Young's Double Slit Experiment (YDSE) Questions in English

(D) The fringe width $\beta$ in an interference experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the slits.
When the chamber is filled with air,the wavelength of light is $\lambda_{air} = \frac{\lambda_0}{\mu_{air}}$,where $\lambda_0$ is the wavelength in vacuum and $\mu_{air} \approx 1.0003$.
When the chamber is evacuated,the refractive index becomes $\mu_{vac} = 1$.
Since $\mu_{vac} < \mu_{air}$,the wavelength of light in the vacuum becomes larger than in air $(\lambda_{vac} > \lambda_{air})$.
As $\beta \propto \lambda$,the fringe width $\beta$ increases when the chamber is evacuated.

(A) Young's double-slit experiment demonstrates the phenomenon of interference of light.
Interference is a characteristic property of waves,where two or more waves superpose to form a resultant wave of greater,lower,or the same amplitude.
Since interference patterns are observed in Young's experiment,it provides direct evidence that light behaves as a wave.
Therefore,the correct option is $A$.

(C) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$.
Given:
- Wavelength $\lambda = 5 \times 10^{-7} \, m$
- Distance between slits $d = 1 \, mm = 10^{-3} \, m$
- Distance to the screen $D = 2 \, m$
Substituting the values:
$\beta = \frac{5 \times 10^{-7} \times 2}{10^{-3}} \, m$
$\beta = 10 \times 10^{-4} \, m = 10^{-3} \, m$
Since $10^{-3} \, m = 1 \, mm$,the separation between bright lines is $1 \, mm$.

(A) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the slit separation.
From the formula,we can see that $\beta \propto \frac{1}{d}$.
If the slit separation $d$ is made $3$ fold (i.e.,$d' = 3d$),the new fringe width $\beta'$ will be $\beta' = \frac{\lambda D}{3d} = \frac{1}{3} \beta$.
Therefore,the fringe width becomes $1/3$ times the original width.

(C) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.
Since the experimental setup ($D$ and $d$) remains unchanged,the fringe width is directly proportional to the wavelength of light: $\beta \propto \lambda$.
Therefore,$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.
Given: $\beta_1 = 1.0 \ mm$,$\lambda_1 = 5000 \ \mathring{A}$,and $\lambda_2 = 6000 \ \mathring{A}$.
Substituting the values: $\beta_2 = \beta_1 \times \frac{\lambda_2}{\lambda_1} = 1.0 \times \frac{6000}{5000} = 1.2 \ mm$.
Thus,the new fringe width is $1.2 \ mm$.

(A) The fringe width $\beta$ in Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.
Given values are:
$\lambda = 6000 \ \mathring{A} = 6000 \times 10^{-10} \, m = 6 \times 10^{-7} \, m$
$D = 25 \, cm = 0.25 \, m$
$d = 1 \, mm = 10^{-3} \, m$
Substituting these values into the formula:
$\beta = \frac{6 \times 10^{-7} \times 0.25}{10^{-3}}$
$\beta = 6 \times 10^{-4} \times 0.25 = 1.5 \times 10^{-4} \, m$
Converting to centimeters:
$\beta = 1.5 \times 10^{-4} \times 10^2 \, cm = 0.015 \, cm$.

(C) The path difference $\Delta x$ between the two waves reaching point $X$ is given by:
$\Delta x = |QX - PX| = |(n + 2)\lambda - n\lambda| = 2\lambda$.
For constructive interference (bright fringe),the condition is $\Delta x = m\lambda$,where $m = 0, 1, 2, ...$.
Here,$\Delta x = 2\lambda$,which corresponds to $m = 2$.
Since the central fringe $(m = 0)$ is at the center,$m = 1$ corresponds to the first bright fringe,and $m = 2$ corresponds to the second bright fringe.
Therefore,the second bright fringe is formed at $X$.

(A) In Young's double slit experiment,the interference pattern is formed by the superposition of light waves originating from two coherent sources (the two slits).
If one of the slits is closed,the condition for interference is no longer satisfied because there is only one source of light.
Consequently,the interference pattern disappears.
Instead,the light from the remaining open slit undergoes diffraction,resulting in a single-slit diffraction pattern on the screen.
Therefore,no interference pattern will exist.

