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

(A) From the given interference pattern,the angular width of a fringe is the distance between two consecutive minima or maxima.
From the figure,the first minimum is at $\theta = 0.75^{\circ}$.
The angular width of a fringe is $\Delta\theta = 0.75^{\circ}$.
For a double slit experiment,the condition for the first minimum is $d \sin\theta = \frac{\lambda}{2}$.
Since $\theta$ is very small,$\sin\theta \approx \theta$ (in radians).
Thus,$d \theta = \frac{\lambda}{2} \Rightarrow d = \frac{\lambda}{2\theta}$.
Convert $\theta$ to radians: $\theta = 0.75^{\circ} = 0.75 \times \frac{\pi}{180} \text{ rad} = \frac{3}{4} \times \frac{\pi}{180} = \frac{\pi}{240} \text{ rad}$.
Given $\lambda = 520 \times 10^{-9} \text{ m}$.
Substituting the values: $d = \frac{520 \times 10^{-9}}{2 \times (\pi / 240)} = \frac{520 \times 10^{-9} \times 120}{\pi} \approx \frac{62400 \times 10^{-9}}{3.14} \approx 1.987 \times 10^{-5} \text{ m}$.
Converting to mm: $d \approx 1.987 \times 10^{-2} \text{ mm} \approx 2 \times 10^{-2} \text{ mm}$.
Therefore,the correct option is $A$.

(C) The position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.
For the $9^{th}$ bright fringe,$y_9 = \frac{9 \lambda D}{d}$.
The position of the $m^{th}$ dark fringe is given by $y'_m = \frac{(2m-1) \lambda D}{2d}$.
For the $2^{nd}$ dark fringe,$y'_2 = \frac{(2 \times 2 - 1) \lambda D}{2d} = \frac{3 \lambda D}{2d}$.
The distance between them is $y_9 - y'_2 = 7.5 \, mm = 7.5 \times 10^{-3} \, m$.
$\frac{9 \lambda D}{d} - \frac{3 \lambda D}{2d} = 7.5 \times 10^{-3}$.
$\frac{(18-3) \lambda D}{2d} = 7.5 \times 10^{-3} \implies \frac{15 \lambda D}{2d} = 7.5 \times 10^{-3}$.
Given $d = 0.5 \times 10^{-3} \, m$ and $D = 1 \, m$.
$\lambda = \frac{7.5 \times 10^{-3} \times 2 \times 0.5 \times 10^{-3}}{15 \times 1} = \frac{7.5 \times 10^{-6}}{15} = 0.5 \times 10^{-6} \, m = 5000 \, \mathring{A}$.

(B) The intensity at any point on the screen is given by $I = I_{\max} \cos^2(\phi/2)$.
Given $I = 0.75 I_{\max}$,we have $0.75 = \cos^2(\phi/2)$,which implies $\cos(\phi/2) = \pm \sqrt{3}/2$.
Thus,$\phi/2 = \pi/6$ or $5\pi/6$,leading to phase differences $\phi = \pi/3$ or $5\pi/3$.
The path difference is $\Delta x = (\lambda/2\pi) \phi = yd/D$.
For $\phi_1 = \pi/3$,$y_1 = (\lambda D / 2\pi d) \times (\pi/3) = \lambda D / 6d$.
For $\phi_2 = 5\pi/3$,$y_2 = (\lambda D / 2\pi d) \times (5\pi/3) = 5\lambda D / 6d$.
Substituting values: $\lambda = 6 \times 10^{-7} \, m$,$D = 1 \, m$,$d = 10^{-3} \, m$.
$y_1 = (6 \times 10^{-7} \times 1) / (6 \times 10^{-3}) = 10^{-4} \, m = 0.1 \, mm$.
$y_2 = 5 \times 0.1 \, mm = 0.5 \, mm$.
The distance between the two points is $\Delta y = y_2 - y_1 = 0.5 - 0.1 = 0.4 \, mm$.

(C) Let $I_0$ be the intensity due to a single slit. The resultant intensity $I$ in a double-slit experiment is given by $I = I_0 + I_0 + 2\sqrt{I_0}\sqrt{I_0}\cos\phi = 2I_0(1 + \cos\phi)$.
Given that the intensity at the point is equal to that of a single slit,we set $I = I_0$.
$I_0 = 2I_0(1 + \cos\phi) \Rightarrow 1 = 2(1 + \cos\phi) \Rightarrow \frac{1}{2} = 1 + \cos\phi$.
$\cos\phi = -\frac{1}{2} \Rightarrow \phi = \frac{2\pi}{3}$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda}\Delta x$.
$\frac{2\pi}{3} = \frac{2\pi}{\lambda}\Delta x \Rightarrow \Delta x = \frac{\lambda}{3}$.
For small angles,the path difference $\Delta x = \frac{dy}{D}$.
Equating the two,$\frac{dy}{D} = \frac{\lambda}{3} \Rightarrow y = \frac{D\lambda}{3d}$.

