MHT CET 2022 Physics Question Paper with Answer and Solution

(C) The truth table for an Ex-$OR$ gate is as follows:
| Input $A$ | Input $B$ | Output $Y = A \oplus B$ |
| :--- | :--- | :--- |
| $0$ | $0$ | $0$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ |
The output equation for an Ex-$OR$ gate is $Y = A \oplus B = (\bar{A} \cdot B) + (A \cdot \bar{B})$.
Key points to remember:
$(1)$ The output is low $(0)$ when both inputs are the same.
$(2)$ The output is high $(1)$ when both inputs are different.

(A) Given inputs are $P = 0$,$Q = 1$,$R = 0$,and $S = 1$.
$1$. The output $X$ is obtained from an $AND$ gate with inputs $P$ and $Q$. Thus,$X = P \cdot Q = 0 \cdot 1 = 0$.
$2$. The input to the $NOT$ gate is the output of an $AND$ gate with inputs $R$ and $S$. Let this be $W$. So,$W = R \cdot S = 0 \cdot 1 = 0$. The output $Y$ is the $NOT$ of $W$,so $Y = \overline{W} = \overline{0} = 1$.
$3$. The output $Z$ is obtained from a $NOR$ gate with inputs $X$ and $Y$. Thus,$Z = \overline{X + Y} = \overline{0 + 1} = \overline{1} = 0$.
Therefore,the outputs are $X = 0$,$Y = 1$,and $Z = 0$.

(D) The Boolean expression for an Exclusive-$OR$ (Ex-$OR$) gate is given by:
$Y = A \oplus B = \bar{A} \cdot B + A \cdot \bar{B}$
The output of an Ex-$OR$ gate is $HIGH$ $(1)$ only when the inputs are at different logic levels. If both inputs are the same ($0,0$ or $1,1$),the output is $LOW$ $(0)$.
Evaluating the truth table:
$1$. For $A=0, B=0$: $Y = 0 \oplus 0 = 0$
$2$. For $A=0, B=1$: $Y = 0 \oplus 1 = 1$
$3$. For $A=1, B=0$: $Y = 1 \oplus 0 = 1$
$4$. For $A=1, B=1$: $Y = 1 \oplus 1 = 0$
Comparing this with the given tables,Table $(S)$ matches this truth table.

(B) The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.
$1$. For $A = 0, B = 1$: $Y = \overline{0 \cdot 1} = \overline{0} = 1$. Thus, $C = 1$.
$2$. For $A = 0, B = 0$: $Y = \overline{0 \cdot 0} = \overline{0} = 1$. Thus, $D = 1$.
$3$. For $A = 1, B = 0$: $Y = \overline{1 \cdot 0} = \overline{0} = 1$. Thus, $E = 1$.
$4$. For $A = 1, B = 1$: $Y = \overline{1 \cdot 1} = \overline{1} = 0$. Thus, $F = 0$.
Therefore, the values are $C = 1, D = 1, E = 1, F = 0$.

(D) The given circuit consists of two $NAND$ gates used as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.
Let the inputs be $A$ and $B$.
The output of the first $NAND$ gate (acting as $NOT$) is $\bar{A}$.
The output of the second $NAND$ gate (acting as $NOT$) is $\bar{B}$.
These two outputs are fed into a third $NAND$ gate.
The final output $Y$ is given by $Y = \overline{(\bar{A} \cdot \bar{B})}$.
Using De Morgan's theorem,$\overline{(\bar{A} \cdot \bar{B})} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.
This is the Boolean expression for an $OR$ gate.
Therefore,the combination acts as an $OR$ gate.

(A) full-wave rectifier converts both halves of the input $AC$ cycle into a single polarity. Without a filter,the output consists of a series of pulsating pulses that are always in the same direction (unidirectional). However,because the magnitude of the current varies with time,it is not a steady or constant $DC$. Therefore,the output is a unidirectional but not steady current.

(A) The correct option is $A$.
In a $p$-type semiconductor,the material is formed by doping a tetravalent semiconductor (like $Si$ or $Ge$) with trivalent impurity atoms (like $B$,$Al$,$Ga$,or $In$).
Since trivalent atoms have three valence electrons,they create a vacancy or 'hole' in the crystal lattice when they bond with the tetravalent atoms.
Therefore,holes become the majority charge carriers,and electrons become the minority charge carriers.
Thus,option $A$ is correct.