(D) The formula for fringe width in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slit and the screen,and $d$ is the distance between the slits.
Given that the new distance between the slits is $d' = \frac{d}{2}$ and the new distance between the slit and the screen is $D' = 2D$.
The new fringe width $\beta'$ will be $\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{(d/2)} = 4 \left( \frac{\lambda D}{d} \right) = 4\beta$.
Therefore,the fringe width will become four times the original value.

(C) Suppose the slit widths are equal,so they produce waves of equal intensity,say $I'$.
The resultant intensity at any point is given by $I_R = 4I' \cos^2 \phi$,where $\phi$ is the phase difference between the waves at the point of observation.
For maximum intensity,$\phi = 0^\circ$,so $I_{\max} = 4I' = I$ ... $(i)$
If one of the slits is closed,the resultant intensity at the same point will be $I'$ only,i.e.,$I' = I_o$ ... $(ii)$
Comparing equation $(i)$ and $(ii)$,we get $I = 4I_o$.

(D) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = 9$.
We know that $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = 9$.
Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = 3$.
Solving for the ratio of amplitudes: $a_1 + a_2 = 3a_1 - 3a_2 \Rightarrow 4a_2 = 2a_1 \Rightarrow \frac{a_1}{a_2} = 2$.
Since intensity $I \propto a^2$,the ratio of intensities of individual sources is $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2 = 2^2 = 4$,which means $I_1 : I_2 = 4 : 1$.
Thus,both statements $(b)$ and $(c)$ are correct.

(C) The distance of the $n^{th}$ bright fringe from the central maximum is given by the formula: $y_n = \frac{n \lambda D}{d}$.
Since $n$,$D$,and $d$ are constant for both cases,we have $y_n \propto \lambda$.
Therefore,the ratio of the distances for the $4^{th}$ fringe is: $\frac{x(\text{Blue})}{x(\text{Green})} = \frac{\lambda(\text{Blue})}{\lambda(\text{Green})} = \frac{4360}{5460}$.
Since $4360 < 5460$,it follows that $x(\text{Blue}) < x(\text{Green})$.

(C) The formula for fringe width $\beta$ in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.
Given values are:
Wavelength $\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$.
Distance of the screen $D = 1.0 \, m$.
Slit separation $d = 0.1 \, mm = 0.1 \times 10^{-3} \, m = 10^{-4} \, m$.
Substituting these values into the formula:
$\beta = \frac{5 \times 10^{-7} \times 1}{10^{-4}} \, m = 5 \times 10^{-3} \, m$.
To convert meters to centimeters, multiply by $100$:
$\beta = 5 \times 10^{-3} \times 10^2 \, cm = 0.5 \, cm$.

(A) The fringe width $\beta$ in Young's double-slit experiment is given by the formula $\beta = \frac{D\lambda}{d}$.
In this problem,the distance between the slits and the screen is denoted by $L$ (instead of $D$),and the fringe width is denoted by $x$ (instead of $\beta$).
Substituting these variables into the formula,we get $x = \frac{L\lambda}{d}$.
Rearranging the equation to solve for the wavelength $\lambda$,we get $\lambda = \frac{xd}{L}$.

(A) In a Young's double slit experiment,the path difference between the two waves reaching the central point on the screen is zero.
Since the path difference $\Delta x = 0$,the phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = 0$.
For constructive interference,the condition is $\Delta x = n\lambda$ where $n = 0, 1, 2, ...$.
Since $n = 0$ satisfies this condition,the central point corresponds to the central maximum,which is always bright.

(A) The fringe width in a Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.
Here,$\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When the apparatus is immersed in a medium of refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Since $D$ and $d$ remain unchanged,the new fringe width $\beta'$ is given by $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Given $\beta = 0.4 \, mm$ and $\mu = 4/3$,we have $\beta' = \frac{0.4}{4/3} = 0.4 \times \frac{3}{4} = 0.3 \, mm$.

(B) The fringe width $\beta$ in Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.
Here,$\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When the experiment is performed in a medium with refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Since the refractive index of water $(\mu \approx 1.33)$ is greater than that of air $(\mu \approx 1.0)$,the wavelength of light decreases in water.
Consequently,the fringe width $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Since $\mu > 1$,the fringe width $\beta'$ will be less than $\beta$. Therefore,the fringe width will decrease.

