MHT CET 2023 Physics Question Paper with Answer and Solution

(C) The fundamental frequency of an open pipe is $f_{o,fund} = \frac{v}{2L_o}$. The first overtone of an open pipe is the second harmonic,given by $f_{o,1} = 2 \times \frac{v}{2L_o} = \frac{v}{L_o}$.
The fundamental frequency of a closed pipe is $f_{c,fund} = \frac{v}{4L_c}$. The first overtone of a closed pipe is the third harmonic,given by $f_{c,1} = 3 \times \frac{v}{4L_c} = \frac{3v}{4L_c}$.
Given that the frequencies of the first overtones are equal:
$\frac{v}{L_o} = \frac{3v}{4L_c}$
Rearranging the terms to find the ratio of the length of the closed pipe $(L_c)$ to the open pipe $(L_o)$:
$\frac{L_c}{L_o} = \frac{3}{4}$
Therefore,the ratio of the length of the closed pipe to the open pipe is $3:4$.

(D) For an open organ pipe,the effective length $(L)$ considering the end correction $(e)$ at both ends is given by:
$L = l + 2e$
Since the end correction for an open pipe is $e = 0.6r$ at each end,the effective length becomes:
$L = l + 2(0.6r) = l + 1.2r$
The fundamental frequency $(f)$ of an open organ pipe is given by the formula:
$f = \frac{V}{2L}$
Substituting the value of $L$ into the frequency formula:
$f = \frac{V}{2(l + 1.2r)}$

(B) When sound waves fall normally on a perfectly reflecting wall, a stationary wave is formed due to the superposition of incident and reflected waves.
At a perfectly reflecting wall, a node is formed because the wall is a rigid boundary.
The nodes are points of zero displacement, while antinodes are points of maximum displacement (maximum amplitude).
The distance of the first antinode from the wall (node) is given by $d = \lambda/4$.
First, calculate the wavelength $\lambda$ using the formula $\lambda = v/f$, where $v = 300 \,ms^{-1}$ and $f = 600 \,Hz$.
$\lambda = 300/600 = 0.5 \,m$.
Now, calculate the distance $d = \lambda/4 = 0.5/4 = 0.125 \,m = 1/8 \,m$.
Thus, the shortest distance from the wall where particles have maximum amplitude is $1/8 \,m$.

(C) Given: Velocity of sound $v = 320 \,m/s$, Frequency $f = 480 \,Hz$, Number of vibrations $N = 180$.
The relationship between velocity, frequency, and wavelength is $v = f \lambda$.
Therefore, the wavelength $\lambda = \frac{v}{f} = \frac{320}{480} = \frac{2}{3} \,m$.
The total distance $D$ covered by the wave after $N$ vibrations is given by $D = N \times \lambda$.
Substituting the values: $D = 180 \times \frac{2}{3} = 120 \,m$.

(D) Stationary waves (also known as standing waves) are formed by the superposition of two identical waves traveling in opposite directions in a bounded medium.
These waves require a medium for propagation and reflection to occur at the boundaries.
Since mechanical waves can propagate through all three states of matter (solids,liquids,and gases),stationary waves can be produced in all these media.
For example,stationary waves are produced in a stretched string (solid),in a column of water (liquid),and in an air column (gas).

(B) The given equation of the harmonic progressive wave is $y=A \sin (100 \pi t+3 x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t+kx)$,we get the wave number $k=3 \ m^{-1}$.
The relationship between wave number $k$ and wavelength $\lambda$ is $k = \frac{2 \pi}{\lambda}$.
Substituting the value of $k$,we have $3 = \frac{2 \pi}{\lambda}$,which gives $\lambda = \frac{2 \pi}{3} \ m$.
The path difference $\Delta x$ and phase difference $\Delta \phi$ are related by the formula $\Delta x = \frac{\lambda}{2 \pi} \times \Delta \phi$.
Given the phase difference $\Delta \phi = \frac{\pi}{3}$ radian.
Substituting the values,we get $\Delta x = \frac{2 \pi / 3}{2 \pi} \times \frac{\pi}{3} = \frac{1}{3} \times \frac{\pi}{3} = \frac{\pi}{9} \ m$.

(C) The mass of the wire is $m = \frac{W}{g} = \frac{50}{10} = 5 \,kg$.
The linear mass density of the wire is $\mu = \frac{m}{L} = \frac{5}{20} = 0.25 \,kg/m$.
At the midpoint of the wire,the tension $T$ is equal to the weight of the lower half of the wire.
The mass of the lower half is $m' = \frac{m}{2} = 2.5 \,kg$.
Thus,$T = m'g = 2.5 \times 10 = 25 \,N$.
The speed of the wave $v$ is given by $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values: $v = \sqrt{\frac{25}{0.25}} = \sqrt{100} = 10 \,ms^{-1}$.

(A) Let the velocity of the pulse at the lower end be $v_1$ and at the top be $v_2$.
Since the frequency $n$ of the wave remains constant,we have $\lambda = \frac{v}{n}$,which implies $\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1}$.
The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Thus,$v \propto \sqrt{T}$.
At the lower end,the tension $T_1$ is due to the block of mass $m_2$,so $T_1 = m_2 g$.
At the top end,the tension $T_2$ is due to the total mass of the rope and the block,so $T_2 = (m_1 + m_2) g$.
Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2)g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.
Taking the reciprocal,we get $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1 + m_2}} = \left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$.

(D) The phase of a wave is given by the argument of the sine function,$\phi = kx - \omega t$.
For the first wave,$\phi_1 = 20x - 30t$.
For the second wave,$\phi_2 = 25x - 40t$.
Given $x = 5 \ cm$ and $t = 2 \ s$:
$\phi_1 = 20(5) - 30(2) = 100 - 60 = 40 \ \text{radians}$.
$\phi_2 = 25(5) - 40(2) = 125 - 80 = 45 \ \text{radians}$.
The phase difference $\Delta \phi = |\phi_2 - \phi_1| = |45 - 40| = 5 \ \text{radians}$.