(B) To obtain an $n$-type semiconductor,pure silicon (group $IV$ element) must be doped with pentavalent (donor) atoms from group $V$ of the periodic table,such as arsenic $(As)$ or phosphorus $(P)$.
To obtain a $p$-type semiconductor,pure silicon must be doped with trivalent (acceptor) atoms from group $III$ of the periodic table,such as boron $(B)$ or gallium $(Ga)$.
Therefore,for $n$-type and $p$-type respectively,we need to dope silicon with arsenic and boron.

(A) The charge carriers that are present in large quantity are called majority charge carriers.
In a $p$-type semiconductor,the majority charge carriers are the holes because the trivalent impurity atoms create vacancies in the valence band.
In an $n$-type semiconductor,the majority charge carriers are the free electrons because the pentavalent impurity atoms provide extra electrons to the conduction band.

(C) In an intrinsic semiconductor,the number of free electrons is very small compared to a conductor at room temperature.
As the temperature increases,more covalent bonds break,leading to an increase in the number of free electrons.
At absolute zero temperature $(T = 0 \ K)$,an intrinsic semiconductor behaves as a perfect insulator because there is insufficient thermal energy to break covalent bonds.
Therefore,there are no free electrons at absolute zero temperature.
Statement $C$ is incorrect because free electrons exist at temperatures above absolute zero.

(A) For an intrinsic semiconductor,the law of mass action states that the product of electron and hole concentration is equal to the square of the intrinsic carrier concentration: $n_e n_h = n_i^2$.
Given:
Intrinsic carrier concentration $n_i = 1.5 \times 10^{16} \ m^{-3}$.
New hole concentration $n_h = 4.5 \times 10^{22} \ m^{-3}$.
Using the formula:
$n_e = \frac{n_i^2}{n_h}$
$n_e = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}}$
$n_e = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}}$
$n_e = 0.5 \times 10^{10} \ m^{-3} = 5 \times 10^9 \ m^{-3}$.

(C) Concept: $A$ $p$-type semiconductor is formed by doping an intrinsic semiconductor with trivalent impurities.
These impurities create acceptor energy levels just above the valence band.
At room temperature,these acceptor levels accept electrons from the valence band,creating a large number of holes in the valence band.
Therefore,in a $p$-type semiconductor,the majority charge carriers are holes and the minority charge carriers are free electrons.

(B) The working principle of these devices is as follows:
$(1)$ Photovoltaic cells convert solar energy directly into electrical energy.
$(2)$ Photo-thermal devices absorb solar radiation to produce heat energy.
$(3)$ $A$ Photodiode is a light-sensitive device that generates current when exposed to light,but it is typically used as a detector rather than a primary solar energy harvester.
$(4)$ An $LED$ (Light Emitting Diode) converts electrical energy into light energy,which is the reverse of a solar cell.
Therefore,devices that primarily work on solar energy are the Photovoltaic cell and Photo-thermal devices.

(C) The width of the central maximum in a single-slit diffraction pattern is given by the formula $\beta = \frac{2\lambda D}{a}$,where $\lambda$ is the wavelength of light,$D$ is the distance of the screen from the slit,and $a$ is the width of the slit.
Since the wavelength of red light $(\lambda_{red})$ is greater than the wavelength of blue light $(\lambda_{blue})$,the width of the diffraction bands $\beta$ is directly proportional to the wavelength $\lambda$.
Therefore,when blue light is replaced by red light,the diffraction bands become broader.

(A) For a single slit diffraction, the condition for the $n^{\text{th}}$ order maximum is given by $a \sin \theta = (n + 1/2) \lambda$.
For small angles, $\sin \theta \approx \tan \theta = x/D$, where $x$ is the distance from the center and $D$ is the distance to the screen.
Thus, $a(x/D) = (n + 1/2) \lambda$.
For the $2^{\text{nd}}$ order maximum, $n = 2$, so $a = \frac{(2 + 1/2) \lambda D}{x} = \frac{2.5 \lambda D}{x}$.
Given: $\lambda = 580 \times 10^{-9} \,m$, $D = 2.5 \,m$, $x = 14.5 \times 10^{-3} \,m$.
Substituting the values: $a = \frac{2.5 \times 580 \times 10^{-9} \times 2.5}{14.5 \times 10^{-3}} \,m$.
$a = \frac{3625 \times 10^{-9}}{14.5 \times 10^{-3}} \,m = 250 \times 10^{-6} \,m = 0.25 \times 10^{-3} \,m \approx 0.26 \times 10^{-3} \,m$.