(D) The condition for the $n^{th}$ bright fringe is $y_n = n \frac{D \lambda_1}{d}$. For $n = 4$ and $\lambda_1 = 6300 \, \mathring{A}$,$y_4 = 4 \frac{D \times 6300}{d}$.
The condition for the $m^{th}$ dark fringe is $y_m = (m - \frac{1}{2}) \frac{D \lambda_2}{d}$. For $m = 5$ and $\lambda_2 = \lambda$,$y_5 = (5 - 0.5) \frac{D \lambda}{d} = 4.5 \frac{D \lambda}{d} = \frac{9}{2} \frac{D \lambda}{d}$.
Since the fringes coincide,$y_4 = y_5$:
$4 \frac{D \times 6300}{d} = \frac{9}{2} \frac{D \lambda}{d}$.
Canceling $\frac{D}{d}$ from both sides:
$4 \times 6300 = 4.5 \times \lambda$.
$\lambda = \frac{4 \times 6300}{4.5} = \frac{25200}{4.5} = 5600 \, \mathring{A}$.

(A) For the bright fringes of two different wavelengths to coincide at a distance $x$ from the central maximum,the condition is given by:
$x = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$
This implies $n_1 \lambda_1 = n_2 \lambda_2$,where $n_1$ and $n_2$ are the orders of the bright fringes.
Rearranging the terms,we get:
$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5200 \mathring{A}}{6500 \mathring{A}} = \frac{4}{5}$
For the least distance,we take the smallest integer values: $n_1 = 4$ and $n_2 = 5$.
Now,substituting these values into the formula for $x$:
$x = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times 6500 \times 10^{-10} \text{ m} \times 1.2 \text{ m}}{2 \times 10^{-3} \text{ m}}$
$x = \frac{4 \times 6500 \times 1.2 \times 10^{-10}}{2 \times 10^{-3}} \text{ m} = 15600 \times 10^{-7} \text{ m} = 1.56 \times 10^{-3} \text{ m}$
$x = 0.156 \text{ cm}$.

(A) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $D$ is the distance between the slits and the screen,$d$ is the separation between the slits,and $\lambda$ is the wavelength of light.
Given the change in screen distance $\Delta D = 5 \times 10^{-2} \ m$ and the change in fringe width $\Delta \beta = 3 \times 10^{-5} \ m$.
The change in fringe width is related to the change in distance by $\Delta \beta = \frac{\lambda \Delta D}{d}$.
Rearranging for $\lambda$,we get $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.
Substituting the given values: $\lambda = \frac{(3 \times 10^{-5} \ m) \times (10^{-3} \ m)}{5 \times 10^{-2} \ m}$.
$\lambda = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} = 0.6 \times 10^{-6} \ m$.
Converting to $\mathring{A}$: $\lambda = 0.6 \times 10^{-6} \times 10^{10} \ \mathring{A} = 6000 \ \mathring{A}$.

(A) Let $d$ be the distance between the slits $S_1$ and $S_2$,and $D$ be the distance between the slits and the screen.
From the geometry of the setup,the angle $\theta$ subtended by the slits at the centre of the screen $C$ is given by $\tan(\theta/2) = \frac{d/2}{D}$.
For small angles,$\tan(\theta/2) \approx \theta/2 = \frac{d}{2D}$,which implies $\theta = \frac{d}{D}$.
The fringe width $\beta$ is defined as $\beta = \frac{\lambda D}{d}$.
Substituting $d/D = \theta$ into the expression for fringe width,we get $\beta = \frac{\lambda}{\theta}$.

(C) The intensity at any point $P$ on the screen is given by $I_P = I_{\max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.
The phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \frac{yd}{D}$,where $y$ is the distance of point $P$ from the centre $O$,$d$ is the slit separation,and $D$ is the distance of the screen from the slits.
At the centre $O$,$y = 0$,so $\phi = 0$ and $I_O = I_{\max}$.
At point $P$,the first maxima is initially observed,so the path difference $\Delta x = \lambda$,which means $\phi = 2\pi$.
As the screen is moved away,$D$ increases. Since $\phi = \frac{2\pi yd}{\lambda D}$,as $D$ increases,$\phi$ decreases from $2\pi$ towards $0$.
As $\phi$ decreases from $2\pi$ to $0$,the value of $\cos^2(\phi/2)$ increases from $\cos^2(\pi) = 1$ to $\cos^2(0) = 1$,passing through $0$ at $\phi = \pi$ (the first minima).
Therefore,the intensity $I_P$ first decreases (as it moves from maxima to minima) and then increases (as it moves from minima towards the central maxima). Thus,the ratio $I_P/I_O$ first decreases and then increases.