(D) At $N.T.P$, the temperature $T_0 = 273 \,K$. The velocity of sound $v_0$ is given by $v_0 = f \lambda_0 = 220 \,Hz \times 1.5 \,m = 330 \,m/s$.
When the temperature increases to $27^{\circ} C$, the new temperature $T = 273 + 27 = 300 \,K$.
The velocity of sound in air is proportional to the square root of the absolute temperature: $\frac{v}{v_0} = \sqrt{\frac{T}{T_0}}$.
Substituting the values: $v = 330 \times \sqrt{\frac{300}{273}} = 330 \times 1.05 = 346.5 \,m/s$.
The new wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{346.5}{220} = 1.575 \,m$.
The increase in wavelength $\Delta \lambda = \lambda - \lambda_0 = 1.575 \,m - 1.5 \,m = 0.075 \,m$.
Rounding to the nearest value provided, we get $0.07 \,m$.

(B) The given equation of the progressive wave is $y = 12 \sin (5t - 4x)$.
The standard form of a progressive wave is $y = A \sin (\omega t - kx)$,where $k = \frac{2\pi}{\lambda}$.
Comparing the given equation with the standard form,we get the wave number $k = 4$.
The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by $\Delta \phi = k \cdot \Delta x$.
Given phase difference $\Delta \phi = 90^{\circ} = \frac{\pi}{2}$ radians.
Substituting the values: $\frac{\pi}{2} = 4 \cdot \Delta x$.
Therefore,$\Delta x = \frac{\pi}{2 \cdot 4} = \frac{\pi}{8}$.

(A) Let the initial length and tension be $l$ and $T$ respectively. The mass per unit length $m$ remains constant.
The fundamental frequency is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
After the changes:
New length $l' = l - 0.40l = 0.6l = \frac{3}{5}l$.
New tension $T' = T + 0.44T = 1.44T = \frac{144}{100}T$.
The new frequency $n'$ is given by $n' = \frac{1}{2l'} \sqrt{\frac{T'}{m}}$.
Taking the ratio of final to initial frequency:
$\frac{n'}{n} = \frac{l}{l'} \sqrt{\frac{T'}{T}} = \frac{l}{0.6l} \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} \times \sqrt{1.44} = \frac{1}{0.6} \times 1.2 = \frac{1.2}{0.6} = 2$.
Therefore,the ratio of final to initial frequency is $2:1$.

(A) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Thus,$n = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
The first overtone of the first wire is $n_1 = 2 \times n_{1, \text{fund}} = 2 \times \frac{1}{2l_1 r_1} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
The second overtone of the second wire is $n_2 = 3 \times n_{2, \text{fund}} = 3 \times \frac{1}{2l_2 r_2} \sqrt{\frac{T}{\pi \rho}} = \frac{3}{2l_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $n_1 = n_2$,we have $\frac{1}{l_1 r_1} = \frac{3}{2l_2 r_2}$.
Rearranging for the ratio of lengths: $\frac{l_1}{l_2} = \frac{2r_2}{3r_1}$.
Since $r_1 = 2r_2$,substituting this gives $\frac{l_1}{l_2} = \frac{2r_2}{3(2r_2)} = \frac{1}{3}$.

(C) The given wave equation is $Y=2 \sin (0.01 x+30 t)$.
Comparing this with the standard wave equation $Y=A \sin (kx+\omega t)$:
We get the wave number $k = 0.01 \ cm^{-1}$ and angular frequency $\omega = 30 \ rad/s$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{30}{0.01} = 3000 \ cm/s$.
Converting the velocity to meters per second: $v = 30 \ m/s$.
The length of the string $L$ is the distance traveled by the wave in time $t = 0.5 \ s$.
$L = v \times t = 30 \ m/s \times 0.5 \ s = 15 \ m$.

(C) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \text{mass per unit length} = \rho \cdot A = \rho \cdot \pi \cdot (\frac{d}{2})^2 = \frac{\rho \pi d^2}{4}$.
Substituting $\mu$,we get $f = \frac{1}{2l} \sqrt{\frac{T}{\frac{\rho \pi d^2}{4}}} = \frac{1}{ld} \sqrt{\frac{T}{\pi \rho}}$.
For wire $A$: $n = \frac{1}{ld} \sqrt{\frac{T}{\pi \rho_1}}$.
For wire $B$: $n = \frac{1}{l(2d)} \sqrt{\frac{2T}{\pi \rho_2}}$.
Equating the frequencies: $\frac{1}{ld} \sqrt{\frac{T}{\pi \rho_1}} = \frac{1}{2ld} \sqrt{\frac{2T}{\pi \rho_2}}$.
Canceling common terms: $\sqrt{\frac{1}{\rho_1}} = \frac{1}{2} \sqrt{\frac{2}{\rho_2}}$.
Squaring both sides: $\frac{1}{\rho_1} = \frac{1}{4} \cdot \frac{2}{\rho_2} = \frac{1}{2\rho_2}$.
Therefore,$2\rho_2 = \rho_1$,which gives $\rho_2 = \frac{\rho_1}{2}$.

(B) Let the frequency of the $1^{\text{st}}$ tuning fork be $n_1$.
The frequencies are in an arithmetic progression with common difference $d = 5 \text{ Hz}$.
The frequency of the $41^{\text{st}}$ tuning fork is given by $n_{41} = n_1 + (41 - 1) \times d$.
$n_{41} = n_1 + 40 \times 5 = n_1 + 200$.
Given that the frequency of the last fork is double the first,$n_{41} = 2n_1$.
Substituting this into the equation: $2n_1 = n_1 + 200$.
Solving for $n_1$: $n_1 = 200 \text{ Hz}$.
Therefore,$n_{41} = 2 \times 200 = 400 \text{ Hz}$.