(B) The condition for the $n^{\text{th}}$ diffraction minimum is given by the path difference $\Delta x = n\lambda$,where $n = 1, 2, 3, \dots$ is the order of the minimum.
For the second minimum,we set $n = 2$,which gives the path difference $\Delta x = 2\lambda$.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the value of $\Delta x$ for the second minimum:
$\phi = \frac{2\pi}{\lambda} (2\lambda) = 4\pi$.
Therefore,the phase difference between the rays coming from the two edges of the slit at the second minimum is $4\pi$.

(B) The angular width of the central maximum in a single-slit diffraction pattern is given by the formula $\theta = \frac{2\lambda}{d}$,where $\lambda$ is the wavelength of the light and $d$ is the width of the slit.
From this relation,it is clear that the angular width $\theta$ is inversely proportional to the slit width $d$ (i.e.,$\theta \propto \frac{1}{d}$).
Therefore,as the slit width $d$ increases,the angular width of the central maximum decreases.

(D) For a single slit diffraction,the condition for minima is $d \sin \theta = n\lambda$ where $n = 1, 2, 3, ...$
For the first minimum $(n=1)$,$d \sin 30^{\circ} = \lambda$.
Since $\sin 30^{\circ} = 0.5$,we have $d(0.5) = \lambda$,which implies $d = 2\lambda$.
The condition for secondary maxima is $d \sin \theta = (n + \frac{1}{2})\lambda$ where $n = 1, 2, 3, ...$
For the first secondary maximum $(n=1)$,$d \sin \theta = (1 + \frac{1}{2})\lambda = \frac{3}{2}\lambda$.
Substituting $d = 2\lambda$ into the equation:
$2\lambda \sin \theta = \frac{3}{2}\lambda$
$\sin \theta = \frac{3}{4}$
$\theta = \sin ^{-1}\left(\frac{3}{4}\right)$.

(C) The linear width of the principal maxima in a single-slit diffraction pattern is given by the formula $\beta = \frac{2 \lambda D}{d}$.
According to the problem,the linear width of the principal maxima is equal to the width of the slit,so $\beta = d$.
Substituting this into the formula,we get $d = \frac{2 \lambda D}{d}$.
Rearranging the equation to solve for $D$,we get $D = \frac{d^{2}}{2 \lambda}$.

(C) The correct option is $C$.
Statement $A$ is false because when the crest of one wave coincides with the crest of another wave,the amplitudes add up,resulting in constructive interference,not destructive interference.
Statement $B$ is true because,by definition,coherent sources are sources that emit light waves of the same frequency and maintain a constant phase difference over time.

(A) In interference,the fringes are equally bright and equally spaced,and the intensity of the bright fringe is four times the intensity of each incident wave.
Resultant intensity at a point is given by $I = I_1 + I_2 + 2\sqrt{I_1}\sqrt{I_2}\cos\delta$,where $I_1$ and $I_2$ are the intensities of the waves from the two sources.
For constructive interference,the crest of one wave coincides with the crest of another wave,or the trough coincides with the trough. Statement $C$ is incorrect because the condition described (crest coinciding with trough) leads to destructive interference.
For constructive interference at the central maxima,the phase difference $\delta = 0$.
If $I_1 = I_2 = I$,then $I_{\max} = I + I + 2\sqrt{I}\sqrt{I}\cos(0) = 4I$.
Thus,statements $A$ and $B$ are correct.

(C) In Young's double slit experiment,the condition for destructive interference (dark fringe) is that the path difference $\Delta x$ must be an odd multiple of half the wavelength: $\Delta x = (2n-1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \ldots$.
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 expression for path difference into the phase difference formula:
$\Delta \phi = \frac{2\pi}{\lambda} \times (2n-1) \frac{\lambda}{2}$.
Simplifying the expression:
$\Delta \phi = (2n-1) \pi$ radians.
Thus,for the $n$th dark fringe,the phase difference is $(2n-1) \pi$.