(C) The intensity at any point in a Young's double-slit experiment is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given $I = \frac{I_{max}}{2}$,we have $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$,which implies $\frac{\phi}{2} = \frac{\pi}{4}$,so $\phi = \frac{\pi}{2}$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. Thus,$\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x$,which gives $\Delta x = \frac{\lambda}{4}$.
For small angles,the path difference $\Delta x = \frac{yd}{D}$.
Equating the two,we get $\frac{yd}{D} = \frac{\lambda}{4}$,so $y = \frac{\lambda D}{4d}$.
Substituting the given values: $\lambda = 5 \times 10^{-7}\, m$,$D = 1\, m$,and $d = 10^{-3}\, m$.
$y = \frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}} = 1.25 \times 10^{-4}\, m$.

(C) Given: Slit separation $d = 0.3\, mm = 0.3 \times 10^{-3}\, m$,wavelength $\lambda = 500\, nm = 500 \times 10^{-9}\, m$.
For constructive interference (maxima),the path difference is given by $d \sin \theta = n \lambda$,where $n$ is an integer.
We need to find the number of maxima in the range $-30^{\circ} < \theta < 30^{\circ}$.
Thus,$\sin \theta < \sin 30^{\circ} = 0.5$.
Substituting the condition $d \sin \theta = n \lambda$ into the inequality:
$n \lambda < d \sin 30^{\circ}$
$n < \frac{d \sin 30^{\circ}}{\lambda} = \frac{0.3 \times 10^{-3} \times 0.5}{500 \times 10^{-9}}$
$n < \frac{0.15 \times 10^{-3}}{5 \times 10^{-7}} = \frac{0.15}{5} \times 10^4 = 0.03 \times 10^4 = 300$.
Since $n$ must be an integer and $|n| < 300$,the possible values for $n$ are $0, \pm 1, \pm 2, \dots, \pm 299$.
The total number of maxima is $299$ (positive) $+ 299$ (negative) $+ 1$ (central maximum at $n=0$) $= 599$.

(C) In a $YDSE$ experiment,the intensity at any point on the screen is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference between the two waves.
At the mid-point of the screen,the path difference due to the geometry is zero.
When a slab of thickness $t$ and refractive index $\mu$ is placed in front of one slit,an additional path difference $\Delta x = (\mu - 1)t$ is introduced.
The corresponding phase difference is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} (\mu - 1)t$.
The intensity at the mid-point is $I = 4I_0 \cos^2(\frac{\pi t}{\lambda}(\mu - 1))$.
As $\mu$ increases from $1$,the term $(\mu - 1)$ increases,causing the phase difference $\phi$ to increase.
The intensity $I$ follows a $\cos^2$ variation,starting from a maximum value $(4I_0)$ at $\mu = 1$ (where $\phi = 0$) and decreasing as $\mu$ increases,following the shape of a squared cosine function.

(A) Let the phase difference introduced by the mica sheet be $\phi$. The intensity at the center of the screen is given by $I = I_0 \cos^2(\phi/2)$. Given $I = 0.75 I_0$,we have $\cos^2(\phi/2) = 3/4$,which implies $\cos(\phi/2) = \pm \sqrt{3}/2$. Thus,$\phi/2 = n\pi \pm \pi/6$,or $\phi = 2n\pi \pm \pi/3$.
For $n=1$,$\phi = 2\pi \pm \pi/3$,giving $\phi = 5\pi/3$ or $7\pi/3$.
For $n=2$,$\phi = 4\pi \pm \pi/3$,giving $\phi = 11\pi/3$ or $13\pi/3$.
For $n=3$,$\phi = 6\pi \pm \pi/3$,giving $\phi = 17\pi/3$ or $19\pi/3$.
The condition that the second minimum is above the center and the third maximum is below the center constrains the value of $\phi$. Based on the shift analysis,the phase difference $\phi$ must satisfy $3\pi < \phi < 4\pi$. Among the options,$\pi/3$ does not fall in the required range.

(A) For points at a large distance $(|x| >> d)$,the path difference between waves from adjacent sources is $\Delta x = d \cos \theta$. For a point on the $x$-axis at a large distance,$\theta = 0$,so $\Delta x = d$.
The total amplitude at a distant point is the sum of the four waves: $A = A_0(1 + e^{ikd} + e^{i2kd} + e^{i3kd})$,where $k = 2\pi / \lambda$.
This is a geometric series with $r = e^{ikd}$. The sum is $A = A_0 \frac{1 - e^{i4kd}}{1 - e^{ikd}}$.
The intensity $I$ is proportional to $|A|^2 = A_0^2 \left| \frac{1 - e^{i4kd}}{1 - e^{ikd}} \right|^2 = A_0^2 \frac{\sin^2(2kd)}{\sin^2(kd/2)}$.
For dark points (intensity = $0$),we need $\sin(2kd) = 0$ but $\sin(kd/2) \neq 0$.
$2kd = n\pi$ (where $n$ is an integer). For $n=1$,$2(2\pi/\lambda)d = \pi \implies d = \lambda / 4$.
Thus,if $d = \lambda / 4$,the intensity is zero,and the points appear dark.