(C) The fundamental frequency of a vibrating string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Mass per unit length $m = \frac{\text{Volume} \times \text{density}}{l} = \frac{(\pi R^2 l) \rho}{l} = \pi R^2 \rho$.
Since $\pi$ and $\rho$ are constant,$m \propto R^2$.
Substituting this into the frequency formula,we get $n \propto \frac{1}{l \sqrt{R^2}} \propto \frac{1}{lR}$.
Given $l_2 = 2l_1$ and $R_2 = 2R_1$,the ratio of the new frequency $n_2$ to the original frequency $n_1$ is:
$\frac{n_2}{n_1} = \frac{l_1 R_1}{l_2 R_2} = \frac{l_1 R_1}{(2l_1)(2R_1)} = \frac{1}{4}$.
Therefore,the new frequency $n_2 = \frac{n}{4}$.

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the wire.
Since the wire carries a block of mass $M$,the tension $T = Mg = V\rho g$,where $V$ is the volume of the block.
Thus,$n \propto \sqrt{T} \propto \sqrt{\rho Vg}$.
When the block is immersed in a liquid of density $\sigma$,the effective weight (tension) becomes $T' = V(\rho - \sigma)g$.
The new frequency $n'$ is given by $n' \propto \sqrt{T'}$.
Taking the ratio: $\frac{n'}{n} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{V(\rho - \sigma)g}{V\rho g}} = \sqrt{\frac{\rho - \sigma}{\rho}}$.
Therefore,$n' = n \left[\frac{\rho - \sigma}{\rho}\right]^{\frac{1}{2}}$.

(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are constant,$f \propto \frac{1}{l}$.
Let $f_1$ and $f_2$ be the frequencies of the two tuning forks corresponding to lengths $l_1 = 0.97 \ m$ and $l_2 = 0.96 \ m$ respectively.
Thus,$f_1 = \frac{k}{0.97}$ and $f_2 = \frac{k}{0.96}$,where $k = \frac{1}{2} \sqrt{\frac{T}{\mu}}$.
Given the beat frequency is $f_2 - f_1 = 5 \ Hz$.
Substituting the values: $\frac{k}{0.96} - \frac{k}{0.97} = 5$.
$k \left( \frac{0.97 - 0.96}{0.96 \times 0.97} \right) = 5$.
$k \left( \frac{0.01}{0.9312} \right) = 5 \implies k = 5 \times 93.12 = 465.6$.
Now,$f_1 = \frac{465.6}{0.97} = 480 \ Hz$.
And $f_2 = \frac{465.6}{0.96} = 485 \ Hz$.

(D) The given wave equation is $Y = 10 \sin \left(\frac{2 \pi t}{30} + \alpha\right)$.
At $t = 0$,the displacement $Y = 5 \text{ cm}$.
Substituting these values: $5 = 10 \sin \left(\frac{2 \pi \times 0}{30} + \alpha\right)$.
$5 = 10 \sin \alpha \implies \sin \alpha = 0.5$.
Since $\sin 30^{\circ} = 0.5$,we have $\alpha = 30^{\circ} = \frac{\pi}{6} \text{ rad}$.
The total phase $\phi$ at any time $t$ is given by $\phi = \frac{2 \pi t}{30} + \alpha$.
At $t = 7.5 \text{ s}$:
$\phi = \frac{2 \pi \times 7.5}{30} + \frac{\pi}{6}$.
$\phi = \frac{15 \pi}{30} + \frac{\pi}{6} = \frac{\pi}{2} + \frac{\pi}{6}$.
$\phi = \frac{3 \pi + \pi}{6} = \frac{4 \pi}{6} = \frac{2 \pi}{3} \text{ rad}$.

(C) The resonant frequencies of a string fixed at both ends are given by $f_n = n f_0$,where $f_0$ is the fundamental frequency (lowest resonant frequency) and $n = 1, 2, 3, \dots$.
Given that there are no resonant frequencies between $315 \,Hz$ and $420 \,Hz$,these two frequencies must be consecutive harmonics,i.e.,$f_n = 315 \,Hz$ and $f_{n+1} = 420 \,Hz$.
Thus,$n f_0 = 315$ and $(n+1) f_0 = 420$.
Subtracting the two equations: $(n+1) f_0 - n f_0 = 420 - 315$,which gives $f_0 = 105 \,Hz$.
Therefore,the lowest resonant frequency is $105 \,Hz$.

(A) For the first stone projected vertically upwards,the maximum height reached is $h_1 = \frac{v^2}{2g}$.
The potential energy at the highest point is $PE_1 = mgh_1 = mg \left( \frac{v^2}{2g} \right) = \frac{mv^2}{2}$.
For the second stone,the angle with the vertical is $60^{\circ}$,which means the angle with the horizontal is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
The maximum height reached by the second stone is $h_2 = \frac{v^2 \sin^2 \theta}{2g} = \frac{v^2 \sin^2 30^{\circ}}{2g} = \frac{v^2 (0.5)^2}{2g} = \frac{v^2}{8g}$.
The potential energy at the highest point is $PE_2 = mgh_2 = mg \left( \frac{v^2}{8g} \right) = \frac{mv^2}{8}$.
The ratio of their potential energies is $\frac{PE_1}{PE_2} = \frac{mv^2/2}{mv^2/8} = \frac{8}{2} = 4:1$.

