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

(B) The position of the $n^{th}$ maxima in a Young's double-slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
For the first maxima,$n = 1$,so $y_1 = \frac{\lambda D}{d}$.
To find the speed at which the first maxima moves,we differentiate the position with respect to time $t$:
$v = \frac{dy_1}{dt} = \frac{\lambda}{d} \frac{dD}{dt}$.
Given values are $\lambda = 500 \, nm = 500 \times 10^{-9} \, m$,$d = 0.4 \, mm = 0.4 \times 10^{-3} \, m$,and $\frac{dD}{dt} = 4 \, m/s$.
Substituting these values into the equation:
$v = \frac{500 \times 10^{-9} \, m}{0.4 \times 10^{-3} \, m} \times 4 \, m/s$.
$v = \frac{500 \times 10^{-6}}{0.4} \times 4 = 500 \times 10^{-6} \times 10 = 5000 \times 10^{-6} \, m/s = 5 \times 10^{-3} \, m/s$.
Converting to $mm/s$:
$v = 5 \, mm/s$.

(A) The distance of the $n^{\text{th}}$ dark fringe from the center in a Young's Double Slit Experiment is given by $y_n = (2n - 1) \frac{\lambda D}{2d}$.
For the $5^{\text{th}}$ dark fringe,$n = 5$,so $y_5 = (2(5) - 1) \frac{\lambda D}{2d} = \frac{9 \lambda D}{2d}$.
Given $y_5 = 4 \, mm = 4 \times 10^{-3} \, m$,$D = 2 \, m$,and $\lambda = 600 \, nm = 600 \times 10^{-9} \, m = 6 \times 10^{-7} \, m$.
Substituting these values into the formula:
$4 \times 10^{-3} = \frac{9 \times (6 \times 10^{-7}) \times 2}{2d}$.
$4 \times 10^{-3} = \frac{9 \times 6 \times 10^{-7}}{d}$.
$d = \frac{54 \times 10^{-7}}{4 \times 10^{-3}} = 13.5 \times 10^{-4} \, m = 1.35 \times 10^{-3} \, m$.
Therefore,$d = 1.35 \, mm$.

(D) The intensity at any point in $YDSE$ is given by $I = I_{0} \cos^{2} \left( \frac{\phi}{2} \right)$,where $I_{0}$ is the maximum intensity and $\phi$ is the phase difference.
Given that the intensity $I = \frac{I_{0}}{2}$,we have:
$\frac{I_{0}}{2} = I_{0} \cos^{2} \left( \frac{\phi}{2} \right)$
$\cos^{2} \left( \frac{\phi}{2} \right) = \frac{1}{2}$
$\cos \left( \frac{\phi}{2} \right) = \frac{1}{\sqrt{2}}$
$\frac{\phi}{2} = \frac{\pi}{4} \implies \phi = \frac{\pi}{2}$.
The path difference $\Delta x$ is related to the phase difference $\phi$ by $\Delta x = \frac{\lambda}{2\pi} \phi$.
Substituting $\phi = \frac{\pi}{2}$,we get $\Delta x = \frac{\lambda}{2\pi} \times \frac{\pi}{2} = \frac{\lambda}{4}$.
The distance $y$ from the central maxima is given by $y = \frac{\Delta x D}{a}$.
Substituting the values $\Delta x = \frac{\lambda}{4}$,$D = 2 \, m$,$a = 2 \times 10^{-3} \, m$,and $\lambda = 500 \times 10^{-9} \, m$:
$y = \frac{\lambda D}{4a} = \frac{500 \times 10^{-9} \times 2}{4 \times 2 \times 10^{-3}}$
$y = \frac{1000 \times 10^{-9}}{8 \times 10^{-3}} = 125 \times 10^{-6} \, m = 125 \, \mu m$.

(B) The formula for fringe width in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.
Here,$\beta = 6 \ mm = 6 \times 10^{-3} \ m$,$d = 1 \ mm = 1 \times 10^{-3} \ m$,and $D = 10 \ m$.
Rearranging the formula to solve for wavelength $\lambda$:
$\lambda = \frac{\beta d}{D} = \frac{(6 \times 10^{-3} \ m) \times (1 \times 10^{-3} \ m)}{10 \ m}$.
$\lambda = \frac{6 \times 10^{-6}}{10} \ m = 6 \times 10^{-7} \ m$.
To convert this to nanometers $(nm)$,we multiply by $10^9$:
$\lambda = 6 \times 10^{-7} \times 10^9 \ nm = 600 \ nm$.
Thus,the value of $x$ is $600$.

(B) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.
Given: $\lambda = 5890 \, \text{Å} = 5890 \times 10^{-10} \, m$, $D = 0.5 \, m$, and $d = 0.5 \, mm = 0.5 \times 10^{-3} \, m$.
Substituting the values: $\beta = \frac{5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}} = 5890 \times 10^{-7} \, m = 589 \times 10^{-6} \, m$.
The distance between the $n^{th}$ and $m^{th}$ bright fringe is $(n-m) \beta$.
For the first and third bright fringe, the distance is $(3-1) \beta = 2 \beta$.
Distance $= 2 \times 589 \times 10^{-6} \, m = 1178 \times 10^{-6} \, m$.