(D) In a $YDSE$ setup,the path difference $\Delta x$ at a point on the screen at a distance $y$ from the central axis is given by $\Delta x = \frac{yd}{D}$.
For a point directly in front of one of the slits,the distance from the central axis is $y = \frac{d}{2}$.
Substituting this into the path difference formula: $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.
For destructive interference (missing wavelengths),the path difference must be an odd multiple of half the wavelength: $\Delta x = (2m + 1) \frac{\lambda}{2}$,where $m = 0, 1, 2, ...$.
Equating the two expressions: $\frac{d^2}{2D} = (2m + 1) \frac{\lambda}{2}$.
Solving for $\lambda$: $\lambda = \frac{d^2}{(2m + 1)D}$.
For $m = 0$,$\lambda = \frac{d^2}{D}$.
For $m = 1$,$\lambda = \frac{d^2}{3D}$.
Thus,both wavelengths $\frac{d^2}{D}$ and $\frac{d^2}{3D}$ are missing.

(D) In $YDSE$ with white light,the path difference for the central fringe is $0$ for all wavelengths. Since all wavelengths overlap at the center,the central fringe appears white.
As we move away from the center,the path difference $\Delta x = d \sin \theta$ increases. For a given point,different wavelengths satisfy the condition for constructive interference at slightly different positions. Since violet light has the shortest wavelength,the first order maximum for violet appears closest to the central fringe.
Because different wavelengths have different positions for their maxima and minima,the fringes overlap significantly,preventing the formation of a completely dark fringe (where intensity would be zero for all wavelengths simultaneously).
Therefore,all the given statements are correct.

(D) In a Young's Double Slit Experiment $(YDSE)$,the fringe width $\beta$ 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.
Since the wavelength of blue light $(\lambda_{blue})$ is smaller than the wavelength of red light $(\lambda_{red})$,the fringe width $\beta$ decreases when the source is changed from red to blue.
As $\beta$ decreases,the consecutive fringes come closer to each other.
Since the fringe width decreases,more fringes can fit within the same space on the screen,which means the number of maxima formed on the screen increases.
Therefore,both statements $(B)$ and $(C)$ are correct.

(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 used,$D$ is the distance between the slits and the screen,and $d$ is the separation between the two slits.
For the fringes to be more closely spaced,the fringe width $\beta$ must decrease.
From the formula,$\beta$ is directly proportional to the wavelength $\lambda$ $(\beta \propto \lambda)$.
Since the wavelength of blue light is less than the wavelength of green light $(\lambda_{\text{blue}} < \lambda_{\text{green}})$,using blue light instead of green light will decrease the fringe width,making the fringes more closely spaced.

(B) In a $YDSE$,the intensity of light coming from the slits is proportional to the width of the slits. If the slits have unequal widths,the amplitudes of the waves from the two slits,$a_1$ and $a_2$,will be unequal $(a_1 \neq a_2)$.
The intensity of the interference pattern is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For destructive interference,the intensity is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Since $I_1 \neq I_2$,$I_{min} \neq 0$. Therefore,the positions of minimum intensity will not be completely dark.

(D) In Young's Double Slit Experiment,the fringe width $\beta$ 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 this formula,it is clear that $\beta \propto \frac{1}{d}$,meaning the fringe width changes in inverse proportion to $d$. This confirms statement $(B)$.
The position of the $n$-th minimum is given by $y_n = (2n - 1) \frac{\lambda D}{2d}$. Since $y_n$ depends on $d$,changing $d$ will shift the positions of all minima on the screen. This confirms statement $(C)$.
Therefore,both statements $(B)$ and $(C)$ are correct.

(D) In Young's Double Slit Experiment,the condition for obtaining a sharp interference pattern is that the source slit must be narrow and the light must be coherent.
When the distance $\ell$ between the source slit and the double slits $S_1S_2$ decreases,the spatial coherence of the light reaching the slits $S_1$ and $S_2$ decreases.
$1$. As $\ell$ decreases,the angular width of the source as seen from the slits increases,which reduces the visibility of the fringes. If $\ell$ becomes too small,the fringe pattern may disappear (Option $A$).
$2$. The reduction in spatial coherence causes the fringes to lose their sharpness (Option $B$).
$3$. The fringe width is given by $\beta = \frac{\lambda D}{d}$. Since $\beta$ depends only on the wavelength $\lambda$,the distance between the slits $d$,and the distance to the screen $D$,it remains unchanged when $\ell$ is varied (Option $C$).
Therefore,all the given statements are correct.

(C) The condition for the $n^{th}$ bright fringe in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
Given that the $3^{rd}$ bright fringe of the known light $(\lambda_1 = 590 \ nm)$ coincides with the $4^{th}$ bright fringe of the unknown light $(\lambda_2 = \lambda)$:
$\frac{3 \lambda_1 D}{d} = \frac{4 \lambda_2 D}{d}$
$3 \lambda_1 = 4 \lambda_2$
$\lambda_2 = \frac{3}{4} \times 590 \ nm$
$\lambda_2 = 0.75 \times 590 \ nm = 442.5 \ nm$.