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
Since $\beta \propto \lambda$,the fringe width is directly proportional to the wavelength of the light used.
Among the given colours,violet light has the minimum wavelength $(\lambda_v \approx 400 \ nm)$ and red light has the maximum wavelength $(\lambda_r \approx 700 \ nm)$.
Therefore,the fringe width will be minimum for violet light.

(C) The fringe width in air is given by $\beta_{air} = \frac{\lambda D}{d} = 0.6 \, mm$.
When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Consequently, the new fringe width $\beta_{medium}$ is given by $\beta_{medium} = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta_{air}}{\mu}$.
Given $\beta_{air} = 0.6 \, mm$ and $\mu = 1.5$, we have:
$\beta_{medium} = \frac{0.6}{1.5} = 0.4 \, mm$.

(C) The intensity of light $I$ is directly proportional to the width of the slit $w$,so $I_1/I_2 = w_1/w_2 = 4/9$.
Since intensity $I \propto a^2$,where $a$ is the amplitude,we have $a_1/a_2 = \sqrt{I_1/I_2} = \sqrt{4/9} = 2/3$.
Let $a_1 = 2k$ and $a_2 = 3k$.
The ratio of maximum intensity to minimum intensity is given by the formula $\frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2}$.
Substituting the values,$\frac{I_{\max}}{I_{\min}} = \frac{(2k + 3k)^2}{(2k - 3k)^2} = \frac{(5k)^2}{(-k)^2} = \frac{25k^2}{k^2} = 25/1$.
Thus,the ratio is $25:1$.

(D) The fringe width $\beta$ in Young's double slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$.
Here,$\lambda = 6000 \ \mathring{A} = 6000 \times 10^{-8} \ cm = 6 \times 10^{-5} \ cm$.
The distance of the screen from the slits is $D = 40 \ cm$.
The fringe width is $\beta = 0.012 \ cm$.
Rearranging the formula to solve for the distance between the slits $d$:
$d = \frac{\lambda D}{\beta} = \frac{6 \times 10^{-5} \ cm \times 40 \ cm}{0.012 \ cm}$.
$d = \frac{240 \times 10^{-5}}{0.012} \ cm = \frac{2.4 \times 10^{-3}}{1.2 \times 10^{-2}} \ cm = 2 \times 10^{-1} \ cm = 0.2 \ cm$.
Therefore,the distance between the slits is $0.2 \ cm$.

(A) The fringe width $\beta$ in Young's double slit experiment is given by $\beta = \frac{D\lambda}{d}$,where $D$ is the distance between the slits and the screen,$\lambda$ is the wavelength of light,and $d$ is the slit separation.
Given that the fringe widths are equal,$\beta_1 = \beta_2$,so $\frac{\beta_1}{\beta_2} = 1$.
We are given the ratios $\frac{\lambda_1}{\lambda_2} = \frac{1}{2}$ and $\frac{d_1}{d_2} = \frac{2}{1}$.
Using the formula $\frac{\beta_1}{\beta_2} = \left(\frac{D_1}{D_2}\right) \left(\frac{\lambda_1}{\lambda_2}\right) \left(\frac{d_2}{d_1}\right)$,we substitute the known values:
$1 = \left(\frac{D_1}{D_2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right)$.
$1 = \left(\frac{D_1}{D_2}\right) \times \frac{1}{4}$.
Therefore,$\frac{D_1}{D_2} = \frac{4}{1}$.

(B) In an interference experiment,the distance between two consecutive bright fringes (maxima) or two consecutive dark fringes (minima) is known as the fringe width,denoted by $\beta$.
The formula for fringe width is derived from the path difference condition for constructive interference,which is $\Delta x = n\lambda = \frac{yd}{D}$.
For consecutive maxima,the separation is given by $\beta = \frac{D\lambda}{d}$,where $D$ is the distance between the screen and the slits,$\lambda$ is the wavelength of light used,and $d$ is the distance between the two slits.
Therefore,the correct option is $B$.

(C) The fringe width $(\beta)$ in Young's double slit experiment is given by the formula $\beta = \frac{D\lambda}{d}$.
From this relation,it is clear that the fringe width is directly proportional to the wavelength of the light used,i.e.,$\beta \propto \lambda$.
Since the wavelength of red light $(\lambda_{red})$ is greater than the wavelength of yellow light $(\lambda_{yellow})$,the fringe width will increase when yellow light is replaced by red light.