(C) Given that the amplitude $A$ is proportional to the slit width $w$,we have $A \propto w$.
Let the widths be $w_1$ and $w_2$,where $w_2 = 3w_1$. Therefore,the amplitudes are $A_1$ and $A_2 = 3A_1$.
The intensity $I$ is proportional to the square of the amplitude,$I \propto A^2$.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + A_2}{|A_1 - A_2|}\right)^2$
Substituting $A_2 = 3A_1$ into the equation:
$\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + 3A_1}{|A_1 - 3A_1|}\right)^2$
$= \left(\frac{4A_1}{2A_1}\right)^2$
$= (2)^2 = 4$
Thus,the ratio is $4: 1$.

(A) The formula for fringe width (fringe separation) in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.
Given values are:
Wavelength $\lambda = 500 \, nm = 500 \times 10^{-9} \, m$.
Distance between slits $d = 2 \, mm = 2 \times 10^{-3} \, m$.
Distance to the screen $D = 1 \, m$.
Substituting these values into the formula:
$\beta = \frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}$
$\beta = 250 \times 10^{-6} \, m$
$\beta = 0.25 \times 10^{-3} \, m$
Since $10^{-3} \, m = 1 \, mm$,we get $\beta = 0.25 \, mm$.

(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 screen and the slits,and $d$ is the distance between the slits.
We know that the wavelength of red light $\lambda_{R}$ is greater than the wavelength of violet light $\lambda_{V}$ (i.e.,$\lambda_{R} > \lambda_{V}$).
Since $\beta$ is directly proportional to $\lambda$ $(\beta \propto \lambda)$,it follows that $\beta_{R} > \beta_{V}$.
When the light source is changed from red to violet,the fringe width decreases,which means the consecutive fringe lines will come closer together.

(C) The position of the $n^{th}$ bright fringe is given by $y = n \frac{D \lambda}{d}$.
For the first bright fringe $(n = 1)$,the position is $y = \frac{D \lambda}{d}$.
Given $D = 1.5 \, m$,$d = 0.3 \, mm = 0.3 \times 10^{-3} \, m$.
For red light,$y_r = 3.5 \, mm = 3.5 \times 10^{-3} \, m$. Thus,$\lambda_r = \frac{y_r d}{D} = \frac{3.5 \times 10^{-3} \times 0.3 \times 10^{-3}}{1.5} = 0.7 \times 10^{-6} \, m = 700 \, nm$.
For violet light,$y_v = 2.0 \, mm = 2.0 \times 10^{-3} \, m$. Thus,$\lambda_v = \frac{y_v d}{D} = \frac{2.0 \times 10^{-3} \times 0.3 \times 10^{-3}}{1.5} = 0.4 \times 10^{-6} \, m = 400 \, nm$.
The difference in wavelengths is $\Delta \lambda = \lambda_r - \lambda_v = 700 \, nm - 400 \, nm = 300 \, nm$.

(D) The distance of the $n^{th}$ bright fringe from the central fringe is given by $y_n = \frac{n \lambda D}{d}$.
The distance between the $4^{th}$ bright fringe on both sides is $2 y_4 = 2 \times \frac{4 \lambda D}{d} = \frac{8 \lambda D}{d}$.
Given: $2 y_4 = 2.4 \, cm = 2.4 \times 10^{-2} \, m$, $D = 1.5 \, m$, $d = 0.3 \, mm = 0.3 \times 10^{-3} \, m$.
Using $\lambda = \frac{c}{f}$, we have $\frac{8 \times c \times D}{f \times d} = 2.4 \times 10^{-2}$.
Substituting the values: $\frac{8 \times (3 \times 10^8) \times 1.5}{f \times (0.3 \times 10^{-3})} = 2.4 \times 10^{-2}$.
$\frac{36 \times 10^8}{f \times 0.3 \times 10^{-3}} = 2.4 \times 10^{-2}$.
$f = \frac{36 \times 10^{11}}{0.3 \times 2.4 \times 10^{-2}} = \frac{36 \times 10^{13}}{0.72} = 50 \times 10^{13} = 5 \times 10^{14} \, Hz$.

(A) Given that the amplitude $A$ is proportional to the slit width $w$,so $A \propto w$.
Since intensity $I$ is proportional to the square of the amplitude,$I \propto A^2 \propto w^2$.
Let the widths be $w_1 = 3w_0$ and $w_2 = w_0$. Then the intensities are $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \left(\frac{w_1}{w_2}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$.
Thus,$I_1 = 9I_2$. Let $I_2 = I$,then $I_1 = 9I$.
The ratio of minimum to maximum intensity is given by $\frac{I_{\min}}{I_{\max}} = \left(\frac{\sqrt{I_1} - \sqrt{I_2}}{\sqrt{I_1} + \sqrt{I_2}}\right)^2$.
Substituting the values: $\frac{I_{\min}}{I_{\max}} = \left(\frac{\sqrt{9I} - \sqrt{I}}{\sqrt{9I} + \sqrt{I}}\right)^2 = \left(\frac{3\sqrt{I} - \sqrt{I}}{3\sqrt{I} + \sqrt{I}}\right)^2 = \left(\frac{2\sqrt{I}}{4\sqrt{I}}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Given the ratio is $x:4$,we have $\frac{x}{4} = \frac{1}{4}$,which implies $x = 1$.