(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.

(D) For the $n^{th}$ secondary maxima in a single-slit diffraction pattern,the path difference $x$ between the rays from the edges of the slit is given by the condition $x = (2n + 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ represents the order of the secondary maxima.
For the first secondary maximum $(n = 1)$,the path difference is $x_1 = (2(1) + 1) \frac{\lambda}{2} = \frac{3\lambda}{2}$.
The phase difference $\phi$ is related to the path difference $x$ by the formula $\phi = \frac{2\pi}{\lambda} x$.
Substituting the value of $x_1$,we get $\phi_1 = \frac{2\pi}{\lambda} \left( \frac{3\lambda}{2} \right) = 3\pi$.

(D) Let the amplitudes of the two waves be $A_1 = 5x$ and $A_2 = x$.
The maximum intensity $I_{max}$ is proportional to the square of the sum of amplitudes: $I_{max} \propto (A_1 + A_2)^2 = (5x + x)^2 = (6x)^2 = 36x^2$.
The minimum intensity $I_{min}$ is proportional to the square of the difference of amplitudes: $I_{min} \propto (A_1 - A_2)^2 = (5x - x)^2 = (4x)^2 = 16x^2$.
The ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{36x^2}{16x^2} = \frac{36}{16} = \frac{9}{4}$.

(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 $\lambda = \frac{d^2}{D}$.

(D) The path difference introduced by the mica film is given by $\Delta x = (\mu - 1)t$.
In Young's double slit experiment,the shift in the fringe pattern $\Delta y$ is related to the path difference by $\Delta y = \frac{D}{d} \Delta x = \frac{D}{d} (\mu - 1)t$.
Since the fringe width is $\beta = \frac{\lambda D}{d}$,we can write the shift as $\Delta y = \frac{\beta}{\lambda} (\mu - 1)t$.
Given that the shift is equal to the fringe width,$\Delta y = \beta$,we have $\beta = \frac{\beta}{\lambda} (\mu - 1)t$.
This simplifies to $(\mu - 1)t = \lambda$.
Substituting the given values,$\mu - 1 = \frac{\lambda}{t} = \frac{6 \times 10^{-5} \text{ cm}}{12 \times 10^{-5} \text{ cm}} = 0.5$.
Therefore,$\mu = 0.5 + 1 = 1.5$.

(B) The angular separation $\theta$ between the fringes is given by the formula $\theta = \frac{\beta}{D}$,where $\beta$ is the fringe width and $D$ is the distance between the slit and the screen.
Since the fringe width $\beta = \frac{D \lambda}{d}$,substituting this into the expression for $\theta$ gives $\theta = \frac{(\frac{D \lambda}{d})}{D} = \frac{\lambda}{d}$.
Here,$\lambda$ is the wavelength of light and $d$ is the slit separation.
As the expression for $\theta$ does not contain the term $D$,the angular separation is independent of the distance between the slit and the screen.
Therefore,when the distance is doubled,the angular separation remains the same.

(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.

(D) The formula for fringe width is given by $\beta = \frac{\lambda D}{d}$,where $d$ is the distance between the two virtual sources.
For the first position: $W_1 = \frac{\lambda D_1}{d}$
For the second position: $W_2 = \frac{\lambda D_2}{d}$
Subtracting the two equations: $W_1 - W_2 = \frac{\lambda}{d}(D_1 - D_2)$
Rearranging for $d$: $d = \frac{\lambda(D_1 - D_2)}{(W_1 - W_2)}$

(B) The intensity in Young's double slit experiment is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,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 $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
Given intensity $K = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
Now,for path difference $\Delta x = \frac{\lambda}{4}$,the phase difference is $\phi' = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $K'$ is $K' = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $K' = 4I_0 \times (\frac{1}{\sqrt{2}})^2 = 4I_0 \times \frac{1}{2} = 2I_0$.
Since $K = 4I_0$,then $2I_0 = \frac{K}{2}$.
Therefore,the intensity is $\frac{K}{2}$.