(D) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$.
Given initial values: $\beta_1 = 1 \times 10^{-4} \ m$,$\lambda_1 = 6.4 \times 10^{-7} \ m$.
New parameters: $D_2 = 2D_1$,$d_2 = \frac{d_1}{2}$,$\lambda_2 = 4.0 \times 10^{-7} \ m$.
Using the ratio: $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \times \frac{D_2}{D_1} \times \frac{d_1}{d_2}$.
Substituting the values: $\frac{\beta_2}{1 \times 10^{-4}} = \left( \frac{4.0 \times 10^{-7}}{6.4 \times 10^{-7}} \right) \times \left( \frac{2D_1}{D_1} \right) \times \left( \frac{d_1}{d_1/2} \right)$.
$\frac{\beta_2}{1 \times 10^{-4}} = \left( \frac{4.0}{6.4} \right) \times 2 \times 2 = \frac{16}{6.4} = 2.5$.
Therefore,$\beta_2 = 2.5 \times 10^{-4} \ m$.

(D) For an interference pattern to be observed,the two light waves must be coherent and have the same frequency and wavelength.
Since a blue filter and a yellow filter transmit light of different wavelengths,the waves originating from the two slits will have different frequencies and wavelengths.
Therefore,the condition for sustained interference is not met,and no stable interference pattern will be formed on the screen.

(D) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the screen and the sources,and $d$ is the distance between the two sources.
From this formula,we can see that the fringe width is directly proportional to the distance $D$,i.e.,$\beta \propto D$.
Given that the initial fringe width is $\beta_1 = 2w$ at distance $D_1 = D$.
If the distance is doubled,$D_2 = 2D$.
Then the new fringe width $\beta_2$ will be: $\beta_2 = \beta_1 \times \frac{D_2}{D_1} = 2w \times \frac{2D}{D} = 4w$.
Therefore,the fringe width becomes $4w$.

(A) The angular width of the fringes in a double-slit experiment is given by the formula $\theta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength of light and $d$ is the distance between the slits.
Since $\theta \propto \lambda$,to increase the angular width $\theta$ by $10\%$,the wavelength $\lambda$ must also be increased by $10\%$.
The required change in wavelength is $\Delta\lambda = 10\% \text{ of } \lambda = \frac{10}{100} \times 5890 \ \mathring{A} = 589 \ \mathring{A}$.
Therefore,the wavelength must be increased by $589 \ \mathring{A}$.

(B) In a Young's double-slit experiment, if the intensity of light from each slit is $I$, the resultant intensity at the central fringe is given by $I_0 = I + I + 2\sqrt{I \cdot I} \cos(0) = 4I$.

When one of the slits is closed, only one source of light remains.

The intensity at the same point due to a single slit is simply $I$.

Since $I_0 = 4I$, we have $I = I_0 / 4$.

Therefore, the intensity at the central point becomes $I_0 / 4$.

(B) The fringe width $\beta$ in a Young's double slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$.
For the fringe width to remain the same,the ratio $\frac{\lambda D}{d}$ must remain constant.
If both $d$ and $D$ are doubled,the new fringe width $\beta'$ becomes: $\beta' = \frac{\lambda (2D)}{(2d)} = \frac{\lambda D}{d} = \beta$.
Therefore,the fringe width remains unchanged when both $d$ and $D$ are doubled.

(B) The distance of the $n^{th}$ bright fringe from the central maxima is given by the formula: $x_n = \frac{n \lambda D}{d}$.
Here,$n = 3$,$\lambda = 6000\ \mathring A = 6000 \times 10^{-10}\ m$,$D = 1.0\ m$,and $d = 0.5\ mm = 0.5 \times 10^{-3}\ m$.
Substituting these values into the formula:
$x_3 = \frac{3 \times 6000 \times 10^{-10} \times 1.0}{0.5 \times 10^{-3}}$
$x_3 = \frac{18000 \times 10^{-10}}{0.5 \times 10^{-3}}$
$x_3 = 36000 \times 10^{-7}\ m = 3.6 \times 10^{-3}\ m = 3.6\ mm$.
Therefore,the separation is $3.6\ mm$.