(B) For bright fringes to coincide,the path difference must be an integer multiple of both wavelengths. Let $n_1$ be the order for $\lambda_1 = 650 \ nm$ and $n_2$ be the order for $\lambda_2 = 520 \ nm$.
The condition for coincidence is $y = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$,which implies $n_1 \lambda_1 = n_2 \lambda_2$.
$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520 \ nm}{650 \ nm} = \frac{4}{5}$.
For the least distance,we take the smallest integers $n_1 = 4$ and $n_2 = 5$.
Now,calculate the position $y$ using $n_1 = 4$,$\lambda_1 = 650 \times 10^{-9} \ m$,$D = 1.5 \ m$,and $d = 0.5 \times 10^{-3} \ m$:
$y = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times 650 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}} \ m$.
$y = \frac{4 \times 650 \times 1.5}{0.5} \times 10^{-6} \ m = 7800 \times 10^{-6} \ m = 7.8 \times 10^{-3} \ m = 7.8 \ mm$.

(D) The angular width of the central maximum in a single slit diffraction is given by $\theta = \frac{2\lambda}{d}$,where $d$ is the slit width.
Given $\theta = 60^o = \frac{\pi}{3}$ radians,but typically for small angles $\sin \theta \approx \theta$. However,using the standard formula $\frac{2\lambda}{d} = 2 \sin^{-1}(\frac{\lambda}{d})$. For $60^o$ angular width,$\frac{\lambda}{d} = \sin(30^o) = 0.5$.
Thus,$\lambda = 0.5 \times d = 0.5 \times 1 \mu m = 0.5 \mu m$.
For Young's double slit experiment,the fringe width is $\beta = \frac{\lambda D}{d'}$,where $d'$ is the slit separation.
Given $\beta = 1 \ cm = 10^{-2} \ m$,$D = 50 \ cm = 0.5 \ m$,and $\lambda = 0.5 \times 10^{-6} \ m$.
$10^{-2} = \frac{0.5 \times 10^{-6} \times 0.5}{d'}$.
$d' = \frac{0.25 \times 10^{-6}}{10^{-2}} = 0.25 \times 10^{-4} \ m = 25 \times 10^{-6} \ m = 25 \mu m$.

(B) The intensity distribution in a $YDSE$ experiment is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.
From the graph,we observe that the minimum intensity $(I_{min})$ is zero. Since $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$,$I_{min} = 0$ implies that $I_1 = I_2$. Thus,the sources must have the same intensity.
Furthermore,for a stable interference pattern to be observed on the screen,the sources $S_1$ and $S_2$ must be coherent. Coherent sources are defined as sources that maintain a constant phase difference over time. Therefore,the statement that $S_1$ and $S_2$ have a constant phase difference is a fundamental requirement for the observed interference pattern.

(A) For Young's Double Slit Experiment,the condition for the $n^{th}$ bright fringe is given by the path difference $\Delta x = d \sin \theta = n \lambda$.
For the first bright fringe $(n = 1)$,the condition becomes $d \sin \theta = \lambda$.
When the apparatus is in air,$\lambda_{air} = \lambda$ and $\theta_1 = 45^{\circ}$,so $d \sin 45^{\circ} = \lambda$.
When the apparatus is immersed in a liquid of refractive index $\mu$,the wavelength changes to $\lambda_{liquid} = \frac{\lambda}{\mu}$. The new angle is $\theta_2 = 30^{\circ}$,so $d \sin 30^{\circ} = \frac{\lambda}{\mu}$.
Dividing the two equations: $\frac{d \sin 45^{\circ}}{d \sin 30^{\circ}} = \frac{\lambda}{\lambda / \mu} = \mu$.
Substituting the values: $\mu = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

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

(C) The intensity at any point in $YDSE$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_{\max} = 4I_0$ and $\phi$ is the phase difference.
Given that the intensity at the center becomes half of the maximum intensity,we have $I = I_{\max}/2 = 2I_0$.
Substituting this into the formula: $2I_0 = 4I_0 \cos^2(\phi/2) \Rightarrow \cos^2(\phi/2) = 1/2 \Rightarrow \phi/2 = \pi/4 \Rightarrow \phi = \pi/2$.
The phase difference introduced by the mica sheet of thickness $t$ and refractive index $\mu$ is $\phi = (2\pi/\lambda)(\mu - 1)t$.
Equating the two expressions for $\phi$: $\pi/2 = (2\pi/\lambda)(3/2 - 1)t$.
$\pi/2 = (2\pi/\lambda)(1/2)t = \pi t / \lambda$.
Solving for $t$,we get $t = \lambda/2$.