(B) The fringe width is given by $\beta = \frac{\lambda D}{d}$.
For the fringe width to be maximum,the slit separation $d$ must be minimum,and for the fringe width to be minimum,the slit separation $d$ must be maximum.
The slit separation is $d(t) = d_{0} + a_{0} \sin \omega t$.
The maximum value of $d$ is $d_{\max} = d_{0} + a_{0}$ and the minimum value of $d$ is $d_{\min} = d_{0} - a_{0}$.
Therefore,the minimum fringe width is $\beta_{\min} = \frac{\lambda D}{d_{0} + a_{0}}$ and the maximum fringe width is $\beta_{\max} = \frac{\lambda D}{d_{0} - a_{0}}$.
The difference between the largest and smallest fringe width is $\beta_{\max} - \beta_{\min} = \frac{\lambda D}{d_{0} - a_{0}} - \frac{\lambda D}{d_{0} + a_{0}}$.
Simplifying this expression: $\beta_{\max} - \beta_{\min} = \lambda D \left( \frac{(d_{0} + a_{0}) - (d_{0} - a_{0})}{d_{0}^{2} - a_{0}^{2}} \right) = \frac{2 \lambda D a_{0}}{d_{0}^{2} - a_{0}^{2}}$.

(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 the light used,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When the source of light changes from orange to blue,the wavelength $\lambda$ decreases because the wavelength of blue light is shorter than that of orange light.
Since $\beta \propto \lambda$,as the wavelength $\lambda$ decreases,the fringe width $\beta$ (the distance between consecutive fringes) also decreases.

(C) The width of the segment on the screen is given by $y = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.
Thus,$y = n \lambda \left(\frac{D}{d}\right)$.
Since the segment length $y$ and the experimental setup parameters $D$ and $d$ remain constant,we have $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 8$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$.
Substituting the values: $8 \times 600 = n_2 \times 400$.
$n_2 = \frac{8 \times 600}{400} = \frac{4800}{400} = 12$.
Therefore,the student will observe $12$ fringes.

(A) Given: Slit separation $d = 0.6 \; mm = 0.6 \times 10^{-3} \; m$. Screen distance $D = 80 \; cm = 0.8 \; m$.
The position of the first dark fringe from the central axis is given by $y = \frac{D \lambda}{2d}$.
According to the problem,the first dark fringe is observed directly opposite to one of the slits,which means its distance from the central axis is $y = \frac{d}{2}$.
Equating the two expressions: $\frac{D \lambda}{2d} = \frac{d}{2}$.
Solving for wavelength $\lambda$: $\lambda = \frac{d^2}{D}$.
Substituting the values: $\lambda = \frac{(0.6 \times 10^{-3})^2}{0.8} = \frac{0.36 \times 10^{-6}}{0.8} = 0.45 \times 10^{-6} \; m$.
Converting to nanometers: $\lambda = 450 \times 10^{-9} \; m = 450 \; nm$.

(B) The angular width of a fringe in a Young's double slit experiment is given by $\theta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength and $d$ is the slit separation.
When the system is immersed in a medium of refractive index $\mu$,the wavelength changes to $\lambda' = \frac{\lambda}{\mu}$.
Therefore,the new angular width $\theta'$ is given by $\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}$.
Given $\theta = 0.35^{\circ}$ and $\mu = 7/5 = 1.4$.
Thus,$\theta' = \frac{0.35^{\circ}}{1.4} = \frac{0.35}{1.4} = \frac{35}{140} = \frac{1}{4}^{\circ}$.
Comparing this with $\frac{1}{\alpha}$,we get $\alpha = 4$.

(B) The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit separation.
When the screen is moved by $\Delta D$,the change in fringe width $\Delta \beta$ is given by $\Delta \beta = \frac{\lambda}{d} \Delta D$.
Given: $\Delta D = 5 \times 10^{-2} \,m$,$\Delta \beta = 3 \times 10^{-3} \,cm = 3 \times 10^{-5} \,m$,and $d = 1 \,mm = 10^{-3} \,m$.
Rearranging the formula for $\lambda$: $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.
Substituting the values: $\lambda = \frac{3 \times 10^{-5} \times 10^{-3}}{5 \times 10^{-2}} = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} = 0.6 \times 10^{-6} \,m$.
Converting to nanometers: $\lambda = 600 \times 10^{-9} \,m = 600 \,nm$.