(B) The total vertical distance fallen by the ball is $(s+h)$.
Since the ball starts from rest,the loss in gravitational potential energy is $mg(s+h)$.
This energy is stored as elastic potential energy in the spring when it is compressed by a distance $h$.
The elastic potential energy stored in the spring is given by $\frac{1}{2} kh^2$.
According to the law of conservation of energy,the loss in potential energy of the ball equals the gain in potential energy of the spring:
$mg(s+h) = \frac{1}{2} kh^2$
Solving for the spring constant $k$:
$k = \frac{2mg(s+h)}{h^2}$

(D) At the lowest point of a vertical circle,the tension $T$ in the string provides the necessary centripetal force and balances the weight of the body.
The equation for tension at the lowest point is $T = \frac{mv^2}{r} + mg$.
To just complete the vertical loop,the minimum velocity $v$ at the lowest point must be $v = \sqrt{5gr}$.
Substituting this value of $v$ into the tension equation:
$T = \frac{m(\sqrt{5gr})^2}{r} + mg$
$T = \frac{m(5gr)}{r} + mg$
$T = 5mg + mg$
$T = 6mg$.
Thus,the apparent weight (tension) at the lowest point is $6mg$.

(C) solar cell is a $P-N$ junction device that converts light energy into electrical energy. Option $C$ states that it converts electrical energy into light,which is incorrect. Therefore,it is the mismatch.

(D) When the diode $D$ is forward biased,it acts as a short circuit (zero resistance). The circuit consists of two resistors of $40 \ \Omega$ and $400 \ \Omega$ in parallel.
Therefore,the effective resistance $R_1$ is given by:
$R_1 = \frac{40 \times 400}{40 + 400} = \frac{16000}{440} = \frac{1600}{44} = \frac{400}{11} \ \Omega$
When the diode $D$ is reverse biased,it acts as an open circuit (infinite resistance). No current flows through the upper branch.
Therefore,the effective resistance $R_2$ is simply the resistance of the lower branch:
$R_2 = 400 \ \Omega$
Now,calculating the ratio $R_1: R_2$:
$\frac{R_1}{R_2} = \frac{400/11}{400} = \frac{1}{11}$
Thus,the ratio $R_1: R_2$ is $1: 11$.

(D) photodiode is operated under reverse bias conditions. In circuit $(B)$,the $p$-side is connected to the negative terminal and the $n$-side to the positive terminal,which is reverse bias. Thus,$(B)$ is correct for a photodiode.
An $LED$ (Light Emitting Diode) is operated under forward bias conditions. In circuit $(D)$,the $p$-side is connected to the positive terminal and the $n$-side to the negative terminal,which is forward bias. Thus,$(D)$ is correct for an $LED$.
Therefore,the correct circuit diagrams for normal operation are $(B)$ and $(D)$.

(D) In a $p-n$ junction diode,when it is in forward bias,the positive terminal of the battery is connected to the $p$-region and the negative terminal to the $n$-region.
This external electric field opposes the internal barrier electric field.
As a result,the majority charge carriers are pushed towards the junction,which reduces the width of the depletion layer.
Therefore,the barrier potential decreases,allowing current to flow easily through the diode.

(D) When a $p-n$ junction is forward biased,the positive terminal of the external battery is connected to the $p$-type region and the negative terminal to the $n$-type region.
This external electric field opposes the internal electric field of the depletion region.
As a result,the effective potential barrier $(V_B)$ decreases.
Due to the reduction in the potential barrier,the majority charge carriers are pushed towards the junction,which causes the width $(X)$ of the depletion region to decrease.
Therefore,both the potential barrier and the width of the depletion region decrease.

(B) junction diode is forward biased when the potential at the $P$-terminal $(V_P)$ is higher than the potential at the $N$-terminal $(V_N)$,i.e.,$V_P > V_N$.
Let us analyze each option:
$A$: $V_P = +5 \ V$,$V_N = +10 \ V$. Here $V_P < V_N$,so it is reverse biased.
$B$: $V_P = -1.0 \ V$,$V_N = -1.5 \ V$. Here $V_P > V_N$ (since $-1.0 > -1.5$),so it is forward biased.
$C$: $V_P = 0 \ V$,$V_N = +1 \ V$. Here $V_P < V_N$,so it is reverse biased.
$D$: $V_P = -2 \ V$,$V_N = 0 \ V$. Here $V_P < V_N$,so it is reverse biased.
Therefore,the correct option is $B$.

(C) In a $P-N$ junction diode,when the reverse bias voltage is increased to a large value,it eventually reaches the breakdown voltage.
At this point,the covalent bonds in the crystal lattice break,leading to a rapid generation of charge carriers.
Consequently,the current through the diode suddenly increases.

(D) The graph shows current $I$ in $mA$ on the $y$-axis and voltage $V$ in $Volt$ on the $x$-axis in the first quadrant,indicating forward bias characteristics.
Among the given options,a Light Emitting Diode $(LED)$ operates in forward bias,where the current increases exponentially with voltage after the threshold voltage is reached,leading to light emission due to the recombination of charge carriers.
Therefore,the graph represents the forward bias characteristics of an $LED$.

(B) In insulators,the energy gap between the valence band and the conduction band is very large (typically $> 3 \ eV$).
At absolute zero temperature,the valence band is completely filled with electrons,and the conduction band is completely empty.
Due to the large energy gap,electrons cannot jump from the valence band to the conduction band even at room temperature,which is why insulators do not conduct electricity.

(C) In insulators,the valence band is completely filled with electrons,and the conduction band is completely empty.
There is a very large energy band gap (typically $E_g > 3 \ eV$) between the valence band and the conduction band.
Because of this large energy gap,electrons cannot easily jump from the valence band to the conduction band,even at room temperature.
Therefore,the band gap is very high and the conduction band is empty.
Thus,the correct option is $C$.

(D) At room temperature,thermal energy is sufficient to excite some electrons from the valence band to the conduction band in a semiconductor.
However,the number of electrons excited is very small compared to the total number of electrons in the valence band.
Therefore,the valence band remains almost completely filled,and the conduction band contains only a small number of electrons,making it partially filled.
Thus,the correct description is that the valence band is almost completely filled and the conduction band is partially filled.