(A) In $YDSE$,the path difference at the central point $(y=0)$ is $\Delta x = 0$ for all wavelengths.
Since the condition for constructive interference is $\Delta x = n\lambda$,for $n=0$,$\Delta x = 0$ is satisfied for all wavelengths $\lambda$ present in white light.
Therefore,all wavelengths produce their zero-order maxima at the central point $(y=0)$.
Since all colors overlap at the center,they combine to form a white fringe.
Thus,Statement-$1$ is True,Statement-$2$ is True,and Statement-$2$ is the correct explanation for Statement-$1$.

(A) The fringe shift $x$ produced by a plate of thickness $t$ and refractive index $\mu$ is given by $x = \frac{(\mu - 1)tD}{d}$,where $D$ is the distance to the screen and $d$ is the slit separation.
Since $t$,$D$,and $d$ are constant for both cases,the shift $x$ is directly proportional to $(\mu - 1)$.
For the first plate: $x \propto (\mu_1 - 1)$,where $\mu_1 = 1.5$.
For the second plate: $\frac{3}{2}x \propto (\mu_2 - 1)$.
Dividing the two expressions: $\frac{(3/2)x}{x} = \frac{\mu_2 - 1}{\mu_1 - 1}$.
$\frac{3}{2} = \frac{\mu_2 - 1}{1.5 - 1}$.
$\frac{3}{2} = \frac{\mu_2 - 1}{0.5}$.
$\mu_2 - 1 = \frac{3}{2} \times 0.5 = 0.75$.
$\mu_2 = 1.75$.

(B) Let the intensity of each source be $I_0$. The maximum intensity is $I_{max} = 4I_0$.
We are looking for points where the intensity $I = \frac{I_{max}}{2} = 2I_0$.
The resultant intensity is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi)$.
Setting $I = 2I_0$,we get $2I_0 = 2I_0(1 + \cos \phi)$,which implies $1 + \cos \phi = 1$,so $\cos \phi = 0$.
This gives the phase difference $\phi = \frac{\pi}{2}$.
The path difference $\Delta x$ is related to the phase difference by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Thus,$\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x$,which gives $\Delta x = \frac{\lambda}{4}$.
For small angles,the path difference is $\Delta x = d \sin \theta \approx d \theta$.
Therefore,$d \theta = \frac{\lambda}{4}$,which gives $\theta = \frac{\lambda}{4d}$.
The angular separation between the points on either side of the central maximum is $\Delta \theta = \theta - (-\theta) = 2\theta = 2 \times \frac{\lambda}{4d} = \frac{\lambda}{2d}$.

(D) The formula for 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 separation between the slits.
According to the problem,the new distance $D' = 2D$ and the new slit separation $d' = \frac{d}{2}$.
Substituting these values into the formula for the new fringe width $\beta'$:
$\beta' = \frac{D' \lambda}{d'} = \frac{(2D) \lambda}{(d/2)} = 4 \left( \frac{D \lambda}{d} \right) = 4\beta$.
Therefore,the fringe width becomes quadrupled.

(C) In Young's double-slit experiment,the fringe width $\beta$ 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.
As the viewing screen is moved farther from the slits,the distance $D$ increases.
Since $\beta \propto D$,an increase in $D$ leads to an increase in the fringe width $\beta$,meaning the interference fringes move farther apart.
Additionally,as the screen moves farther away,the intensity of light reaching the screen decreases due to the inverse square law and the spreading of the beam,making the pattern less bright.
Therefore,the pattern becomes less bright and the fringes move farther apart.

(B) The position of the $n^{th}$ maxima for wavelength $\lambda_1$ is given by $y_n = \frac{n \lambda_1 D}{d}$.
The position of the $m^{th}$ maxima for wavelength $\lambda_2$ is given by $y_m = \frac{m \lambda_2 D}{d}$.
Since the maxima coincide exactly in front of one of the slits,the distance from the central fringe is $y = \frac{d}{2}$.
Thus,$\frac{n \lambda_1 D}{d} = \frac{d}{2} \implies n \lambda_1 = \frac{d^2}{2D} = \frac{(3 \times 10^{-3})^2}{2 \times 1.5} = \frac{9 \times 10^{-6}}{3} = 3 \times 10^{-6} \ m = 30000 \ \mathring{A}$.
Similarly,$m \lambda_2 = 30000 \ \mathring{A}$.
For option $B$: $n=5, m=6$. Then $\lambda_1 = \frac{30000}{5} = 6000 \ \mathring{A}$ and $\lambda_2 = \frac{30000}{6} = 5000 \ \mathring{A}$.
Both wavelengths $6000 \ \mathring{A}$ and $5000 \ \mathring{A}$ lie within the given range $4500 \ \mathring{A} < \lambda < 7000 \ \mathring{A}$.
Therefore,the correct values are $n=5, m=6, \lambda_1 = 6000 \ \mathring{A}$.

(B) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
Given intensity $I = k$ at $\Delta x = \lambda$,we have $k = I_{max} \cos^2(2\pi/2) = I_{max} \cos^2(\pi) = I_{max}(1)^2 = I_{max}$. So,$I_{max} = k$.
Now,for path difference $\Delta x = \lambda/4$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity at this point is $I = I_{max} \cos^2(\phi/2) = k \cos^2(\pi/4) = k \cdot (1/\sqrt{2})^2 = k/2$.