(D) The formula for fringe width is $\beta = \frac{D \lambda}{d}$,where $D$ is the distance between the screen and the slits,$\lambda$ is the wavelength,and $d$ is the slit separation.
Given for the first case:
$\lambda_1 = 5000 \,\mathring A$,$d_1 = d$,$\beta_1 = 0.5 \,mm$.
So,$\beta_1 = \frac{D \lambda_1}{d} = 0.5 \,mm$ ... $(I)$
For the second case:
$\lambda_2 = 6000 \,\mathring A$,$d_2 = 2d$,$\beta_2 = ?$
So,$\beta_2 = \frac{D \lambda_2}{d_2} = \frac{D \times 6000 \,\mathring A}{2d}$ ... $(II)$
Dividing $(II)$ by $(I)$:
$\frac{\beta_2}{\beta_1} = \frac{D \times 6000 \,\mathring A / 2d}{D \times 5000 \,\mathring A / d} = \frac{6000}{2 \times 5000} = \frac{6}{10} = 0.6$
Therefore,$\beta_2 = 0.6 \times \beta_1 = 0.6 \times 0.5 \,mm = 0.3 \,mm$.

(B) The fringe width 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 distance between the two slits.
When the entire arrangement is placed in a medium with refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Consequently,the new fringe width $\beta'$ becomes $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Given $\beta = 12 \ mm$ and $\mu = \frac{4}{3}$,the new fringe width is $\beta' = \frac{12}{4/3} = 12 \times \frac{3}{4} = 9 \ mm$.

(C) The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.
Since $D$ and $d$ are constant,we have $\beta \propto \lambda$.
Therefore,$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.
Given $\beta_1 = 7.2\,mm$,$\lambda_1 = 560\,nm$,and $\beta_2 = 8.1\,mm$.
Substituting the values: $\frac{8.1}{7.2} = \frac{\lambda_2}{560}$.
$\lambda_2 = \frac{8.1}{7.2} \times 560 = \frac{9}{8} \times 560$.
$\lambda_2 = 9 \times 70 = 630\,nm$.

(D) In Young's double slit experiment,the intensity at any point on the screen is the sum of the intensities due to individual wavelengths because they are incoherent with respect to each other.
For a single wavelength with slit intensity $I_{0}$,the maximum intensity is given by $I_{\max} = (\sqrt{I_{0}} + \sqrt{I_{0}})^2 = 4I_{0}$.
Since the source emits two wavelengths,$400 \,nm$ and $800 \,nm$,each with slit intensity $I_{0}$,the maximum intensity for each wavelength is $4I_{0}$.
At the central maximum,both wavelengths produce their respective maximum intensities simultaneously.
Therefore,the total maximum intensity is $I_{\text{total}} = 4I_{0} + 4I_{0} = 8I_{0}$.

(C) The resultant intensity $I$ when two waves with amplitudes $A_{1}$ and $A_{2}$ interfere with a phase difference $\phi$ is given by $I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} \cos \phi$.
Since intensity $I \propto A^{2}$,we can write $I = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2} \cos \phi$.
Given $A_{1} = A$ and $A_{2} = 2A$,the resultant intensity is:
$I = A^{2} + (2A)^{2} + 2(A)(2A) \cos \phi = A^{2}(1 + 4 + 4 \cos \phi) = A^{2}(5 + 4 \cos \phi) \quad \dots (i)$.
The maximum intensity $I_{0}$ occurs when $\cos \phi = 1$:
$I_{0} = A^{2}(5 + 4(1)) = 9A^{2} \implies A^{2} = \frac{I_{0}}{9} \quad \dots (ii)$.
Substituting $(ii)$ into $(i)$,we get:
$I = \frac{I_{0}}{9}(5 + 4 \cos \phi)$.

(B) When electrons are accelerated through a potential of $V$ volts,the momentum $p$ gained by the electrons is $p = \sqrt{2meV}$,where $m$ is the mass of the electron,$e$ is the charge,and $V$ is the potential.
The de Broglie wavelength $\lambda$ associated with these electrons is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.
The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $D$ is the distance to the screen and $d$ is the slit separation.
Substituting $\lambda$,we get $\beta = \frac{hD}{d\sqrt{2meV}}$. Thus,$\beta \propto \frac{1}{\sqrt{V}}$.
If the potential is doubled $(V_f = 2V_i)$,the new fringe width $\beta_f$ is $\beta_f = \frac{\beta_i}{\sqrt{2}} = \frac{\omega}{\sqrt{2}} \approx 0.707\omega$.
Therefore,the width is close to $0.7\omega$.

(B) Accelerated electrons exhibit wave nature and therefore they form an interference pattern. The fringe width $\beta$ of the pattern is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the de Broglie wavelength associated with the electrons.
The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meU}}$.
Substituting this into the fringe width formula,we get $\beta = \frac{hD}{d\sqrt{2meU}}$.
Since $h, D, d, m,$ and $e$ are constants,we have $\beta \propto \frac{1}{\sqrt{U}}$.
Let $\beta_i$ be the initial fringe width with potential $U$ and $\beta_f$ be the final fringe width with potential $4U$.
Then,$\frac{\beta_f}{\beta_i} = \sqrt{\frac{U}{4U}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$\beta_f = \frac{1}{2} \beta_i$,which means the fringe width becomes half of its initial value.