(D) The total width of the visible region $W$ remains constant for a given experimental setup.
For $n$ fringes of wavelength $\lambda$,the width is given by $W = n \times \beta$,where $\beta = \frac{D\lambda}{d}$ is the fringe width.
Thus,$W = n \times \frac{D\lambda}{d}$.
Since $W$,$D$,and $d$ are constant,we have $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 62$,$\lambda_1 = 5893 \ \mathring{A}$,and $\lambda_2 = 4358 \ \mathring{A}$.
Substituting the values: $62 \times 5893 = n_2 \times 4358$.
$n_2 = \frac{62 \times 5893}{4358} \approx 83.83$.
Rounding to the nearest whole number,we get $n_2 = 84$.

(B) The angular fringe width is given by $\theta = \frac{\lambda}{d}$,which implies $\theta \propto \lambda$.
When the system is dipped in water,the wavelength of light changes to $\lambda_w = \frac{\lambda_a}{\mu_w}$,where $\mu_w$ is the refractive index of water.
Since $\theta \propto \lambda$,the new angular width $\theta_w$ is given by $\theta_w = \frac{\theta_a}{\mu_w}$.
Given $\theta_a = 0.20^o$ and $\mu_w = \frac{4}{3}$,we have $\theta_w = \frac{0.20}{4/3} = 0.20 \times \frac{3}{4} = 0.15^o$.

(A) The formula for the position of the $n^{th}$ bright fringe is given by $x_n = \frac{n \lambda D}{d}$.
Given values are:
Slit separation $d = 1 \,mm = 1 \times 10^{-3} \,m$.
Distance to screen $D = 1 \,m$.
Position of $10^{th}$ fringe $x_{10} = 5 \,mm = 5 \times 10^{-3} \,m$.
Order of fringe $n = 10$.
Substituting these values into the formula:
$5 \times 10^{-3} = \frac{10 \times \lambda \times 1}{1 \times 10^{-3}}$
$\lambda = \frac{5 \times 10^{-3} \times 10^{-3}}{10} = 5 \times 10^{-7} \,m$.
Converting to $\mathring A$:
$\lambda = 5 \times 10^{-7} \times 10^{10} \, \mathring A = 5000 \, \mathring A$.

(B) The formula for the position of the $n^{th}$ bright fringe (maxima) in Young's double slit experiment is given by $x_n = \frac{n \lambda D}{d}$.
Given values are:
$\lambda = 5000\;\mathring{A} = 5000 \times 10^{-10}\;m = 5 \times 10^{-7}\;m$
$d = 0.2\;mm = 0.2 \times 10^{-3}\;m = 2 \times 10^{-4}\;m$
$D = 200\;cm = 2\;m$
For the third maximum, $n = 3$.
Substituting these values into the formula:
$x_3 = \frac{3 \times (5 \times 10^{-7}\;m) \times (2\;m)}{2 \times 10^{-4}\;m}$
$x_3 = \frac{30 \times 10^{-7}}{2 \times 10^{-4}}\;m = 15 \times 10^{-3}\;m = 1.5 \times 10^{-2}\;m = 1.5\;cm$.
Therefore, the third maximum is at $x = 1.5\;cm$.

(D) The distance of the $n^{th}$ dark fringe from the central fringe is given by the formula:
$x_n = \frac{(2n - 1) \lambda D}{2d}$
For the second dark fringe $(n = 2)$:
$x_2 = \frac{(2 \times 2 - 1) \lambda D}{2d} = \frac{3 \lambda D}{2d}$
Given: $x_2 = 1 \, mm = 1 \times 10^{-3} \, m$,$D = 1 \, m$,and $d = 0.90 \, mm = 0.9 \times 10^{-3} \, m$.
Substituting the values:
$1 \times 10^{-3} = \frac{3 \times \lambda \times 1}{2 \times 0.9 \times 10^{-3}}$
$1 \times 10^{-3} = \frac{3 \lambda}{1.8 \times 10^{-3}}$
$3 \lambda = 1.8 \times 10^{-6}$
$\lambda = 0.6 \times 10^{-6} \, m = 6 \times 10^{-7} \, m = 6 \times 10^{-5} \, cm$.