(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.

(D) Concept: In a biprism experiment,the position of the $n^{\text{th}}$ dark fringe from the central axis is given by the formula: $y_n = (2n - 1) \frac{\lambda D}{2d}$.
Given that the $4^{\text{th}}$ dark band is formed opposite to one of the slits,the distance of the slit from the central axis is $y = \frac{d}{2}$.
For $n = 4$,the position is $y_4 = (2(4) - 1) \frac{\lambda D}{2d} = \frac{7 \lambda D}{2d}$.
Equating the two expressions for $y_4$: $\frac{7 \lambda D}{2d} = \frac{d}{2}$.
Solving for $\lambda$: $7 \lambda D = d^2 \implies \lambda = \frac{d^2}{7D}$.
Thus,option $(D)$ is correct.

(B) 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 slit and the eyepiece,and $d$ is the distance between the two coherent sources.
Initially,$\beta = \frac{\lambda D}{d}$.
In the new situation,the distance between the sources becomes $d' = 2d$ and the distance between the slit and the eyepiece becomes $D' = 2D$.
The new fringe width $\beta'$ is given by $\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{(2d)} = \frac{\lambda D}{d}$.
Therefore,$\beta' = \beta$,which means the fringe width remains unchanged.

(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) When a thin transparent plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the interfering waves,the additional optical path introduced is given by $\Delta x = (\mu - 1)t$.
According to the problem,this path difference is equal to half the wavelength of light,i.e.,$\Delta x = \frac{\lambda}{2}$.
Equating the two expressions,we get $(\mu - 1)t = \frac{\lambda}{2}$.
Solving for $t$,we find $t = \frac{\lambda}{2(\mu - 1)}$.

(C) The optical path length in a medium of refractive index $\mu$ for a physical path length $L$ is defined as $\Delta = \mu L$.
For the first ray,the optical path is $\Delta_1 = \mu_1 L_1$.
For the second ray,the optical path is $\Delta_2 = \mu_2 L_2$.
The path difference between the two rays is $\Delta x = |\Delta_1 - \Delta_2| = |\mu_1 L_1 - \mu_2 L_2|$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2 \pi}{\lambda} \Delta x$.
Therefore,the phase difference is $\frac{2 \pi}{\lambda} |\mu_1 L_1 - \mu_2 L_2|$.
Comparing this with the given options,the correct expression is $\frac{2 \pi}{\lambda} [\mu_1 L_1 - \mu_2 L_2]$.

(B) Let the intensities of the two waves be $I_1$ and $I_2$.
Given: $I_1/I_2 = 49/16$.
Since intensity $I \propto A^2$,the ratio of amplitudes is $\sqrt{I_1}/\sqrt{I_2} = A_1/A_2 = \sqrt{49}/\sqrt{16} = 7/4$.
The ratio of maximum intensity 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 the values:
$\frac{I_{\max}}{I_{\min}} = \left(\frac{7 + 4}{7 - 4}\right)^2 = \left(\frac{11}{3}\right)^2 = \frac{121}{9}$.
Thus,the ratio is $121:9$.

(B) The condition for destructive interference is that the path difference $\Delta x$ must be an odd multiple of $\frac{\lambda}{2}$.
Given path difference $\Delta x = 31.5 \lambda = \frac{63}{2} \lambda = 63 \left( \frac{\lambda}{2} \right)$.
Since $63$ is an odd integer,the path difference is an odd multiple of $\frac{\lambda}{2}$.
Therefore,the point corresponds to destructive interference,which results in a dark point.

(C) In a series combination,the charge on each capacitor is the same,so $Q_1 = Q_2$.
Let the potential at point $D$ be $V$.
The charge on capacitor $C_1$ is $Q_1 = C_1(V_1 - V)$.
The charge on capacitor $C_2$ is $Q_2 = C_2(V - V_2)$.
Equating the charges: $C_1(V_1 - V) = C_2(V - V_2)$.
Expanding the terms: $C_1 V_1 - C_1 V = C_2 V - C_2 V_2$.
Rearranging to solve for $V$: $C_1 V_1 + C_2 V_2 = V(C_1 + C_2)$.
Therefore,the potential at point $D$ is $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.