \, \mathrm{m}^2}{4 \times 10^{-7} \, \mathrm{m}} = 10 \, \mathrm{m} $

Therefore, the Fresnel distance is approximately 10 meters.

Given:

Central maxima in air: $ Y = 2 \text{ cm} $

10th maxima in air: $ Y = 5 \text{ cm} $

Refractive index of liquid (µ): $ 1.5 = \frac{3}{2} $

When the apparatus is immersed in a liquid, the wavelength changes due to the refractive index. The relationship between wavelengths and refractive index is:

$ \mu = \frac{c}{v} = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in medium}} = \frac{\lambda_{\text{vacuum}}}{\lambda_{\text{medium}}^{\prime}} $

From this, the new wavelength in the medium is:

$ \lambda_{\text{medium}}^{\prime} = \frac{2}{3} \lambda_{\text{vacuum}} $

Considering path difference along the X-axis:

The path difference for maxima is given by:

$ \Delta x = 0, \lambda, 2\lambda, \ldots, 10\lambda $

For maxima along the Y-axis, the relationship is:

$ \Delta Y = \frac{\Delta x D}{d} = \frac{n \lambda D}{d} \Rightarrow \Delta Y \propto \lambda $

Central maxima remains unchanged because it corresponds to $ n = 0 $:

$ Y' = 2 \text{ cm} $

For the 10th maxima:

We know:

$ \frac{\Delta Y}{\Delta Y^{\prime}} = \frac{\lambda}{\lambda_{\text{medium}}^{\prime}} $

By substituting the given values:

$ \frac{5 - 2}{Y' - 2} = \frac{3}{2} $

Solving for $ Y' $:

$ 3 = \frac{3}{2} (Y' - 2) $

$ Y' - 2 = 2 $

$ Y' = 4 \text{ cm} $

Thus, the new coordinates when the apparatus is immersed in the liquid are $ 2 \text{ cm} $ and $ 4 \text{ cm} $.

On introduction of a transparent sheet,

$\Delta x=(\mu-1) t$

if the central maxima occupies position of the nth fringe, then $(\mu-1) t=n \lambda=d \sin \theta$

$\begin{aligned} \Rightarrow \quad \sin \theta & =\frac{(\mu-1)}{d} t \\ & =\frac{(1.25-1) \times 1.5 \times 10^{-7}}{3 \times 10^{-7}}=0.125 \end{aligned}$

Therefore, central maxima has angular position, $\theta=\sin ^{-1}(0.125)=7.18^{\circ} \approx 7.2^{\circ}$

for small angles, $\sin \theta \approx \tan \theta$

$\begin{aligned} \tan \theta & =\frac{y}{D} \\ \therefore \quad \frac{y}{D} & =\frac{(\mu-1) t}{d} \Rightarrow y=\frac{D(\mu-1)}{d} t \end{aligned}$

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

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

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

$ = {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 ....

${\lambda _{obs}} = {\lambda _{source}}\sqrt {{{1 + {v \over C}} \over {1 - {v \over C}}}} $

For $v < < C$,

${{670.7} \over {670}} = 1 + {v \over C}$

$ \Rightarrow v = {{0.7} \over {670}} \times 3 \times {10^8}$ m/s

$ \Rightarrow v \simeq 3.13 \times {10^5}$ m/s

${{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} + {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}} } \over {{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} - {I_1} - {I_2} + 2\sqrt {{I_1}{I_2}} }}$

$ = {{2({I_1} + {I_2})} \over {4\sqrt {{I_1}{I_2}} }}$

$ = {{\left( {{{{I_1}} \over {{I_2}}} + 1} \right)} \over {2\sqrt {{{{I_1}} \over {{I_2}}}} }} = {{4 + 1} \over {2 \times 2}} = {5 \over 4}$

So $x = 4$

When electric field vector is completely removed and incident on Brewster's angle then only refraction takes place.

${I_P} = I + 9I + 2\sqrt {I \times 9I} \cos {\pi \over 2} = 10I$

${I_Q} = I + 9I + 2\sqrt {I \times 9I} \cos \pi = 4I$

So, ${I_P} - {I_Q} = 6I$

${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2} = {\left( {{5 \over 1}} \right)^2}$

$ = {{25} \over 1}$

Distance between the two maxima is equal to fringe width which is given as

$ \begin{gathered} \beta=\frac{D \lambda}{d} \Rightarrow \beta \propto \frac{\lambda}{d} \\ \Rightarrow \frac{\beta_2}{\beta_1}=\frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2}=\frac{4 \lambda_1}{\lambda_1} \times \frac{d_1}{d_{1 / 2}}=4 \times 2=8 \\ \end{gathered} $

$ \Rightarrow \quad \beta_2=8 \beta_1 $

In Young's double slit experiment,

$ \begin{aligned} & n_1=12, \lambda_1=6000 \mathop {\rm{A}}\limits^{\rm{o}}=6 \times 10^{-7} \mathrm{~m} \\ & \lambda_2=4800 \mathop {\rm{A}}\limits^{\rm{o}}=4.8 \times 10^{-7} \mathrm{~m} \end{aligned} $

Then, $n_2=$ ?

In Young's double slit experiment, position of $n$th fringe,

$ \begin{aligned} & & x_n & =\frac{n D \lambda}{d} \\ \Rightarrow & & x_n & =\frac{n_1 D \lambda_1}{d}=\frac{n_2 D \lambda_2}{d} \\ \Rightarrow & & n_1 \lambda_1 & =n_2 \lambda_2 \\ \Rightarrow & & n_2 & =\frac{n_1 \lambda_1}{\lambda_2}=\frac{12 \times 6 \times 10^{-7}}{4.8 \times 10^{-7}}=15 \end{aligned} $