(D) The formula for fringe width $(\beta)$ in Young's double slit experiment is given by: $\beta = \frac{\lambda D}{d}$.
Here, $\beta = 4 \, mm = 4 \times 10^{-3} \, m$, $\lambda = 4 \times 10^{-7} \, m$, and $d = 0.1 \, mm = 0.1 \times 10^{-3} \, m$.
Substituting these values into the formula:
$4 \times 10^{-3} = \frac{(4 \times 10^{-7}) \times D}{0.1 \times 10^{-3}}$
$D = \frac{(4 \times 10^{-3}) \times (0.1 \times 10^{-3})}{4 \times 10^{-7}}$
$D = \frac{0.4 \times 10^{-6}}{4 \times 10^{-7}} = 0.1 \times 10^1 = 1 \, m$.
Therefore, the distance between the screen and the slit is $1 \, m$.

(A) The formula for fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.
Here,the fringe width $\beta = 0.06 \ cm = 0.06 \times 10^{-2} \ m = 6 \times 10^{-4} \ m$.
The distance between the sources $d = 1 \ mm = 10^{-3} \ m$.
The distance between the screen and the source $D = 1 \ m$.
Substituting these values into the formula:
$6 \times 10^{-4} = \frac{\lambda \times 1}{10^{-3}}$
$\lambda = 6 \times 10^{-4} \times 10^{-3} \ m$
$\lambda = 6 \times 10^{-7} \ m$
To convert this into $\mathring{A}$,we multiply by $10^{10}$:
$\lambda = 6 \times 10^{-7} \times 10^{10} \ \mathring{A} = 6000 \ \mathring{A}$.

(B) When a mica sheet of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the interfering beams,an additional optical path difference is created.
The optical path difference introduced is $\Delta x = (\mu - 1)t$.
Let the fringe pattern shift by a distance $y$ on the screen. The path difference at a point $y$ on the screen is given by $\Delta x = \frac{yd}{D}$,where $d$ is the distance between the slits and $D$ is the distance between the slits and the screen.
Equating the two expressions for path difference:
$(\mu - 1)t = \frac{yd}{D}$
Solving for $y$:
$y = \frac{D}{d}(\mu - 1)t$
Therefore,the fringe pattern is displaced by a distance of $\frac{D}{d}(\mu - 1)t$. Thus,option $B$ is correct.

(D) The total width of the interference pattern $(W)$ remains constant for a given experimental setup.
The width of the pattern is given by $W = n_1 \beta_1 = n_2 \beta_2$, where $\beta = \frac{D\lambda}{d}$.
Thus, $n_1 \lambda_1 = n_2 \lambda_2$.
Given: $n_1 = 92$, $\lambda_1 = 5898 \ \text{\AA}$, and $\lambda_2 = 5461 \ \text{\AA}$.
Substituting the values: $92 \times 5898 = n_2 \times 5461$.
$n_2 = \frac{92 \times 5898}{5461} \approx 99.37$.
Since the number of fringes must be an integer, we consider $n_2 = 99$.

(D) In Young's double-slit experiment,interference fringes are formed due to the superposition of light waves from two coherent sources.
For clear interference fringes,the light source must be monochromatic (having a single wavelength).
$A$ torch emits white light,which consists of a continuous spectrum of many different wavelengths.
Each wavelength produces its own interference pattern with a different fringe width.
These patterns overlap in such a way that the bright and dark fringes of different colors wash each other out,resulting in a general illumination on the screen.
Therefore,no distinct interference fringes will be observed.

(B) In Young's double-slit experiment,the interference pattern is formed due to the superposition of coherent light waves originating from two slits.
When a thin metal plate is placed in the path of one of the interfering beams,it blocks that beam completely.
Since interference requires the superposition of two coherent waves,blocking one beam prevents the overlapping of waves.
Without the overlapping of two waves,no interference pattern can be formed,and therefore,the fringes disappear.

