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

(A) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $I_{max} = K$ and $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given path difference $\Delta x = \frac{\lambda}{6}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
Substituting this into the intensity formula: $I = K \cos^2(\frac{\pi/3}{2}) = K \cos^2(\frac{\pi}{6})$.
Since $\cos(\frac{\pi}{6}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $I = K (\frac{\sqrt{3}}{2})^2 = K (\frac{3}{4}) = \frac{3K}{4}$.

(D) The fringe width is given by $W = \frac{\lambda D}{d} = X$ (given).
The position of the $n^{\text{th}}$ bright fringe from the central maximum is $y_n = n \frac{\lambda D}{d} = nX$.
For the $4^{\text{th}}$ bright fringe,$y_4 = 4X$.
The position of the $n^{\text{th}}$ dark fringe from the central maximum is $y'_n = (2n - 1) \frac{\lambda D}{2d} = (2n - 1) \frac{X}{2}$.
For the $6^{\text{th}}$ dark fringe,$y'_6 = (2(6) - 1) \frac{X}{2} = \frac{11X}{2} = 5.5X$.
Since the fringes are on opposite sides of the central bright band,the total distance is the sum of the magnitudes of their positions:
Total distance $= y_4 + y'_6 = 4X + 5.5X = 9.5X$.

(A) The position of the $n^{\text{th}}$ maximum in Young's double-slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
For the $n^{\text{th}}$ maximum with wavelength $\lambda_1$,the distance is $y_1 = \frac{n \lambda_1 D}{d} \quad (i)$.
For the $(\frac{n}{3})^{\text{th}}$ maximum with wavelength $\lambda_2$,the distance is $y_2 = \frac{(\frac{n}{3}) \lambda_2 D}{d} \quad (ii)$.
Taking the ratio of $(i)$ and $(ii)$:
$\frac{y_1}{y_2} = \frac{\frac{n \lambda_1 D}{d}}{\frac{n \lambda_2 D}{3d}} = \frac{n \lambda_1 D}{d} \times \frac{3d}{n \lambda_2 D} = \frac{3 \lambda_1}{\lambda_2}$.
Thus,the ratio $\frac{y_1}{y_2}$ is $\frac{3 \lambda_1}{\lambda_2}$.

(C) For the $n^{\text{th}}$ bright fringe,the position is given by $y_n = \frac{n \lambda D}{d}$.
For the $10^{\text{th}}$ bright fringe $(n=10)$:
$y_{10} = \frac{10 \lambda D}{d} = \frac{10 \times \lambda \times 1.2}{0.6 \times 10^{-3}} = (20 \times 10^3) \lambda \quad \dots(i)$
For the $n^{\text{th}}$ dark fringe,the position is given by $y'_n = \frac{(2n-1) \lambda D}{2d}$.
For the $3^{\text{rd}}$ dark fringe $(n=3)$:
$y'_3 = \frac{(2 \times 3 - 1) \lambda D}{2d} = \frac{5 \lambda D}{2d} = \frac{5 \times \lambda \times 1.2}{2 \times 0.6 \times 10^{-3}} = (5 \times 10^3) \lambda \quad \dots(ii)$
Given that the distance between them is $8.85 \ mm$:
$y_{10} - y'_3 = 8.85 \times 10^{-3} \ m$
$(20 \times 10^3) \lambda - (5 \times 10^3) \lambda = 8.85 \times 10^{-3}$
$(15 \times 10^3) \lambda = 8.85 \times 10^{-3}$
$\lambda = \frac{8.85 \times 10^{-3}}{15 \times 10^3} = 0.59 \times 10^{-6} \ m = 5.9 \times 10^{-7} \ m$
$\lambda = 5900 \ Å$.

(B) Let the point $P$ be exactly in front of slit $S_1$. The distance $S_1P = D$ and $S_2P = \sqrt{D^2 + d^2}$.
Using the binomial expansion for $S_2P$:
$S_2P = D(1 + \frac{d^2}{D^2})^{1/2} \approx D(1 + \frac{d^2}{2D^2}) = D + \frac{d^2}{2D}$.
The path difference $\Delta x$ at point $P$ is:
$\Delta x = S_2P - S_1P = (D + \frac{d^2}{2D}) - D = \frac{d^2}{2D}$.
For the $n^{\text{th}}$ dark fringe,the condition for path difference is:
$\Delta x = (2n - 1) \frac{\lambda}{2}$.
Given $\lambda = \frac{d^2}{3D}$,we substitute this into the condition:
$\frac{d^2}{2D} = (2n - 1) \frac{d^2}{6D}$.
Dividing both sides by $\frac{d^2}{D}$:
$\frac{1}{2} = \frac{2n - 1}{6}$.
$3 = 2n - 1 \Rightarrow 2n = 4 \Rightarrow n = 2$.