(A) When a glass plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the beams,the path difference introduced is $\Delta x = (\mu - 1)t$.
The shift in the central maximum is given by $\Delta y = \frac{D}{d}(\mu - 1)t$.
For the intensity at the original central position to remain unchanged,the new path difference at that point must correspond to a condition of constructive interference,i.e.,$\Delta x = n\lambda$,where $n$ is an integer.
Since the intensity remains the same,the path difference must be an integer multiple of the wavelength. For the minimum thickness,we take the smallest non-zero integer $n = 1$.
$(\mu - 1)t = 1 \cdot \lambda$
Given $\mu = 1.5$,we have $(1.5 - 1)t = \lambda$.
$0.5t = \lambda \implies t = 2\lambda$.

(D) The formula for fringe width in Young's double-slit experiment is $\beta = \frac{\lambda D}{d}$,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.
Since the wavelength of blue light $(\lambda_{b})$ is less than the wavelength of yellow light $(\lambda_{y})$,i.e.,$\lambda_{b} < \lambda_{y}$.
As $\beta \propto \lambda$,the fringe width $\beta$ will decrease when yellow light is replaced by blue light.

(B) The width of the segment on the screen is constant. The width of a fringe is given by $\beta = \frac{\lambda D}{d}$.
For a segment containing $n$ fringes,the total width $W = n \beta = n \frac{\lambda D}{d}$.
Since the segment width $W$ remains the same for both wavelengths,we have:
$n_1 \lambda_1 = n_2 \lambda_2$
Given $n_1 = 12$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$:
$12 \times 600 = n_2 \times 400$
$n_2 = \frac{12 \times 600}{400}$
$n_2 = 18$
Therefore,$18$ fringes will be observed.

(B) The condition for maxima in $YDSE$ is given by $\Delta x = d \sin \theta = n \lambda$,where $n$ is an integer.
Given $d = 700 \, nm$ and $\lambda = 200 \, nm$.
Substituting these values: $700 \sin \theta = n \times 200$.
Thus,$\sin \theta = \frac{200n}{700} = \frac{2n}{7}$.
Since the maximum value of $\sin \theta$ is $1$,we have $|\frac{2n}{7}| \le 1$,which implies $|n| \le \frac{7}{2} = 3.5$.
Therefore,the possible integer values for $n$ are $n = 0, \pm 1, \pm 2, \pm 3$.
Counting these values: $0, 1, -1, 2, -2, 3, -3$,we get a total of $7$ maxima.

(C) The path difference at point $P$ is given by $\Delta x = \frac{xd}{D}$.
The phase difference $\phi$ at point $P$ is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \frac{xd}{D}$.
Since the fringe width $\beta = \frac{\lambda D}{d}$,we can write $\phi = \frac{2\pi x}{\beta}$.
The intensity at any point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Assuming $I_1 = I_2 = I$,we have $I = 2I + 2I \cos \phi = 4I \cos^2 \frac{\phi}{2}$.
Given the intensity at the central maximum is $I_0 = 4I$,we substitute $I = \frac{I_0}{4}$ and $\phi = \frac{2\pi x}{\beta}$:
$I_P = 4 \left( \frac{I_0}{4} \right) \cos^2 \left( \frac{2\pi x / \beta}{2} \right) = I_0 \cos^2 \frac{\pi x}{\beta}$.

(B) In $YDSE$,the position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$.
For the fringes to coincide,their positions must be equal: $y_n = y_m$.
Therefore,$n \lambda_1 = m \lambda_2$.
Substituting the given values: $n(6000) = m(5500)$.
This simplifies to $\frac{n}{m} = \frac{5500}{6000} = \frac{11}{12}$.
Since $n$ and $m$ must be integers,the smallest values are $n = 11$ and $m = 12$.

(B) Given: $\lambda = 6.5 \times 10^{-7} \, m$,$d = 1 \, mm = 10^{-3} \, m$,$D = 1 \, m$.
The position of the $n^{th}$ bright fringe is given by $x_{n, bright} = n \frac{\lambda D}{d}$.
For the $5^{th}$ bright fringe $(n=5)$:
$x_{5, bright} = 5 \times \frac{6.5 \times 10^{-7} \times 1}{10^{-3}} = 32.5 \times 10^{-4} \, m = 3.25 \, mm$.
The position of the $n^{th}$ dark fringe is given by $x_{n, dark} = (2n - 1) \frac{\lambda D}{2d}$.
For the $3^{rd}$ dark fringe $(n=3)$:
$x_{3, dark} = (2 \times 3 - 1) \times \frac{6.5 \times 10^{-7} \times 1}{2 \times 10^{-3}} = 5 \times \frac{6.5 \times 10^{-4}}{2} = 16.25 \times 10^{-4} \, m = 1.625 \, mm$.
The distance between the $5^{th}$ bright fringe and the $3^{rd}$ dark fringe is:
$\Delta x = |x_{5, bright} - x_{3, dark}| = |3.25 - 1.625| \, mm = 1.625 \, mm \approx 1.63 \, mm$.