(A) In a Young's double slit experiment,the resultant intensity $I$ at any point on the screen is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $I_1 = I_2 = I_0$.
Thus,$I = 2 I_0 + 2 I_0 \cos \phi = 4 I_0 \cos^2(\phi / 2)$.
$1$. The minimum intensity occurs when $\cos^2(\phi / 2) = 0$,so $I_{\min} = 0$.
$2$. The maximum intensity occurs when $\cos^2(\phi / 2) = 1$,so $I_{\max} = 4 I_0$.
$3$. The average intensity $I_{\text{avg}}$ over the entire screen is the average of $I_{\max}$ and $I_{\min}$ for a sinusoidal interference pattern,which is $I_{\text{avg}} = \frac{I_{\max} + I_{\min}}{2} = \frac{4 I_0 + 0}{2} = 2 I_0$.
Therefore,the values are $0, 4 I_0, 2 I_0$.

(B) In Young's double slit experiment,the fringe width (fringe separation) $\beta$ is given by the formula:
$\beta = \frac{\lambda D}{d}$
Since the wavelength $\lambda$ is related to the frequency $f$ by $\lambda = \frac{c}{f}$,where $c$ is the speed of light,we can substitute this into the formula:
$\beta = \frac{cD}{fd}$
Assuming the distance between the slits $d$ and the distance to the screen $D$ remain constant,the fringe width is inversely proportional to the frequency:
$\beta \propto \frac{1}{f}$
Given that the initial fringe separation $\beta_1 = 1.0 \,mm$ at frequency $f_1 = f$,when the frequency is doubled $(f_2 = 2f)$,the new fringe separation $\beta_2$ will be:
$\beta_2 = \beta_1 \times \frac{f_1}{f_2} = 1.0 \,mm \times \frac{f}{2f} = 0.5 \,mm$
Therefore,the new separation of the bright fringes is $0.5 \,mm$.

(A) Let the intensity of each individual source be $I_0$. The maximum intensity is $I_{max} = 4I_0$.
Given that the intensity at a point is $75 \%$ of the maximum intensity,we have $I = 0.75 \times 4I_0 = 3I_0$.
Using the formula for intensity in interference,$I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference:
$3I_0 = 4I_0 \cos^2(\phi/2) \implies \cos^2(\phi/2) = 3/4 \implies \cos(\phi/2) = \sqrt{3}/2$.
Thus,$\phi/2 = \pi/6$,which gives the phase difference $\phi = \pi/3$.
The relation between phase difference $\phi$ and path difference $\Delta x$ is $\phi = (2\pi/\lambda) \Delta x$. Therefore,$\Delta x = (\lambda/2\pi) \times \phi = (\lambda/2\pi) \times (\pi/3) = \lambda/6$.
In Young's double slit experiment,the path difference at a distance $y$ from the central fringe is $\Delta x = yd/D$.
Equating the two expressions for path difference: $yd/D = \lambda/6$.
Solving for $y$: $y = (\lambda D) / (6d)$.
Substituting the given values: $\lambda = 600 \times 10^{-9} \,m$,$D = 1 \,m$,$d = 0.1 \times 10^{-3} \,m$.
$y = (600 \times 10^{-9} \times 1) / (6 \times 0.1 \times 10^{-3}) = (600 \times 10^{-9}) / (0.6 \times 10^{-3}) = 1000 \times 10^{-6} \,m = 1 \times 10^{-3} \,m = 1 \,mm$.

(A) The formula for fringe width $(\beta)$ in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of the light source,$D$ is the distance between the slits and the screen,and $d$ is the separation between the two slits.
From the relation $\beta \propto \frac{1}{d}$,it is clear that the fringe width is inversely proportional to the separation between the slits.
Therefore,if we decrease the separation of the slits $(d)$,the fringe width $(\beta)$ will increase.
Thus,the correct option is $A$.

(B) In Young's double-slit experiment,the two slits act as coherent sources of light.
Coherent sources are sources that emit light waves of the same frequency and maintain a constant phase difference.
To achieve this,both slits must be illuminated by a single monochromatic source of light.
Therefore,the correct statement is that both slits receive light from a single monochromatic source of light.
Thus,the correct option is $(b)$.

(B) The position of the $n^{\text{th}}$ bright fringe from the central maximum is given by $y_n = \frac{n \lambda D}{d}$.
For the $n^{\text{th}}$ red bright band: $y_{R,n} = \frac{n \lambda_R D}{d}$.
For the $(n+1)^{\text{th}}$ blue bright band: $y_{B,n+1} = \frac{(n+1) \lambda_B D}{d}$.
Given that these two bands coincide,we have $y_{R,n} = y_{B,n+1}$.
Therefore,$\frac{n \lambda_R D}{d} = \frac{(n+1) \lambda_B D}{d}$.
Canceling $D/d$ from both sides,we get $n \lambda_R = (n+1) \lambda_B$.
Substituting the given values: $n(7.8 \times 10^{-5}) = (n+1)(5.2 \times 10^{-5})$.
$7.8n = 5.2n + 5.2$.
$2.6n = 5.2$.
$n = 2$.