(C) In Young's double slit experiment,the fringe width $W$ is given by the formula $W = \frac{\lambda D}{d}$,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.
Since the wavelength of violet light $(\lambda_{violet} \approx 400 \ nm)$ is significantly smaller than the wavelength of sodium light $(\lambda_{sodium} \approx 589 \ nm)$,the fringe width $W$ will decrease when violet light is used instead of sodium light.

(C) In Young's double-slit experiment,dark fringes are formed due to destructive interference.
For destructive interference,the path difference $\Delta x$ must be an odd multiple of half-wavelength,i.e.,$\Delta x = (2n-1) \frac{\lambda}{2}$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the value of $\Delta x$: $\Delta \phi = \frac{2\pi}{\lambda} \times (2n-1) \frac{\lambda}{2} = (2n-1) \pi$.
Thus,the phase difference for a dark fringe is $(2n-1) \pi$.

(B) Let the intensities of the two individual waves be $I_0$. The resultant intensity $I$ is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi) = 4I_0 \cos^2(\phi/2)$.
- The minimum intensity $I_{min}$ occurs when $\phi = \pi$,so $I_{min} = 4I_0 \cos^2(\pi/2) = 0$.
- The path difference $\Delta x = \lambda/4$ corresponds to a phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
- The intensity $I_1$ at this point is $I_1 = 4I_0 \cos^2(\pi/4) = 4I_0 (1/\sqrt{2})^2 = 2I_0$.
- The ratio $I_{min} / I_1 = 0 / 2I_0 = 0$.

(C) For any point in the interference pattern,the intensity $I$ is given by $I = I_{\max} \cos^2 \left(\frac{\phi}{2}\right)$.
Given that $I = \frac{I_{\max}}{4}$,we have $\frac{I_{\max}}{4} = I_{\max} \cos^2 \left(\frac{\phi}{2}\right)$.
This simplifies to $\cos^2 \left(\frac{\phi}{2}\right) = \frac{1}{4}$,which implies $\cos \left(\frac{\phi}{2}\right) = \frac{1}{2}$.
Thus,$\frac{\phi}{2} = 60^{\circ} = \frac{\pi}{3}$,so the phase difference $\phi = \frac{2\pi}{3}$.
The phase difference is related to the path difference $\Delta x$ by $\phi = \left(\frac{2\pi}{\lambda}\right) \Delta x$.
Substituting $\Delta x = d \sin \theta$,we get $\frac{2\pi}{3} = \left(\frac{2\pi}{\lambda}\right) d \sin \theta$.
Solving for $\sin \theta$,we find $\sin \theta = \frac{\lambda}{3d}$,which means $\theta = \sin^{-1} \left(\frac{\lambda}{3d}\right)$.

(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.
Given that the new distance between slits $d' = 10d$ and the new distance from the screen $D' = \frac{D}{2}$.
The new fringe width $\beta'$ is given by $\beta' = \frac{D' \lambda}{d'} = \frac{(\frac{D}{2}) \lambda}{10d} = \frac{D \lambda}{20d}$.
Since $\beta = \frac{D \lambda}{d}$,we have $\beta' = \frac{\beta}{20}$.
Thus,the fringe width becomes $\frac{1}{20}$ times the original fringe width.

(C) Given: Fringe width $W = 2 \, mm$.
The distance of the $n^{\text{th}}$ bright fringe from the centre of the screen is $y_n = nW$.
The distance of the $n^{\text{th}}$ dark fringe from the centre of the screen is $y'_n = (n - 0.5)W$.
For the $13^{\text{th}}$ bright fringe, $n = 13$: $y_{13} = 13 \times 2 = 26 \, mm$.
For the $4^{\text{th}}$ dark fringe, $n = 4$: $y'_4 = (4 - 0.5) \times 2 = 3.5 \times 2 = 7 \, mm$.
The separation between them is $y_{13} - y'_4 = 26 \, mm - 7 \, mm = 19 \, mm$.