(D) The power gain of an amplifier is defined as the product of voltage gain and current gain.
$\text{Power Gain} = \text{Voltage Gain} \times \text{Current Gain}$
Therefore,the ratio of power gain to voltage gain is equal to the current gain.
In a common emitter configuration,the current gain is denoted by $\beta$.
Thus,$\frac{\text{Power Gain}}{\text{Voltage Gain}} = \beta$.

(C) In a transistor,the emitter-base junction is always forward-biased to allow charge carriers to move into the base.
In an $n-p-n$ transistor,the emitter is $n$-type,so it injects electrons (majority carriers) into the base.
In a $p-n-p$ transistor,the emitter is $p$-type,so it injects holes (majority carriers) into the base.
Therefore,the primary difference in operation is the type of charge carrier injected by the emitter into the base region: $p-n-p$ injects holes,while $n-p-n$ injects electrons.

(D) We know the relationship between current gains $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.
Alternatively,we can use the fundamental current relation $I_E = I_C + I_B$.
Dividing by $I_C$,we get $\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$.
This implies $\frac{1}{\alpha} = 1 + \frac{1}{\beta}$.
Rearranging the terms,we get $\frac{1}{\alpha} - \frac{1}{\beta} = 1$.
Substituting the given values: $\frac{1}{0.98} - \frac{1}{49} \approx 1.0204 - 0.0204 = 1$.

(D) It is given that $80 \%$ of the emitted electrons reach the collector, which means the collector current $I_c$ is $80 \%$ of the emitter current $I_e$.
$I_c = 0.80 \times I_e$
Given $I_c = 28 \,mA$, we have $28 = 0.80 \times I_e$.
$I_e = \frac{28}{0.80} = 35 \,mA$.
Using the relation for transistor currents, $I_e = I_c + I_b$, we can find the base current $I_b$.
$I_b = I_e - I_c = 35 \,mA - 28 \,mA = 7 \,mA$.

(B) The circuit consists of an $OR$ gate with inputs $A$ and $C$, where $C$ is the output of gate $G$ with inputs $A$ and $B$. Thus, the output is $Y = A + C = A + (A \text{ gate } B)$.
Let's test the options for gate $G$:
$1$. If $G$ is $AND$, then $C = A \cdot B$. The output $Y = A + (A \cdot B)$.
For $(A, B) = (0, 0)$, $Y = 0 + (0 \cdot 0) = 0$.
For $(A, B) = (0, 1)$, $Y = 0 + (0 \cdot 1) = 0$.
For $(A, B) = (1, 0)$, $Y = 1 + (1 \cdot 0) = 1$.
For $(A, B) = (1, 1)$, $Y = 1 + (1 \cdot 1) = 1$.
This matches the given truth table perfectly. Therefore, $G$ must be an $AND$ gate.

(D) The output of the circuit is given by $Y = A + C$, where $C$ is the output of gate $G$ with inputs $A$ and $B$. Thus, $C = A \text{ (gate } G) B$.
Let us test the options for gate $G$:
If $G$ is $NOR$, then $C = \overline{A+B}$.
The output $Y = A + \overline{A+B}$.
Let's verify this for all inputs:
$1$. For $A=0, B=0$: $C = \overline{0+0} = 1$. Then $Y = 0 + 1 = 1$. (Matches)
$2$. For $A=0, B=1$: $C = \overline{0+1} = 0$. Then $Y = 0 + 0 = 0$. (Matches)
$3$. For $A=1, B=0$: $C = \overline{1+0} = 0$. Then $Y = 1 + 0 = 1$. (Matches)
$4$. For $A=1, B=1$: $C = \overline{1+1} = 0$. Then $Y = 1 + 0 = 1$. (Matches)
Since all values match the given truth table, the gate $G$ is $NOR$.

(C) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y_1 = A + B$.
This output $Y_1$ is connected to both inputs of a $NAND$ gate. Let the inputs of the $NAND$ gate be $X_1 = Y_1$ and $X_2 = Y_1$.
The output of the $NAND$ gate is $Y = \overline{X_1 \cdot X_2} = \overline{Y_1 \cdot Y_1} = \overline{Y_1}$.
Since $Y_1 = A + B$,the final output is $Y = \overline{A + B}$.
This is the Boolean expression for a $NOR$ gate.

Input $(A, B)$	Output of $OR$ gate $(Y_1 = A+B)$	Final Output $(Y = \overline{Y_1})$
$0, 0$	$0$	$1$
$0, 1$	$1$	$0$
$1, 0$	$1$	$0$
$1, 1$	$1$	$0$

The truth table matches that of a $NOR$ gate.

(D) $NAND$ gate is a combination of an $AND$ gate followed by a $NOT$ gate.
First,the $AND$ operation is performed on inputs $A$ and $B$,which results in $A \cdot B$.
Then,the $NOT$ operation is applied to this result,which inverts the output.
Therefore,the output $Y$ of a $NAND$ gate is given by the Boolean expression $Y = \overline{A \cdot B}$.

(C) The given circuit consists of a $NAND$ gate and a $NOT$ gate,whose outputs are fed into an $OR$ gate.
$1$. The inputs $A$ and $B$ are fed into a $NAND$ gate. The output of the $NAND$ gate is $\overline{A \cdot B}$.
$2$. The input $C$ is fed into a $NOT$ gate. The output of the $NOT$ gate is $\overline{C}$.
$3$. These two outputs are then fed into an $OR$ gate. The $OR$ operation of two inputs $X$ and $Z$ is $X+Z$.
$4$. Therefore,the final output $Y$ is the $OR$ sum of the two previous outputs: $Y = \overline{A \cdot B} + \overline{C}$.