(C) For constructive interference at the hole,the path difference $\Delta x$ must be an integer multiple of the wavelength $\lambda$.
$\Delta x = d \sin \theta \approx d \frac{y}{D} = n \lambda$
Given: $y = 1.0 \, mm = 10^{-3} \, m$,$d = 0.5 \, mm = 0.5 \times 10^{-3} \, m$,$D = 50 \, cm = 0.5 \, m$.
Substituting the values:
$n \lambda = \frac{d \cdot y}{D} = \frac{(0.5 \times 10^{-3} \, m) \times (1.0 \times 10^{-3} \, m)}{0.5 \, m} = 1.0 \times 10^{-6} \, m = 1000 \, nm$.
So,$\lambda = \frac{1000}{n} \, nm$.
For $n = 1$,$\lambda = 1000 \, nm$ (not in range).
For $n = 2$,$\lambda = 500 \, nm$ (in range).
For $n = 3$,$\lambda = 333.3 \, nm$ (not in range).
Thus,the wavelength with strong intensity is $500 \, nm$.

(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Given $\Delta x = \frac{\lambda}{6}$,we have $\Delta \phi = \frac{2 \pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
The intensity $I$ at any point is given by $I = I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right)$,where $I_0$ is the maximum intensity.
Substituting the value of $\Delta \phi$,we get $I = I_0 \cos^2 \left( \frac{\pi/3}{2} \right) = I_0 \cos^2 \left( \frac{\pi}{6} \right)$.
Since $\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$,we have $I = I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = I_0 \times \frac{3}{4}$.
Therefore,the ratio $\frac{I}{I_0} = \frac{3}{4}$.

(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.
According to the problem,the slit separation $d$ is made $3$ folds,so the new slit separation $d' = 3d$.
The new fringe width $\beta'$ is given by: $\beta' = \frac{\lambda D}{d'} = \frac{\lambda D}{3d}$.
Substituting the original formula: $\beta' = \frac{1}{3} \left( \frac{\lambda D}{d} \right) = \frac{\beta}{3}$.
Therefore,the fringe width becomes $\frac{1}{3}$ times the original fringe width.

(D) The formula for fringe width $\beta$ 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 $\beta'$ is given by:
$\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{d/2} = 4 \left( \frac{\lambda D}{d} \right) = 4\beta$.
Therefore,the fringe width becomes four times the original fringe width.

(C) The position of the $n^{\text{th}}$ maximum is given by $y_n = \frac{n \lambda D}{d}$.
For the $5^{\text{th}}$ maximum $(n=5)$,$y_5 = \frac{5 \lambda D}{d}$.
The position of the $n^{\text{th}}$ minimum is given by $y'_n = \frac{(2n-1) \lambda D}{2d}$.
For the $3^{\text{rd}}$ minimum $(n=3)$,$y'_3 = \frac{(2(3)-1) \lambda D}{2d} = \frac{5 \lambda D}{2d}$.
The separation between them is $\Delta y = y_5 - y'_3 = \frac{5 \lambda D}{d} - \frac{5 \lambda D}{2d} = \frac{5 \lambda D}{2d}$.
Since the fringe width $\beta = \frac{\lambda D}{d}$,the separation is $\Delta y = 2.5 \beta$.
Therefore,the separation is $2.5$ times the fringe width.

(B) 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 the light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
The wavelength of violet light is shorter than the wavelength of blue light $(\lambda_{\text{violet}} < \lambda_{\text{blue}})$.
Since the fringe width $\beta$ is directly proportional to the wavelength $\lambda$ $(\beta \propto \lambda)$,a decrease in wavelength results in a decrease in the fringe width.
Therefore,when the source is changed from blue to violet,the width of the fringes decreases.

(D) In Young's double slit experiment,the condition for the $n^{th}$ minimum is given by the path difference $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$.
For the third minimum,we substitute $n = 3$ into the formula.
$\Delta x = (2(3) - 1) \frac{\lambda}{2} = (6 - 1) \frac{\lambda}{2} = \frac{5 \lambda}{2}$.
Therefore,the path difference for the third minimum is $\frac{5 \lambda}{2}$.

(B) The condition for the $n^{th}$ order maximum in Young's double slit experiment is given by $y = \frac{n \lambda D}{d}$.
Since the point of observation $y$ remains the same,the product $n \lambda$ must be constant for both sources.
Therefore,$n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 10$,$\lambda_1 = 7000 \; \mathring{A}$,and $\lambda_2 = 5000 \; \mathring{A}$.
Substituting the values: $10 \times 7000 = n_2 \times 5000$.
$n_2 = \frac{10 \times 7000}{5000} = \frac{70000}{5000} = 14$.
Thus,the $14^{th}$ order maximum will be obtained at the same point.