(B) The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
For a path difference $\Delta x = \frac{\lambda}{4}$,the phase difference is $\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity $I$ at any point in an interference pattern 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 $\Delta \phi = \frac{\pi}{2}$ into the formula:
$\frac{I}{I_0} = \cos^2 \left( \frac{\pi/2}{2} \right) = \cos^2 \left( \frac{\pi}{4} \right)$.
Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$,we have $\frac{I}{I_0} = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$.

(D) Given: Slit separation $d = 0.6 \, mm = 0.6 \times 10^{-3} \, m$, distance to screen $D = 1.2 \, m$, wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, m$, and position on screen $y = 3 \, mm = 3 \times 10^{-3} \, m$.
The path difference $\Delta x$ at a point $y$ is given by $\Delta x = \frac{d \cdot y}{D}$.
Substituting the values: $\Delta x = \frac{(0.6 \times 10^{-3} \, m) \times (3 \times 10^{-3} \, m)}{1.2 \, m} = \frac{1.8 \times 10^{-6}}{1.2} \, m = 1.5 \times 10^{-6} \, m$.
The phase difference $\Delta \phi$ is related to path difference by $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \Delta x$.
Substituting the values: $\Delta \phi = \frac{2 \pi}{6000 \times 10^{-10}} \times 1.5 \times 10^{-6} = \frac{2 \pi \times 1.5 \times 10^{-6}}{6 \times 10^{-7}} = \frac{3 \pi \times 10^{-6}}{6 \times 10^{-7}} = 0.5 \pi \times 10 = 5 \pi \, \text{radian}$.

(C) The formula for fringe width $(W)$ in Young's double slit experiment is given by:
$W = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the separation between the slits.
According to the problem,the slit separation is doubled,so the new separation $d' = 2d$.
We want the fringe width to remain the same,so $W' = W$.
Substituting the values into the formula:
$W' = \frac{\lambda D'}{d'} = \frac{\lambda D'}{2d}$
Since $W = W'$,we have:
$\frac{\lambda D}{d} = \frac{\lambda D'}{2d}$
By canceling $\lambda$ and $d$ from both sides,we get:
$D = \frac{D'}{2}$
Therefore,$D' = 2D$.
Thus,the distance of the screen from the slits should be doubled.

(C) The wavelength of light in a medium is given by $\lambda_{\text{liquid}} = \frac{\lambda_{\text{air}}}{\mu}$.
Given $\lambda_{\text{air}} = 6300 \,Å = 6300 \times 10^{-10} \,m$ and $\mu = 1.33$.
So, $\lambda_{\text{liquid}} = \frac{6300 \times 10^{-10}}{1.33} \,m$.
The fringe width $W$ is given by $W = \frac{\lambda_{\text{liquid}} \times D}{d}$.
Here, $D = 1.33 \,m$ and $d = 1 \,mm = 10^{-3} \,m$.
Substituting the values:
$W = \frac{(6300 \times 10^{-10} / 1.33) \times 1.33}{10^{-3}}$
$W = \frac{6300 \times 10^{-10}}{10^{-3}} = 6300 \times 10^{-7} \,m = 6.3 \times 10^{-4} \,m$.

(B) The formula for fringe width is $W = \frac{\lambda D}{d}$.
From this relation,we see that $W \propto \lambda$,$W \propto D$,and $W \propto \frac{1}{d}$.
To make the fringes more closely spaced,the fringe width $W$ must decrease.
Since $W \propto \lambda$,decreasing the wavelength $\lambda$ will decrease the fringe width.
The wavelength of blue light is shorter than that of green light $(\lambda_{\text{blue}} < \lambda_{\text{green}})$.
Therefore,using blue light instead of green light will cause the fringes to be more closely spaced.

(A) The position of the $n^{th}$ bright fringe (maximum) from the central maxima in Young's double slit experiment is given by the formula: $y_n = \frac{n \lambda D}{d}$.
For the fifth maximum $(n = 5)$ with wavelength $\lambda_1$,the distance is $y_1 = \frac{5 \lambda_1 D}{d}$.
For the fifth maximum $(n = 5)$ with wavelength $\lambda_2$,the distance is $y_2 = \frac{5 \lambda_2 D}{d}$.
Taking the ratio of the two distances:
$\frac{y_1}{y_2} = \frac{\frac{5 \lambda_1 D}{d}}{\frac{5 \lambda_2 D}{d}} = \frac{\lambda_1}{\lambda_2}$.