(D) The circuit consists of a $NAND$ gate followed by an $AND$ gate.
In the provided diagram,the output of the $NAND$ gate is connected to only one input terminal of the $AND$ gate.
An $AND$ gate is a multi-input logic gate that requires at least two input signals to perform its logical operation.
Since the $AND$ gate in this circuit is receiving only one input,it cannot function as a standard logic gate.
Therefore,the gate is not operational.

(C) The given circuit consists of an $OR$ gate and a $NAND$ gate whose outputs are fed into an $AND$ gate.
Let the inputs be $A$ and $B$.
The output of the $OR$ gate is $(A + B)$.
The output of the $NAND$ gate is $(\overline{A \cdot B})$.
These are the inputs to the final $AND$ gate.
Therefore,the final output $Y$ is given by:
$Y = (A + B) \cdot (\overline{A \cdot B})$
Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$.
$Y = (A + B) \cdot (\bar{A} + \bar{B})$
$Y = A \cdot \bar{A} + A \cdot \bar{B} + B \cdot \bar{A} + B \cdot \bar{B}$
Since $A \cdot \bar{A} = 0$ and $B \cdot \bar{B} = 0$,we have:
$Y = 0 + A \cdot \bar{B} + B \cdot \bar{A} + 0$
$Y = A \cdot \bar{B} + \bar{A} \cdot B$
This is the Boolean expression for an $XOR$ gate.

(D) The circuit consists of one $NAND$ gate,one $NOT$ gate,one $NOR$ gate,and one final $NOR$ gate.
Let the output of the top $NAND$ gate be $Y_1 = \overline{A \cdot B}$.
Let the output of the bottom branch ($NOT$ gate followed by $NOR$ gate) be $Y_2 = \overline{\overline{A} + B}$.
The final output $Y$ is the output of a $NOR$ gate with inputs $Y_1$ and $Y_2$,so $Y = \overline{Y_1 + Y_2} = \overline{(\overline{A \cdot B}) + (\overline{\overline{A} + B})}$.
Testing the options:
For $A=1, B=1$:
$Y_1 = \overline{1 \cdot 1} = 0$
$Y_2 = \overline{\overline{1} + 1} = \overline{0 + 1} = 0$
$Y = \overline{0 + 0} = 1$.
Thus,for $A=1, B=1$,the output $Y$ is $1$.

(C) $NAND$ gate has two inputs,say $A$ and $B$,and its output is given by $Y = \overline{A \cdot B}$.
If the two inputs are shorted,then $A = B$. Let this common input be $A$.
Substituting this into the $NAND$ expression,we get $Y = \overline{A \cdot A}$.
Since $A \cdot A = A$ in Boolean algebra,the expression becomes $Y = \overline{A}$.
The expression $Y = \overline{A}$ represents the operation of a $NOT$ gate.
Therefore,when the inputs of a $NAND$ gate are shorted,it functions as a $NOT$ gate.

(C) The given circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate.
$1$. The input $A$ passes through a $NOT$ gate to become $\overline{A}$,and input $B$ goes directly to the first $AND$ gate. The output of this $AND$ gate is $\overline{A} \cdot B$.
$2$. The input $B$ passes through a $NOT$ gate to become $\overline{B}$,and input $A$ goes directly to the second $AND$ gate. The output of this $AND$ gate is $A \cdot \overline{B}$.
$3$. These two outputs are then fed into an $OR$ gate. The final output $Y$ is the sum of these two expressions: $Y = (\overline{A} \cdot B) + (A \cdot \overline{B})$.
This is the Boolean expression for an $XOR$ gate.

(D) The given circuit consists of two $NOR$ gates whose outputs are fed into an $OR$ gate. Let the inputs be $A$ and $B$. The output of the top $NOR$ gate is $C = \overline{A+B}$ and the output of the bottom $NOR$ gate is $D = \overline{A+B}$. These are fed into an $OR$ gate,so the final output is $Y = C + D = \overline{A+B} + \overline{A+B} = \overline{A+B}$.
The truth table for this combination is:

$A$	$B$	$C = \overline{A+B}$	$D = \overline{A+B}$	$Y = C + D$
$0$	$0$	$1$	$1$	$1$
$0$	$1$	$0$	$0$	$0$
$1$	$0$	$0$	$0$	$0$
$1$	$1$	$0$	$0$	$0$

Comparing this with the truth table of standard logic gates,the output $Y$ is $1$ only when both $A$ and $B$ are $0$,which is the characteristic behavior of a $NOR$ gate.

(B) In a half-wave rectifier, the output pulse occurs once per input cycle. Therefore, the output frequency is equal to the input frequency, which is $50 \,Hz$.
In a full-wave rectifier, the output pulse occurs twice per input cycle (once for each half-cycle). Therefore, the output frequency is double the input frequency, which is $2 \times 50 \,Hz = 100 \,Hz$.
Thus, the output frequencies are $50 \,Hz$ and $100 \,Hz$ respectively.

(C) The maximum efficiency of a full wave rectifier is given by $x = 81.2 \%$.
The maximum efficiency of a half wave rectifier is given by $y = 40.6 \%$.
Comparing these two values,we find that $x = 2 \times 40.6 \% = 2y$.
Therefore,the correct relation is $x = 2y$.

(B) An intrinsic semiconductor is a pure semiconductor without any significant dopant species present.
In an intrinsic semiconductor,charge carriers are generated solely due to thermal excitation.
When a covalent bond breaks due to thermal energy,an electron is excited from the valence band to the conduction band,leaving behind a hole in the valence band.
Since each electron-hole pair is created simultaneously,the number of electrons in the conduction band $(n_{e})$ must be equal to the number of holes in the valence band $(n_{h})$.
Therefore,for an intrinsic semiconductor,$n_{h} = n_{e}$.

