MHT CET 2023 Physics Question Paper with Answer and Solution

(B) In the diffraction of light by a single slit,the linear width of the central (principal) maximum is given by the formula:
$W_{c} = \frac{2 \lambda D}{d}$
According to the problem,the linear width of the principal maximum is equal to the width of the slit,so $W_{c} = d$.
Substituting this into the formula:
$d = \frac{2 \lambda D}{d}$
Rearranging the equation to solve for $D$:
$D = \frac{d^2}{2 \lambda}$

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

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

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

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

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

(B) The condition for the $n^{th}$ dark fringe in a single slit diffraction pattern is given by $d \sin \theta = n \lambda$. For small angles, $\sin \theta \approx \tan \theta = \frac{y_n}{D}$.
Thus, the position of the $n^{th}$ dark fringe is $y_n = \frac{n \lambda D}{d}$.
Given: $\lambda = 600 \,nm = 600 \times 10^{-9} \,m$, $D = 2 \,m$, $d = 1 \,mm = 10^{-3} \,m$.
For the first dark fringe $(n=1)$:
$y_1 = \frac{1 \times 600 \times 10^{-9} \times 2}{10^{-3}} = 1200 \times 10^{-6} \,m = 1.2 \times 10^{-3} \,m = 1.2 \,mm$.
The distance between the first dark fringe on either side of the central bright fringe is $2 y_1 = 2 \times 1.2 \,mm = 2.4 \,mm$.

(A) The formula for fringe width $W$ in a Biprism experiment is given by $W = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slit and the screen,and $d$ is the slit separation.
Since the fringe width $W$ and the wavelength $\lambda$ are constant for both setups,we have $W = \frac{\lambda D_1}{d_1} = \frac{\lambda D_2}{d_2}$.
This implies that $\frac{D_1}{d_1} = \frac{D_2}{d_2}$,or $\frac{D_1}{D_2} = \frac{d_1}{d_2}$.
Given the ratio of slit separation $\frac{d_1}{d_2} = \frac{2}{3}$,it follows that the ratio of the distances is $\frac{D_1}{D_2} = \frac{2}{3}$.

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

(D) The shift in the central fringe due to the introduction of a transparent sheet of thickness $t$ and refractive index $\mu$ is given by the formula: $\Delta x = \frac{(\mu - 1)tD}{d}$.
Given that the central fringe shifts to the position of the $25^{th}$ bright fringe,the shift is equal to $25\beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.
Equating the shift: $\frac{(\mu - 1)tD}{d} = 25 \frac{\lambda D}{d}$.
This simplifies to: $(\mu - 1)t = 25\lambda$.
Given values: $t = 2.9 \times 10^{-3} \ cm = 2.9 \times 10^{-5} \ m$ and $\lambda = 5800 \ \mathring{A} = 5800 \times 10^{-10} \ m$.
Substituting the values: $\mu - 1 = \frac{25 \times 5800 \times 10^{-10}}{2.9 \times 10^{-5}}$.
$\mu - 1 = \frac{145000 \times 10^{-10}}{2.9 \times 10^{-5}} = \frac{1.45 \times 10^{-5}}{2.9 \times 10^{-5}} = 0.5$.
Therefore,$\mu = 1 + 0.5 = 1.50$.

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

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

(C) For single slit diffraction, the condition for the $n^{th}$ minimum is given by $d \sin \theta = n \lambda$.
For small angles, $\sin \theta \approx \tan \theta = \frac{y}{D}$.
For the first minimum, $n = 1$, so $d \frac{y}{D} = \lambda$.
Rearranging for the slit width $d$, we get $d = \frac{\lambda D}{y}$.
Given: $\lambda = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$, $D = 2 \text{ m}$, and $y = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}$.
Substituting the values: $d = \frac{(5 \times 10^{-7} \text{ m}) \times (2 \text{ m})}{5 \times 10^{-3} \text{ m}} = 2 \times 10^{-4} \text{ m}$.
Converting to centimeters: $d = 2 \times 10^{-4} \times 10^2 \text{ cm} = 0.02 \text{ cm}$.

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

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

(D) In a single-slit diffraction pattern,the central maximum is the brightest and widest. The width of the central maximum is $2\lambda D/a$,while the width of the secondary maxima is $\lambda D/a$. As the order of the fringes increases,the intensity decreases significantly. Therefore,the diffraction fringes obtained by a single slit have unequal width and unequal intensity.

(D) The condition for the $n^{th}$ maximum in a single slit diffraction pattern is given by $d \sin \theta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, ...$
For the first maximum $(n=1)$,the position $x$ on the screen is given by $x = \frac{3 \lambda D}{2 d}$.
Given: $\lambda_1 = 590 \,nm$,$\lambda_2 = 596 \,nm$,$D = 1.5 \,m$,$d = 2 \times 10^{-6} \,m$.
The separation between the positions of the first maximum is $\Delta x = x_2 - x_1 = \frac{3 D}{2 d} (\lambda_2 - \lambda_1)$.
Substituting the values: $\Delta x = \frac{3 \times 1.5}{2 \times 2 \times 10^{-6}} \times (596 - 590) \times 10^{-9} \,m$.
$\Delta x = \frac{4.5}{4 \times 10^{-6}} \times 6 \times 10^{-9} \,m$.
$\Delta x = 1.125 \times 10^6 \times 6 \times 10^{-9} \,m = 6.75 \times 10^{-3} \,m = 6.75 \,mm$.

(C) The shift in the fringe pattern $\Delta x$ is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.
The width of a bright fringe is given by: $\beta = \frac{\lambda D}{d}$.
Given that the shift is equal to the fringe width,we have $\Delta x = \beta$.
Therefore,$\frac{(\mu - 1) t D}{d} = \frac{\lambda D}{d}$.
Simplifying this,we get $(\mu - 1) t = \lambda$.
Rearranging for the refractive index $\mu$: $\mu = \frac{\lambda}{t} + 1$.
Substituting the given values: $\lambda = 6 \times 10^{-5} \ cm$ and $t = 12 \times 10^{-5} \ cm$.
$\mu = \frac{6 \times 10^{-5}}{12 \times 10^{-5}} + 1 = 0.5 + 1 = 1.5$.

(D) The energy difference between two orbits in a hydrogen atom is given by the Rydberg formula for wavenumber: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Since the frequency $f$ is related to wavelength $\lambda$ by $f = \frac{C}{\lambda}$,we can write $\frac{f}{C} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Rearranging for frequency,we get $f = RC \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given $n_1 = 2$ and $n_2 = 3$,we substitute these values into the equation:
$f = RC \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = RC \left( \frac{1}{4} - \frac{1}{9} \right)$.
Calculating the fraction: $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.
Thus,the frequency is $f = \frac{5RC}{36}$.