(C) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.
For the first point,the path difference is $\Delta x_1 = \frac{\lambda}{4}$.
Thus,$\phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The resultant intensity $I$ is given by $I = 4I_0 \cos^2(\frac{\phi}{2}) = 2I_0(1 + \cos \phi)$.
For $\phi_1 = \frac{\pi}{2}$,$I_1 = 2I_0(1 + \cos(\frac{\pi}{2})) = 2I_0(1 + 0) = 2I_0$.
For the second point,the path difference is $\Delta x_2 = \frac{\lambda}{3}$.
Thus,$\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
For $\phi_2 = \frac{2\pi}{3}$,$I_2 = 2I_0(1 + \cos(\frac{2\pi}{3})) = 2I_0(1 - \frac{1}{2}) = 2I_0(\frac{1}{2}) = I_0$.
Therefore,$\frac{I_1 + I_2}{I_0} = \frac{2I_0 + I_0}{I_0} = \frac{3I_0}{I_0} = 3$.

(A) The fringe shift $y$ due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by the formula: $y = \frac{D}{\text{d}} (\mu - 1) t = \frac{\beta}{\lambda} (\mu - 1) t$.
Given for the first plate,$\mu_1 = 1.44$,the displacement is $y_1 = y = \frac{\beta}{\lambda} (1.44 - 1) t = 0.44 t \frac{\beta}{\lambda}$.
For the second plate,$\mu_2 = 1.66$,the new displacement $y_2$ is: $y_2 = \frac{\beta}{\lambda} (1.66 - 1) t = 0.66 t \frac{\beta}{\lambda}$.
Taking the ratio of the two displacements: $\frac{y_2}{y_1} = \frac{0.66}{0.44} = \frac{66}{44} = \frac{3}{2}$.
Therefore,$y_2 = \frac{3}{2} y$.

(B) The intensity at a point due to interference is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For a path difference $\Delta x$,the phase difference is $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x_1 = \frac{\lambda}{4}$,the phase difference is $\phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
For path difference $\Delta x_2 = \frac{\lambda}{6}$,the phase difference is $\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
Assuming equal intensity of the interfering waves,$I_1 = I_2 = I_0$,the intensity formula becomes $I = 2I_0(1 + \cos \phi) = 4I_0 \cos^2(\frac{\phi}{2})$.
For $\phi_1 = \frac{\pi}{2}$,$I_1 = 2I_0(1 + \cos 90^{\circ}) = 2I_0(1 + 0) = 2I_0$.
For $\phi_2 = \frac{\pi}{3}$,$I_2 = 2I_0(1 + \cos 60^{\circ}) = 2I_0(1 + 0.5) = 2I_0(1.5) = 3I_0$.
The ratio of intensities is $\frac{I_1}{I_2} = \frac{2I_0}{3I_0} = \frac{2}{3}$.

(D) Given path difference $\Delta x = 3 \mu m = 3 \times 10^{-6} \text{ m}$.
The wavelength $\lambda = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$.
For constructive interference (bright band),the condition is $\Delta x = n \lambda$,where $n = 0, 1, 2, 3, \dots$.
Substituting the values: $3 \times 10^{-6} = n \times (5 \times 10^{-7})$.
$n = \frac{3 \times 10^{-6}}{5 \times 10^{-7}} = \frac{30}{5} = 6$.
Since $n = 6$ is an integer,the point $Q$ corresponds to the $6^{\text{th}}$ bright band.

(B) The distance of the $n^{\text{th}}$ maximum from the central maximum in Young's double slit experiment is given by the formula: $d = \frac{n \lambda D}{a}$,where $n$ is the order of the maximum,$\lambda$ is the wavelength,$D$ is the distance to the screen,and $a$ is the slit separation.
Since $D$ and $a$ are constant,we have $d \propto n \lambda$.
For the $8^{\text{th}}$ maximum with wavelength $\lambda_1$: $d_1 = 8 \lambda_1 \times (\frac{D}{a})$.
For the $6^{\text{th}}$ maximum with wavelength $\lambda_2$: $d_2 = 6 \lambda_2 \times (\frac{D}{a})$.
Taking the ratio $\frac{d_2}{d_1}$:
$\frac{d_2}{d_1} = \frac{6 \lambda_2}{8 \lambda_1} = \frac{3 \lambda_2}{4 \lambda_1}$.