(A) The angular fringe width $\alpha$ is given by the formula $\alpha = \frac{\lambda}{d}$,where $\lambda$ is the wavelength of light and $d$ is the distance between the two coherent sources.
Rearranging the formula to solve for $d$,we get $d = \frac{\lambda}{\alpha}$.
Given values are $\lambda = 6280 \; \mathring{A} = 6280 \times 10^{-10} \; m$ and $\alpha = 1^{\circ}$.
First,convert the angular width from degrees to radians: $\alpha = 1^{\circ} = \frac{\pi}{180} \text{ rad} \approx \frac{3.14}{180} \text{ rad}$.
Substituting these values into the formula:
$d = \frac{6280 \times 10^{-10} \times 180}{3.14}$.
$d = \frac{6280 \times 10^{-10} \times 180}{3.14} = 2000 \times 10^{-10} \times 180 = 360000 \times 10^{-10} \; m$.
$d = 3.6 \times 10^{-5} \; m$.
To convert meters to millimeters,multiply by $10^3$:
$d = 3.6 \times 10^{-5} \times 10^3 \; mm = 3.6 \times 10^{-2} \; mm = 0.036 \; mm$.

(D) The position of the $n^{th}$ bright fringe (maxima) is $y_n = n\beta$,where $\beta$ is the fringe width.
For the first maxima $(n=1)$,$y_{max,1} = 1\beta$.
The position of the $n^{th}$ dark fringe (minima) is $y'_n = (n - 0.5)\beta$.
For the fifth minima $(n=5)$,$y_{min,5} = (5 - 0.5)\beta = 4.5\beta$.
The distance between the first maxima and the fifth minima is given as $y_{min,5} - y_{max,1} = 4.5\beta - 1\beta = 3.5\beta = 7\,mm$.
Therefore,the fringe width $\beta = \frac{7\,mm}{3.5} = 2\,mm = 2 \times 10^{-3}\,m$.
The formula for fringe width is $\beta = \frac{\lambda D}{d}$,where $D = 50\,cm = 0.5\,m$ and $d = 0.15\,mm = 0.15 \times 10^{-3}\,m$.
Rearranging for wavelength $\lambda = \frac{\beta d}{D} = \frac{(2 \times 10^{-3}\,m)(0.15 \times 10^{-3}\,m)}{0.5\,m} = \frac{0.3 \times 10^{-6}}{0.5}\,m = 0.6 \times 10^{-6}\,m = 600 \times 10^{-9}\,m = 600\,nm$.

(A) Given:
Distance between slits and screen,$D = 1\,m$.
Wavelength of light,$\lambda = 600\,nm = 600 \times 10^{-9}\,m$.
Position of the $n^{\text{th}}$ bright fringe,$y_n = 5\,cm = 5 \times 10^{-2}\,m$.
Order of the fringe,$n = 5$.
The formula for the position of the $n^{\text{th}}$ bright fringe is given by:
$y_n = \frac{n \lambda D}{d}$
Substituting the given values:
$5 \times 10^{-2} = \frac{5 \times 600 \times 10^{-9} \times 1}{d}$
Rearranging for $d$:
$d = \frac{5 \times 600 \times 10^{-9}}{5 \times 10^{-2}}$
$d = 600 \times 10^{-7}\,m = 60 \times 10^{-6}\,m$
Since $1\,\mu m = 10^{-6}\,m$,we have:
$d = 60\,\mu m$.

(C) The condition for the first minimum at point $P$ is given by the path difference $\Delta x = A_2 P - A_1 P = \frac{\lambda}{2}$.
From the geometry of the figure,$A_1 P = D$ and $A_2 P = \sqrt{D^2 + a^2}$.
Thus,$\sqrt{D^2 + a^2} - D = \frac{\lambda}{2}$.
Using the binomial approximation for $a \ll D$,we have $\sqrt{D^2 + a^2} = D(1 + \frac{a^2}{D^2})^{1/2} \approx D(1 + \frac{a^2}{2D^2}) = D + \frac{a^2}{2D}$.
Substituting this into the path difference equation: $(D + \frac{a^2}{2D}) - D = \frac{\lambda}{2}$.
This simplifies to $\frac{a^2}{2D} = \frac{\lambda}{2}$,which gives $a = \sqrt{\lambda D}$.
Given $\lambda = 800 \, nm = 800 \times 10^{-9} \, m = 8 \times 10^{-7} \, m$ and $D = 5 \, cm = 0.05 \, m$.
$a = \sqrt{8 \times 10^{-7} \times 0.05} = \sqrt{40 \times 10^{-9}} = \sqrt{400 \times 10^{-10}} = 20 \times 10^{-5} \, m = 0.2 \times 10^{-3} \, m = 0.2 \, mm$.

(C) The resultant intensity in a Young's double slit experiment is given by $I = 4I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)$,where $\Delta\phi$ is the phase difference.
Phase difference $\Delta\phi$ is related to path difference $\Delta x$ by $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x_1 = \frac{\lambda}{4}$,the phase difference is $\Delta\phi_1 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
Thus,$I_1 = 4I_0 \cos^2\left(\frac{\pi/2}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{4}\right) = 4I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = 2I_0$.
For path difference $\Delta x_2 = \frac{\lambda}{3}$,the phase difference is $\Delta\phi_2 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$.
Thus,$I_2 = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = 4I_0 \cdot \left(\frac{1}{2}\right)^2 = I_0$.
Therefore,$\frac{I_1 + I_2}{I_0} = \frac{2I_0 + I_0}{I_0} = 3$.