(C) Given: Density of $Si$ atoms $= 4 \times 10^{28} \text{ atoms}/m^3$.
After doping with $1 \text{ ppm}$ (part per million) of $Sb$ (antimony),the concentration of $Sb$ atoms is:
$\text{Number of } Sb \text{ atoms} = \frac{4 \times 10^{28}}{10^6} = 4 \times 10^{22} \text{ atoms}/m^3$.
Since $Sb$ is a pentavalent impurity,each $Sb$ atom donates one free electron to the crystal lattice.
Therefore,the total number of free electrons available will be $4 \times 10^{22} \ m^{-3}$.

(A) The resistivity of a semiconductor is given by $\rho = \frac{1}{\sigma} = \frac{1}{e(n_e \mu_e + n_h \mu_h)}$.
When a semiconductor is doped with impurity atoms,the concentration of charge carriers ($n_e$ or $n_h$) increases significantly.
Since the conductivity $\sigma$ is directly proportional to the carrier concentration,the conductivity increases.
Because resistivity $\rho$ is the reciprocal of conductivity,an increase in conductivity leads to a decrease in resistivity.
Therefore,doping generally decreases the resistivity of a semiconductor.

(B) In a transistor,the current gain parameters are defined as $\beta = \frac{I_C}{I_B}$ and $\alpha = \frac{I_C}{I_E}$.
Since $I_E = I_B + I_C$,we can write $\alpha = \frac{I_C}{I_B + I_C}$.
Dividing the numerator and denominator by $I_C$,we get $\alpha = \frac{1}{\frac{I_B}{I_C} + 1} = \frac{1}{\frac{1}{\beta} + 1} = \frac{\beta}{1 + \beta}$.
From this,we can derive $\frac{1}{\alpha} = \frac{1 + \beta}{\beta} = \frac{1}{\beta} + 1$,which implies $\frac{1}{\beta} = \frac{1}{\alpha} - 1$.
Comparing these with the given options,the relation $\alpha = \frac{\beta}{1 - \beta}$ is incorrect.

(B) semiconductor in which the conductivity is solely due to the thermal excitation of electrons from the valence band to the conduction band,resulting in the breaking of covalent bonds,is known as an intrinsic semiconductor. In this state,the number of free electrons is equal to the number of holes $(n_e = n_h)$. Therefore,the correct option is $B$.

(C) The correct answer is $C$.
$1$. LEDs are highly energy-efficient compared to conventional light sources.
$2$. LEDs have a very long operational lifespan when manufactured with high quality.
$3$. The brightness of light emitted by an $LED$ can be easily controlled by varying the forward current passing through it. Therefore,the statement that it cannot be controlled is incorrect.
$4$. The light emitted by LEDs is monochromatic and does not fade over time.

(D) light emitting diode $(LED)$ is a heavily doped $p-n$ junction diode which emits spontaneous radiation when forward biased.
In the forward biased condition,the potential barrier is reduced,allowing electrons from the $n$-region and holes from the $p$-region to cross the junction.
These charge carriers recombine at the junction,releasing energy in the form of photons (light).
Therefore,an $LED$ is always operated in the forward biased condition to emit light.

(C) The position of the $n^{\text{th}}$ bright fringe is given by $y_n = \frac{n \lambda_1 D}{d}$.
The position of the $m^{\text{th}}$ dark fringe is given by $y'_m = \frac{(2m - 1) \lambda_2 D}{2d}$.
Given that the $5^{\text{th}}$ bright band coincides with the $6^{\text{th}}$ dark band,we set $n = 5$ and $m = 6$:
$\frac{5 \lambda_1 D}{d} = \frac{(2 \times 6 - 1) \lambda_2 D}{2d}$
$\frac{5 \lambda_1 D}{d} = \frac{11 \lambda_2 D}{2d}$
Canceling $D$ and $d$ from both sides:
$5 \lambda_1 = \frac{11 \lambda_2}{2}$
$\frac{\lambda_1}{\lambda_2} = \frac{11}{5 \times 2} = \frac{11}{10}$

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

(C) The angular width of the central maximum in a single-slit diffraction experiment 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 slit width.
From this relation,it is clear that the width of the central maximum is inversely proportional to the slit width $(a)$.
Therefore,when the slit width $a$ is decreased,the width of the central maximum increases.

(C) In a single-slit diffraction pattern,the intensity of the principal maximum is proportional to the square of the slit width $(a^2)$ and inversely proportional to the square of the wavelength. However,the total energy incident on the slit also increases as the slit width increases. Specifically,the intensity at the central maximum is given by $I_0 \propto a^2$. When the slit width $a$ is doubled to $2a$,the amplitude of the wavelets increases by a factor of $2$,and since intensity is proportional to the square of the amplitude,the intensity becomes $I \propto (2a)^2 = 4I_0$. However,the width of the central maximum decreases by half $(w = \frac{2\lambda D}{a})$,concentrating the energy into a smaller region. For a standard single-slit diffraction experiment where the source is coherent and the slit width is increased,the intensity of the central maximum increases by a factor of $4$.

(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \Delta x$.
For the first point,$\Delta x_1 = 0$,so $\Delta \phi_1 = 0$. The intensity is $I_1 = I_{max} \cos^2(\frac{\Delta \phi_1}{2}) = I_{max} \cos^2(0) = I_{max}$.
For the second point,$\Delta x_2 = \frac{\lambda}{2}$,so $\Delta \phi_2 = \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{2} = \pi$. The intensity is $I_2 = I_{max} \cos^2(\frac{\pi}{2}) = 0$.
The ratio of intensities is $\frac{I_1}{I_2} = \frac{I_{max}}{0} = \infty$.
Thus,the ratio is $\infty: 1$.