(A) The intensity at any point in an interference pattern 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 the path difference $\Delta x = \frac{\lambda}{4}$,the phase difference $\phi$ is calculated as $\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
Substituting this into the intensity formula: $I = I_0 \cos^2 \left( \frac{\pi/2}{2} \right) = I_0 \cos^2 \left( \frac{\pi}{4} \right)$.
Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$,we have $I = I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{I_0}{2}$.
Therefore,the ratio $\frac{I_0}{I} = \frac{I_0}{I_0/2} = 2$,which is $2: 1$.

(B) The condition for constructive interference (bright band) is path difference $\Delta x = n\lambda$,where $n = 0, 1, 2, \dots$
The condition for destructive interference (dark band) is path difference $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$
Given path difference $\Delta x = \frac{87}{2} \lambda = 87 \left( \frac{\lambda}{2} \right)$.
Comparing this with the condition for the dark band: $(2n - 1) \frac{\lambda}{2} = 87 \frac{\lambda}{2}$.
$2n - 1 = 87$
$2n = 88$
$n = 44$.
Therefore,the point corresponds to the $44^{\text{th}}$ dark band.

(B) The path difference $\Delta x$ at a point $P$ on the screen at a distance $y$ from the center is given by $\Delta x = d \sin \theta \approx d \tan \theta = d \left( \frac{y}{D} \right)$.
For a dark fringe,the path difference must be an odd multiple of half the wavelength: $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$.
The first dark fringe corresponds to $n = 1$,so $\Delta x = \frac{\lambda}{2}$.
The problem states that the first dark fringe is observed directly opposite to one of the slits. The distance of the slits from the center is $d/2$. Thus,$y = d/2$.
Substituting these values into the path difference formula:
$\frac{d}{2} \cdot \frac{d}{D} = \frac{\lambda}{2}$
$\frac{d^2}{D} = \lambda$
Therefore,the wavelength of the light is $\lambda = \frac{d^2}{D}$.

(C) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$.
Since the number of fringes per unit length $n$ is inversely proportional to the fringe width $\beta$,we have $n \propto \frac{1}{\beta}$.
Therefore,$n \propto \frac{1}{\lambda}$,which implies $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 15$ fringes/$cm$,$\lambda_1 = 5600$ Å,and $\lambda_2 = 7000$ Å.
Substituting the values: $15 \times 5600 = n_2 \times 7000$.
$n_2 = \frac{15 \times 5600}{7000} = \frac{15 \times 56}{70} = \frac{15 \times 8}{10} = \frac{120}{10} = 12$.
Thus,$12$ fringes per $cm$ will be obtained.

(A) The fringe width $\omega$ in Young's double slit experiment is given by $\omega = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance of the screen from the slits, and $d$ is the slit separation.
Given the change in screen distance $\Delta D = 5 \times 10^{-2} \, m$ and the change in fringe width $\Delta \omega = 3 \times 10^{-5} \, m$, the relationship is $\Delta \omega = \frac{\lambda (\Delta D)}{d}$.
Rearranging for $\lambda$: $\lambda = \frac{d \cdot \Delta \omega}{\Delta D}$.
Substituting the values: $\lambda = \frac{(10^{-3} \, m)(3 \times 10^{-5} \, m)}{5 \times 10^{-2} \, m} = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} = 0.6 \times 10^{-6} \, m$.
Converting to $\text{\AA}$: $\lambda = 0.6 \times 10^{-6} \times 10^{10} \, \text{\AA} = 6000 \, \text{\AA}$.

(B) The angular fringe width $\theta$ is given by the formula:
$\theta = \frac{\lambda}{d}$
Given,in air,$\theta = 0.20$ radians.
When the entire system is immersed in water,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$,where $\mu$ is the refractive index of water.
The new angular fringe width $\theta'$ is:
$\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}$
Given $\mu = \frac{4}{3}$,we have:
$\theta' = \frac{0.20}{4/3} = 0.20 \times \frac{3}{4} = 0.15$ radians.
Thus,the angular width of the fringes in water is $0.15$ radians.