(B) The position of the $n^{th}$ bright fringe is given by $x_n = \frac{n D \lambda}{d}$.
Given: $n = 3$,$D = 1.4 \,m$,$d = 0.28 \,mm = 0.28 \times 10^{-3} \,m$,and $x_3 = 0.9 \,cm = 0.9 \times 10^{-2} \,m$.
Substituting these values into the formula:
$0.9 \times 10^{-2} = \frac{3 \times 1.4 \times \lambda}{0.28 \times 10^{-3}}$
$\lambda = \frac{0.9 \times 10^{-2} \times 0.28 \times 10^{-3}}{3 \times 1.4}$
$\lambda = \frac{0.252 \times 10^{-5}}{4.2} = 0.06 \times 10^{-5} \,m = 6 \times 10^{-7} \,m$
Converting to $\mathring{A}$: $\lambda = 6 \times 10^{-7} \times 10^{10} \, \mathring{A} = 6000 \, \mathring{A}$.

(A) The distance between two consecutive dark fringes is known as the fringe width, denoted by $\beta$.
The formula for fringe width is $\beta = \frac{\lambda D}{d}$.
Given:
$\lambda = 6000 \,\mathring{A} = 6000 \times 10^{-10} \,m = 6 \times 10^{-7} \,m$
$D = 1 \,m$
$d = 0.6 \,mm = 0.6 \times 10^{-3} \,m$
Substituting the values:
$\beta = \frac{6 \times 10^{-7} \times 1}{0.6 \times 10^{-3}}$
$\beta = \frac{6}{0.6} \times 10^{-7+3}$
$\beta = 10 \times 10^{-4} \,m = 10^{-3} \,m$
$\beta = 1 \,mm$.

(D) For constructive interference (maxima),the path difference $\Delta x$ must be an integer multiple of the wavelength $\lambda$,i.e.,$\Delta x = n\lambda$,where $n = 0, 1, 2, \dots$.
For the first maxima $(n=1)$,the path difference is $\Delta x = 1 \times 6320 \ \mathring{A} = 6320 \ \mathring{A}$.
The relationship between path difference $\Delta x$ and phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For the first maxima,substituting $\Delta x = \lambda$,we get $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi \ \text{radian}$.
Since both conditions are satisfied,the correct option is $(d)$.

(D) When a thin glass plate of thickness $t$ and refractive index $\mu$ is placed in the path of one of the beams,an additional path difference is introduced.
This path difference is given by $\Delta x = (\mu - 1)t$.
Due to this additional path difference,the entire interference pattern shifts by a distance $y = \frac{D}{d}(\mu - 1)t$ towards the side where the glass plate is placed.
However,the fringe width $\beta = \frac{\lambda D}{d}$ depends only on the wavelength $\lambda$,the distance between the slits $d$,and the distance to the screen $D$.
Since these parameters remain unchanged,the fringe width remains constant.
Therefore,the correct option is $(d)$.

(A) The formula for fringe width in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
From the relation $\beta \propto \lambda$,it is clear that the fringe width $\beta$ is directly proportional to the wavelength $\lambda$ of the light used.
Therefore,if the wavelength $\lambda$ increases,the fringe width $\beta$ also increases.
Thus,option $A$ is correct.

(B) The formula for fringe width $\beta$ in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.
Given:
- Wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, m = 6 \times 10^{-7} \, m$.
- Distance between slits $d = 4 \, mm = 4 \times 10^{-3} \, m$.
- Distance of screen from slits $D = 2 \, m$.
Substituting these values into the formula:
$\beta = \frac{6 \times 10^{-7} \times 2}{4 \times 10^{-3}}$
$\beta = \frac{12 \times 10^{-7}}{4 \times 10^{-3}}$
$\beta = 3 \times 10^{-4} \, m$
Converting meters to millimeters:
$\beta = 3 \times 10^{-4} \times 10^3 \, mm = 0.3 \, mm$.
Therefore,the correct option is $B$.

(C) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
From the relation $\beta \propto \lambda$,it is clear that the fringe width is directly proportional to the wavelength of the light used.
Among the given colours,the wavelength of red light is the longest and the wavelength of blue light is the shortest.
Since blue light has the shortest wavelength,the fringe width will be the least for blue light.

(C) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the separation between the two slits.
Given that the fringe width $\beta$ remains constant,we have $\frac{\lambda D}{d} = \text{constant}$.
If the slit separation $d$ is doubled $(d' = 2d)$,let the new distance be $D'$.
Then,$\frac{\lambda D}{d} = \frac{\lambda D'}{2d}$.
Solving for $D'$,we get $D' = 2D$.
Therefore,the distance of the screen from the slits should be doubled.