(C) The total phase difference $\Delta \phi$ at point $P$ is given by the sum of the initial phase difference between the sources and the phase difference due to the path difference.
Let $\phi_1$ be the initial phase difference between the sources $A$ and $B$. Given $\phi_1 = 54^{\circ}$.
Converting to radians: $\phi_1 = 54 \times \frac{\pi}{180} = 0.3 \pi \text{ rad}$.
Let $\phi_2$ be the phase difference due to the path difference $\Delta x = PB - PA = 2.5 \lambda$.
The formula for phase difference due to path difference is $\phi_2 = \frac{2 \pi}{\lambda} \times \Delta x$.
Substituting the values: $\phi_2 = \frac{2 \pi}{\lambda} \times 2.5 \lambda = 5 \pi \text{ rad}$.
Since source $A$ is ahead in phase, the total phase difference is $\Delta \phi = \phi_2 + \phi_1 = 5 \pi + 0.3 \pi = 5.3 \pi \text{ rad}$.

(B) The formula for fringe width in a biprism experiment is given by $Z = \frac{\lambda D}{d}$.
Since the wavelength $\lambda$ and the distance between the coherent sources $d$ are kept constant,the ratio $\frac{\lambda}{d}$ is a constant value.
Therefore,we can write $\frac{Z}{D} = \frac{\lambda}{d} = \text{constant}$.
This implies that for different values of $D$ and $Z$,the ratio remains the same:
$\frac{Z_1}{D_1} = \frac{Z_2}{D_2} = \frac{Z_3}{D_3} = \frac{Z_4}{D_4}$.

(B) In Young's double-slit experiment,the intensity of interference fringes is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Initially,$I_1 = I_2 = I_0$,so $I_{max} = 4I_0$ and $I_{min} = 0$.
When one slit is covered,the intensity of light from that slit becomes $I_1' = 0.5 I_0$,while $I_2 = I_0$.
Now,the new maximum intensity is $I_{max}' = (\sqrt{0.5 I_0} + \sqrt{I_0})^2 = I_0(0.5 + 1 + 2\sqrt{0.5}) \approx 2.91 I_0$,which is less than $4I_0$.
The new minimum intensity is $I_{min}' = (\sqrt{I_0} - \sqrt{0.5 I_0})^2 = I_0(1 + 0.5 - 2\sqrt{0.5}) \approx 0.086 I_0$,which is greater than $0$.
Therefore,the intensity of the bright fringe decreases and the intensity of the dark fringe increases.

(A) According to Malus' Law,the intensity of light transmitted through a polarizer is given by $I = I_{max} \cos^2 \theta$,where $\theta$ is the angle between the transmission axes of the two polarizers.
Initially,the principal planes are parallel,so $\theta = 0^{\circ}$ and the intensity is $I_0 = I_{max}$.
When crystal $B$ is rotated by $45^{\circ}$,the new angle between the transmission axes becomes $\theta = 45^{\circ}$.
Substituting the values into the formula:
$I = I_0 \cos^2(45^{\circ})$
$I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2$
$I = I_0 \left(\frac{1}{2}\right) = \frac{I_0}{2}$.

(A) According to Brewster's law,the refractive index $n$ is given by $\tan \theta_{B} = n$,where $\theta_{B}$ is the Brewster angle.
Given $\theta_{B} = 60^{\circ}$,we have $n = \tan 60^{\circ} = \sqrt{3}$.
Now,using Snell's law,$\frac{\sin i}{\sin r} = n$,where $i = 45^{\circ}$ is the angle of incidence and $r$ is the angle of refraction.
Substituting the values,$\sin r = \frac{\sin 45^{\circ}}{n} = \frac{1/\sqrt{2}}{\sqrt{3}} = \frac{1}{\sqrt{6}}$.
Therefore,the angle of refraction is $r = \sin^{-1}\left(\frac{1}{\sqrt{6}}\right)$.

(C) Given: Angle of prism $= A$,Angle of incidence $i = 2A$.
For the light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally.
Let $r$ be the angle of refraction at the first surface. In the triangle formed by the ray inside the prism,the angle at the silvered surface is $90^{\circ}$. Thus,the angle of refraction $r$ is related to the prism angle $A$ by $r + A = 90^{\circ}$ is incorrect here; rather,from the geometry,$r = A$.
Using Snell's law at the first surface:
$\mu = \frac{\sin i}{\sin r} = \frac{\sin 2A}{\sin A}$
Using the trigonometric identity $\sin 2A = 2 \sin A \cos A$:
$\mu = \frac{2 \sin A \cos A}{\sin A} = 2 \cos A$

(D) In a single slit diffraction experiment,the angular width of the central maximum is given by $\theta = \frac{2\lambda}{a}$,where $a$ is the slit width. If the slit width $a$ is doubled $(a' = 2a)$,the new angular width becomes $\theta' = \frac{2\lambda}{2a} = \frac{\theta}{2}$. Thus,the angular width is halved.
Regarding intensity,the intensity of the central maximum is proportional to the square of the slit width $(I \propto a^2)$. If the slit width is doubled,the new intensity $I'$ becomes $I' \propto (2a)^2 = 4a^2 = 4I$. Therefore,the intensity increases by a factor of $4$.

(A) For the $n^{\text{th}}$ minimum in a single-slit diffraction pattern,the condition is $a \sin \theta = n \lambda$,where $n = 1, 2, 3, \dots$
For the $n^{\text{th}}$ secondary maximum,the condition is $a \sin \theta = (2n + 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$
Given for the first minimum $(n=1)$: $a \sin 30^{\circ} = \lambda \Rightarrow a(1/2) = \lambda \Rightarrow a = 2\lambda$.
For the first secondary maximum $(n=1)$: $a \sin \theta = (2(1) + 1) \frac{\lambda}{2} = \frac{3\lambda}{2}$.
Substituting $a = 2\lambda$ into the equation: $(2\lambda) \sin \theta = \frac{3\lambda}{2}$.
$\sin \theta = \frac{3}{4} \Rightarrow \theta = \sin^{-1}\left(\frac{3}{4}\right)$.

(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 $\frac{I}{I_0} = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$.