(B) 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 separation between the two slits.
Given that the slit separation $d$ is made threefold,i.e.,$d' = 3d$,while $\lambda$ and $D$ remain constant.
The new fringe width $\beta'$ will be: $\beta' = \frac{\lambda D}{d'} = \frac{\lambda D}{3d} = \frac{1}{3} \beta$.
Therefore,the fringe width becomes $\frac{1}{3}$ times the original fringe width.

(D) In Young's double slit experiment,the condition for constructive interference (maxima) is given by the path difference $\Delta x = n \lambda$,where $n = 0, 1, 2, ...$ and $\lambda$ is the wavelength of light.
For the central maximum,$n = 0$,and for the first maximum,$n = 1$.
Given the wavelength $\lambda = 6300 Å$,the path difference for the first maximum is $\Delta x = 1 \times 6300 Å = 6300 Å$.

(D) Let $I'$ be the intensity of each individual wave. The resultant intensity $I$ is given by the formula:
$I = 4I' \cos^2 \left( \frac{\phi}{2} \right)$
where $\phi$ is the phase difference.
The maximum intensity $I_0$ occurs when $\cos^2 \left( \frac{\phi}{2} \right) = 1$,so $I_0 = 4I'$.
Given the path difference $\Delta x = \frac{\lambda}{4}$,the phase difference $\phi$ is calculated as:
$\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
Substituting this into the intensity formula:
$I = 4I' \cos^2 \left( \frac{\pi/2}{2} \right) = 4I' \cos^2 \left( \frac{\pi}{4} \right) = 4I' \times \left( \frac{1}{\sqrt{2}} \right)^2 = 4I' \times \frac{1}{2} = 2I'$.
Therefore,the ratio is:
$\frac{I}{I_0} = \frac{2I'}{4I'} = \frac{1}{2}$.

(B) The position of the $n^{\text{th}}$ bright fringe (maximum) in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
For the first case,the $n^{\text{th}}$ maximum with wavelength $\lambda_1$ is at distance $y_1 = \frac{n \lambda_1 D}{d}$.
For the second case,the $(n/2)^{\text{th}}$ maximum with wavelength $\lambda_2$ is at distance $y_2 = \frac{(n/2) \lambda_2 D}{d} = \frac{n \lambda_2 D}{2d}$.
Now,taking the ratio of $y_1$ and $y_2$:
$\frac{y_1}{y_2} = \frac{(n \lambda_1 D / d)}{(n \lambda_2 D / 2d)} = \frac{n \lambda_1 D}{d} \times \frac{2d}{n \lambda_2 D} = \frac{2 \lambda_1}{\lambda_2}$.
Therefore,the ratio is $\frac{2 \lambda_1}{\lambda_2}$.

(C) The fringe width $\beta$ in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance of the screen from the slits, and $d$ is the separation between the slits.
Initially, $\beta_1 = \frac{\lambda D_1}{d_1}$, where $D_1 = 75 \,cm$.
When the slit separation is doubled, $d_2 = 2d_1$. To keep the fringe width constant $(\beta_2 = \beta_1)$, we must have $\frac{\lambda D_2}{d_2} = \frac{\lambda D_1}{d_1}$.
Substituting $d_2 = 2d_1$, we get $\frac{D_2}{2d_1} = \frac{D_1}{d_1}$, which implies $D_2 = 2D_1$.
$D_2 = 2 \times 75 \,cm = 150 \,cm$.
The screen must be moved from $75 \,cm$ to $150 \,cm$, which means it should be moved $150 \,cm - 75 \,cm = 75 \,cm$ away from the slits.

(A) The position of the $n^{\text{th}}$ bright fringe (maximum) in Young's double slit experiment is given by $Y_n = \frac{n \lambda D}{d}$,where $D$ is the distance between the slits and the screen,and $d$ is the distance between the slits.
For the $10^{\text{th}}$ maximum with wavelength $\lambda_1$,the distance is $Y_1 = \frac{10 \lambda_1 D}{d}$.
For the $5^{\text{th}}$ maximum with wavelength $\lambda_2$,the distance is $Y_2 = \frac{5 \lambda_2 D}{d}$.
Taking the ratio of $Y_1$ to $Y_2$:
$\frac{Y_1}{Y_2} = \frac{10 \lambda_1 D / d}{5 \lambda_2 D / d} = \frac{10 \lambda_1}{5 \lambda_2} = \frac{2 \lambda_1}{\lambda_2}$.