(C) The fringe width for a wavelength $\lambda$ is given by $\omega = \frac{\lambda D}{d}$.
Given: $\lambda_1 = 800 \, nm = 8 \times 10^{-7} \, m$,$\lambda_2 = 600 \, nm = 6 \times 10^{-7} \, m$,$D = 7 \, m$,and $d = 0.35 \, mm = 3.5 \times 10^{-4} \, m$.
Calculating fringe widths:
$\omega_1 = \frac{8 \times 10^{-7} \times 7}{3.5 \times 10^{-4}} = 16 \times 10^{-3} \, m = 16 \, mm$.
$\omega_2 = \frac{6 \times 10^{-7} \times 7}{3.5 \times 10^{-4}} = 12 \times 10^{-3} \, m = 12 \, mm$.
The bright fringes coincide at a distance $y$ from the central maximum where $y = n_1 \omega_1 = n_2 \omega_2$,where $n_1$ and $n_2$ are integers.
To find the shortest distance,we calculate the Least Common Multiple $(LCM)$ of $\omega_1$ and $\omega_2$.
$LCM(16, 12) = 48 \, mm$.
Thus,the shortest distance is $48 \, mm$.

(C) The linear 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.
When the screen is moved away from the plane of the slits,the distance $D$ increases.
Since $\beta \propto D$,an increase in $D$ leads to an increase in the linear fringe width $\beta$.
Therefore,the linear separation of the fringes increases.

(C) Given: $\lambda_1 = 7000 \; \mathring{A} = 7000 \times 10^{-10} \; m$,$\lambda_2 = 5500 \; \mathring{A} = 5500 \times 10^{-10} \; m$,$d = 2.5 \; mm = 2.5 \times 10^{-3} \; m$,$D = 150 \; cm = 1.5 \; m$.
The condition for the coincidence of bright fringes is $y_n = y_m$,where $y_n = \frac{n \lambda_1 D}{d}$ and $y_m = \frac{m \lambda_2 D}{d}$.
Equating the two,we get $n \lambda_1 = m \lambda_2$,which implies $\frac{n}{m} = \frac{\lambda_2}{\lambda_1} = \frac{5500}{7000} = \frac{11}{14}$.
For the least distance,we take the smallest integer values $n = 11$ and $m = 14$.
The distance $y$ is given by $y = \frac{n \lambda_1 D}{d} = \frac{11 \times 7000 \times 10^{-10} \times 1.5}{2.5 \times 10^{-3}}$.
$y = \frac{11 \times 7000 \times 1.5}{2.5} \times 10^{-7} = \frac{115500}{2.5} \times 10^{-7} = 46200 \times 10^{-7} = 462 \times 10^{-5} \; m$.
Comparing this with $n \times 10^{-5} \; m$,we get $n = 462$.

(C) The formula for fringe width $(\beta)$ in a 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 distance between the slits.
Since $D$ and $d$ remain constant,the fringe width is directly proportional to the wavelength: $\beta \propto \lambda$.
Therefore,we can write the ratio as $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$.
Given $\beta_1 = 2\,mm$,$\lambda_1 = 400\,nm$,and $\lambda_2 = 600\,nm$.
Substituting these values: $\frac{\beta_2}{2\,mm} = \frac{600\,nm}{400\,nm} = \frac{3}{2}$.
Solving for $\beta_2$: $\beta_2 = 2\,mm \times 1.5 = 3\,mm$.

(C) The resultant intensity $I$ for two waves of equal intensity $I_0$ with a phase difference $\phi$ is given by the formula: $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi) = 4I_0 \cos^2(\phi/2)$.
For point $P$,the phase difference is $\phi_1 = \pi/3$. The intensity $I_P$ is:
$I_P = 2I_0(1 + \cos(\pi/3)) = 2I_0(1 + 0.5) = 3I_0$.
For point $Q$,the phase difference is $\phi_2 = \pi/2$. The intensity $I_Q$ is:
$I_Q = 2I_0(1 + \cos(\pi/2)) = 2I_0(1 + 0) = 2I_0$.
The ratio of intensities at $P$ and $Q$ is:
$\frac{I_P}{I_Q} = \frac{3I_0}{2I_0} = \frac{3}{2}$ or $3:2$.

(A) Given the ratio of amplitudes of light from the two slits is $\frac{A_1}{A_2} = \frac{2}{1}$.
The intensity $I$ is proportional to the square of the amplitude,$I \propto A^2$.
The formula for the ratio of maximum intensity to minimum intensity is given by:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.
Substituting the given values:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \left( \frac{3}{1} \right)^2 = \frac{9}{1}$.
Therefore,the ratio of maximum to minimum intensity is $9:1$.