(A) The path difference $\delta$ at a point on the screen directly in front of one of the slits is given by $\delta = \frac{b^2}{2d}$.
For destructive interference (missing wavelengths),the path difference must be an odd multiple of $\frac{\lambda}{2}$,i.e.,$\delta = (2n-1)\frac{\lambda}{2}$ where $n = 1, 2, 3, ...$.
Setting $\delta = \frac{b^2}{2d}$,we get $\frac{b^2}{2d} = (2n-1)\frac{\lambda}{2}$.
For $n=1$,$\frac{b^2}{2d} = \frac{\lambda_1}{2} \implies \lambda_1 = \frac{b^2}{d}$.
For $n=2$,$\frac{b^2}{2d} = \frac{3\lambda_2}{2} \implies \lambda_2 = \frac{b^2}{3d}$.
Thus,the missing wavelengths are $\frac{b^2}{d}$ and $\frac{b^2}{3d}$.

(C) The resultant intensity $I_R$ in a double slit experiment is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Given that the intensity from each slit is $I_1 = I_2 = I$ and the phase difference $\phi = 60^{\circ}$.
Substituting these values into the formula:
$I_R = I + I + 2\sqrt{I \cdot I} \cos 60^{\circ}$
$I_R = 2I + 2I \cdot (1/2)$
$I_R = 2I + I = 3I$.
Therefore,the resultant intensity is $3I$.

(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 $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}$.
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 $I = I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = I_0 \cdot \frac{3}{4}$.
Therefore,$\frac{I}{I_0} = \frac{3}{4}$.

(C) In a Young's double-slit experiment,the condition for constructive or destructive interference at any point on the screen is based on the path difference between the waves from the two slits.
Let the slits be at $S_1$ and $S_2$. For any point $P(x, y)$ on the screen,the path difference $\Delta = S_2P - S_1P = n\lambda$ (for maxima) or $(n + 1/2)\lambda$ (for minima).
The locus of points having a constant path difference from two fixed points (the slits) is a hyperbola.
Therefore,the interference fringes formed on a screen are hyperbolic in shape.

(A) The position of the $n^{th}$ maximum in Young's double slit experiment is given by the formula $x_n = \frac{n \lambda D}{d}$,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 distance between the two slits.
For the seventh maximum $(n=7)$ with wavelength $\lambda_1$,the distance from the central maximum is $d_1 = \frac{7 \lambda_1 D}{d}$.
For the seventh maximum $(n=7)$ with wavelength $\lambda_2$,the distance from the central maximum is $d_2 = \frac{7 \lambda_2 D}{d}$.
Taking the ratio of the two distances:
$\frac{d_1}{d_2} = \frac{\frac{7 \lambda_1 D}{d}}{\frac{7 \lambda_2 D}{d}} = \frac{\lambda_1}{\lambda_2}$.

(A) The fringe width $\beta$ for Young's double-slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$.
Given values are:
$\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \, m = 6 \times 10^{-7} \, m$
$D = 1 \, m$
$d = 0.6 \, mm = 0.6 \times 10^{-3} \, m$
Substituting these values into the formula:
$\beta = \frac{6 \times 10^{-7} \times 1}{0.6 \times 10^{-3}}$
$\beta = \frac{6}{0.6} \times 10^{-4} = 10 \times 10^{-4} = 10^{-3} \, m$
Since $10^{-3} \, m = 1 \, mm$,the distance between two consecutive dark fringes is $1 \, mm$.

(B) The path difference at the central point $O$ on the screen due to the source $S$ being at a distance $y$ from the central axis is given by $\Delta x = \frac{a y}{h}$.
For constructive interference (maximum brightness) at the central point,the path difference must be an integer multiple of the wavelength: $\Delta x = n \lambda$,where $n = 0, 1, 2, \dots$.
Thus,$\frac{a y}{h} = n \lambda \Rightarrow y = \frac{n \lambda h}{a}$.
The velocity of the source is $v = \frac{dy}{dt}$.
The rate of change of the order $n$ is given by $\frac{dn}{dt} = \frac{a}{h \lambda} \frac{dy}{dt} = \frac{a v}{h \lambda}$.
This represents the frequency of the fringes passing through the central point,which corresponds to the frequency of change in brightness.

(D) For constructive interference,the path difference $\Delta x$ for the $n^{th}$ bright fringe is given by $\Delta x = n\lambda$,where $n$ is the order of the fringe.
For the third bright fringe,$n = 3$,so the path difference is $\Delta x = 3\lambda$.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Substituting the value of $\Delta x$,we get $\phi = \frac{2\pi}{\lambda} \times 3\lambda = 6\pi$.
Therefore,the phase difference is $6\pi$.