(A) The intensity $I$ in Young's double slit experiment is given by the formula: $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual slit and $\phi$ is the phase difference.
Maximum intensity $K$ occurs when $\cos^2(\frac{\phi}{2}) = 1$,so $K = 4I_0$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the path difference $\Delta x = \frac{\lambda}{3}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
Substituting this into the intensity formula: $I = K \cos^2(\frac{2\pi/3}{2}) = K \cos^2(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I = K (\frac{1}{2})^2 = \frac{K}{4}$.

(B) In Young's double-slit experiment,the path difference $\Delta x$ at a point exactly in front of one slit is given by $\Delta x = \frac{d^2}{2D}$.
For a minimum (destructive interference) to occur at this point,the path difference must be an odd multiple of $\frac{\lambda}{2}$,i.e.,$\Delta x = (2n-1)\frac{\lambda}{2}$ where $n = 1, 2, 3, \ldots$.
Equating the two expressions: $\frac{d^2}{2D} = (2n-1)\frac{\lambda}{2}$.
This simplifies to $\lambda = \frac{d^2}{(2n-1)D}$.
For $n=1, 2, 3, \ldots$,the values of $(2n-1)$ are $1, 3, 5, \ldots$.
Thus,$\lambda$ is proportional to $\frac{1}{D}, \frac{1}{3D}, \frac{1}{5D}, \ldots$.
Therefore,the possible wavelengths are inversely proportional to $D, 3D, 5D, \ldots$.

(D) In Young's double-slit experiment,the formation of interference fringes requires the superposition of coherent light waves originating from two separate slits.
When one of the slits is covered with black opaque paper,light can only pass through the remaining single slit.
Since there is no longer a second source to interfere with the light from the first,the condition for interference is lost.
Consequently,no interference pattern (fringes) will be observed on the screen.

(B) The formula for fringe width in $YDSE$ is given by $W = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance between the slits and the screen, and $d$ is the separation between the slits.
Let the original fringe width be $W_1 = \frac{\lambda D_1}{d_1} = W$.
According to the problem, the new distance $D_2 = D_1 + 0.25 D_1 = 1.25 D_1$ and the new separation $d_2 = \frac{d_1}{2}$.
The new fringe width $W_2$ is given by $W_2 = \frac{\lambda D_2}{d_2}$.
Substituting the values, we get $W_2 = \frac{\lambda (1.25 D_1)}{(d_1 / 2)} = 1.25 \times 2 \times \frac{\lambda D_1}{d_1}$.
Since $W = \frac{\lambda D_1}{d_1}$, we have $W_2 = 2.5 W$.

(B) The position of the $n^{\text{th}}$ dark fringe from the central maximum is given by $y_n = (n - 0.5) \frac{\lambda D}{d}$.
For the $4^{\text{th}}$ dark band,$n = 4$,so $y_4 = (4 - 0.5) \frac{\lambda D}{d} = 3.5 \frac{\lambda D}{d}$.
Since the dark band is formed opposite to one of the slits,its distance from the central axis is $y = \frac{d}{2}$.
Equating the two expressions: $\frac{d}{2} = 3.5 \frac{\lambda D}{d}$.
Rearranging for $\lambda$: $\lambda = \frac{d^2}{2 \times 3.5 D} = \frac{d^2}{7 D}$.

(C) In Young's double slit experiment,the position of the $n^{\text{th}}$ dark fringe from the central bright fringe is given by the formula:
$y_n = (n - \frac{1}{2}) \beta$
where $n = 1, 2, 3, \dots$ represents the order of the dark fringe.
Thus,the distance is $(n - 0.5) \beta$.

(D) The path difference is given as $\Delta x = \frac{57}{2} \lambda = 28.5 \lambda$.
For constructive interference (bright band),the path difference must be an integral multiple of $\lambda$,i.e.,$\Delta x = n \lambda$ where $n = 1, 2, 3, \dots$. Since $28.5 \lambda$ is not an integer multiple,it is not a bright band.
For destructive interference (dark band),the path difference is given by $\Delta x = (n - \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$.
Equating the two: $28.5 \lambda = (n - 0.5) \lambda$.
Solving for $n$: $n - 0.5 = 28.5$,which gives $n = 29$.
Therefore,the point corresponds to the $29^{\text{th}}$ dark band.

(B) The position of the $n^{\text{th}}$ maximum in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$,where $n$ is the order of the maximum,$\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the slit separation.
For the $6^{\text{th}}$ maximum with wavelength $\lambda_1$,the distance is $d_1 = \frac{6 \lambda_1 D}{d}$.
For the $4^{\text{th}}$ maximum with wavelength $\lambda_2$,the distance is $d_2 = \frac{4 \lambda_2 D}{d}$.
Taking the ratio of $d_1$ to $d_2$:
$\frac{d_1}{d_2} = \frac{6 \lambda_1 D / d}{4 \lambda_2 D / d} = \frac{6 \lambda_1}{4 \lambda_2} = \frac{3}{2} \frac{\lambda_1}{\lambda_2}$.

(B) The fringe width $\beta$ in a medium of refractive index $\mu$ is given by $\beta = \frac{\lambda_w D}{d}$.
Here, the wavelength in water is $\lambda_w = \frac{\lambda_{air}}{\mu} = \frac{6300 \times 10^{-10} \,m}{1.33}$.
Given: $D = 1.33 \,m$, $d = 1 \,mm = 10^{-3} \,m$, and $\mu = 1.33$.
Substituting these values into the formula:
$\beta = \frac{(6300 \times 10^{-10} / 1.33) \times 1.33}{10^{-3}} \,m$.
$\beta = \frac{6300 \times 10^{-10}}{10^{-3}} \,m$.
$\beta = 6300 \times 10^{-7} \,m = 6.3 \times 10^{-4} \,m$.

(B) 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 separation between the slits.
To make the fringes more closely spaced,the fringe width $\beta$ must be decreased.
From the formula,$\beta \propto \lambda$ and $\beta \propto \frac{1}{d}$.
Since the wavelength of blue light is shorter than that of green light $(\lambda_{blue} < \lambda_{green})$,using blue light will decrease the fringe width $\beta$.
Therefore,the correct option is $B$.

(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 distance between the two slits.
When the experiment is performed in water,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$,where $\mu$ is the refractive index of water $(\mu > 1)$.
Since $\mu > 1$,the new wavelength $\lambda'$ is smaller than the original wavelength $\lambda$ in air.
Consequently,the new fringe width $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Since $\mu > 1$,the fringe width $\beta'$ decreases.

(C) In Young's Double Slit Experiment,the fringe-width $(z)$ is given by the formula:
$z = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen (or eye-piece),and $d$ is the distance between the two slits.
Since $\lambda$ and $d$ are kept constant,we have $z \propto D$.
This represents a linear relationship passing through the origin,which corresponds to a straight line graph.
Looking at the given options,graph $(C)$ represents a direct linear relationship $(z \propto D)$.
Therefore,the correct graph is $(C)$.

(D) Given: Slit separation $d = 3 \,mm = 3 \times 10^{-3} \,m$,distance to screen $D = 2 \,m$,wavelengths $\lambda_1 = 480 \,nm = 480 \times 10^{-9} \,m$ and $\lambda_2 = 600 \,nm = 600 \times 10^{-9} \,m$.
The position of the $n^{th}$ order bright fringe is given by $y_n = n \frac{\lambda D}{d}$.
For the $5^{th}$ order bright fringe of the first wavelength: $y_5^{(1)} = 5 \frac{\lambda_1 D}{d}$.
For the $5^{th}$ order bright fringe of the second wavelength: $y_5^{(2)} = 5 \frac{\lambda_2 D}{d}$.
The separation between these fringes is $\Delta y = y_5^{(2)} - y_5^{(1)} = \frac{5D}{d} (\lambda_2 - \lambda_1)$.
Substituting the values: $\Delta y = \frac{5 \times 2}{3 \times 10^{-3}} \times (600 - 480) \times 10^{-9} \,m$.
$\Delta y = \frac{10}{3 \times 10^{-3}} \times 120 \times 10^{-9} \,m = \frac{1200}{3} \times 10^{-6} \,m = 400 \times 10^{-6} \,m = 4 \times 10^{-4} \,m$.