MHT CET 2026 Physics Question Paper with Answer and Solution

(A) The fringe width $\beta$ in a biprism experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance of the screen from the slits, and $d$ is the distance between the virtual sources.
Initial state: $\lambda_1 = 5000 \text{ Å}$, $D_1 = 100 \text{ cm}$, $\beta_1 = \frac{\lambda_1 D_1}{d}$.
Final state: $\lambda_2 = 6400 \text{ Å}$, $D_2 = D_1 + x$ (or $D_1 - x$), $\beta_2 = \frac{\lambda_2 D_2}{d}$.
We are given that $\beta_2 = \beta_1 - 0.20 \beta_1 = 0.80 \beta_1$.
Substituting the expressions: $\frac{\lambda_2 D_2}{d} = 0.80 \frac{\lambda_1 D_1}{d}$.
$\lambda_2 D_2 = 0.80 \lambda_1 D_1 \implies 6400 \times D_2 = 0.80 \times 5000 \times 100$.
$6400 \times D_2 = 400000 \implies D_2 = \frac{400000}{6400} = 62.5 \text{ cm}$.
Since the new distance $D_2 = 62.5 \text{ cm}$ is less than the initial distance $D_1 = 100 \text{ cm}$, the screen must be moved towards the source by $100 - 62.5 = 37.5 \text{ cm}$.

(C) $1$. When unpolarised light of intensity $I_0$ passes through the first polaroid $A$, the intensity of the transmitted light becomes $I_1 = I_0/2$.
$2$. The pass-axis of polaroid $A$ is at $0^\circ$ and polaroid $B$ is at $90^\circ$. Polaroid $C$ is placed at an angle of $45^\circ$ to $A$.
$3$. According to Malus's Law, $I = I_{max} \cos^2 \theta$. The intensity after passing through $C$ is $I_2 = I_1 \cos^2(45^\circ) = (I_0/2) \times (1/\sqrt{2})^2 = I_0/4$.
$4$. The angle between the pass-axes of $C$ and $B$ is $90^\circ - 45^\circ = 45^\circ$. The intensity after passing through $B$ is $I_3 = I_2 \cos^2(45^\circ) = (I_0/4) \times (1/\sqrt{2})^2 = I_0/8$.

(A) According to Brewster's law, when light is incident at Brewster's angle $\Phi$, the reflected ray and the refracted ray are perpendicular to each other.
Let $i$ be the angle of incidence, $r$ be the angle of refraction, and $i_p$ be the Brewster's angle $\Phi$.
At Brewster's angle, $i = \Phi$.
From Snell's law, $\mu = \frac{\sin \Phi}{\sin r}$.
Also, Brewster's law states that $\mu = \tan \Phi = \frac{\sin \Phi}{\cos \Phi}$.
Comparing the two expressions for $\mu$, we get $\sin r = \cos \Phi = \sin(90^\circ - \Phi)$, which implies $r = 90^\circ - \Phi$.
The angle between the reflected ray and the normal is $i = \Phi$.
The angle between the refracted ray and the normal is $r = 90^\circ - \Phi$.
The angle between the reflected ray and the refracted ray is $180^\circ - (i + r) = 180^\circ - (\Phi + 90^\circ - \Phi) = 180^\circ - 90^\circ = 90^\circ$.

(B) When unpolarized light of intensity $I$ passes through the first polarizer, the intensity becomes $I_1 = I/2$.
For subsequent polarizers, Malus's Law states that $I_{out} = I_{in} \cos^2(\theta)$, where $\theta = 45^\circ$.
Since $\cos^2(45^\circ) = (1/\sqrt{2})^2 = 1/2$, the intensity after each subsequent polarizer is halved.
After the first polarizer, there are $(n-1)$ additional polarizers.
The intensity after $n$ polarizers is $I_n = (I/2) \times (1/2)^{n-1} = I / 2^n$.
Given $I_n = I/64$, we have $I/2^n = I/64$.
Therefore, $2^n = 64 = 2^6$, which gives $n = 6$.

(B) When unpolarized light of intensity $I_0$ passes through the first polaroid $P_1$, the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.
According to Malus's Law, the intensity of light passing through a polaroid is $I = I_{incident} \cos^2 \theta$, where $\theta$ is the angle between the polarization direction of incident light and the pass axis of the polaroid.
For $P_2$, the angle between its axis and the axis of $P_1$ is $60^\circ$. Thus, the intensity after $P_2$ is $I_2 = I_1 \cos^2 60^\circ = \frac{I_0}{2} \cos^2 60^\circ$.
For $P_3$, the angle between its axis ($90^\circ$ to $P_1$) and the axis of $P_2$ ($60^\circ$ to $P_1$) is $\theta = 90^\circ - 60^\circ = 30^\circ$.
Thus, the final intensity after $P_3$ is $I_3 = I_2 \cos^2 30^\circ = (\frac{I_0}{2} \cos^2 60^\circ) \cos^2 30^\circ = \frac{I_0}{2} \cos^2 60^\circ \cos^2 30^\circ$.

(A) Let the intensity of the unpolarised light be $I_0 = 64 \text{ Wm}^{-2}$.
After passing through the first polarizer, the intensity becomes $I_1 = \frac{I_0}{2} = 32 \text{ Wm}^{-2}$.
Let $\theta$ be the angle between the transmission axes of the first and second polarizers.
The intensity after the second polarizer is $I_2 = I_1 \cos^2 \theta = 32 \cos^2 \theta$.
The angle between the first and third polarizer is $90^\circ$ (crossed).
Therefore, the angle between the second and third polarizer is $(90^\circ - \theta)$.
The intensity after the third polarizer is $I_3 = I_2 \cos^2(90^\circ - \theta) = I_2 \sin^2 \theta$.
Substituting $I_2$, we get $I_3 = (32 \cos^2 \theta) \sin^2 \theta = 32 (\sin \theta \cos \theta)^2 = 32 \left( \frac{\sin 2\theta}{2} \right)^2 = 8 \sin^2 2\theta$.
Given $I_3 = 6 \text{ Wm}^{-2}$, so $8 \sin^2 2\theta = 6$, which means $\sin^2 2\theta = \frac{6}{8} = \frac{3}{4}$.
Thus, $\sin 2\theta = \frac{\sqrt{3}}{2}$, which implies $2\theta = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right)$.
Therefore, $\theta = \frac{1}{2} \sin^{-1} \left( \frac{\sqrt{3}}{2} \right)$.

(B) According to Brewster's Law, the refractive index of the glass $(\mu)$ is given by $\mu = \tan \theta$, where $\theta$ is the polarising angle.
We know that the refractive index is also defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the medium: $\mu = \frac{v_a}{v_g}$.
Since the frequency $(f)$ of light remains constant when it travels from one medium to another, the speed of light is related to wavelength $(\lambda)$ by $v = f \lambda$.
Therefore, $\mu = \frac{f \lambda_a}{f \lambda_g} = \frac{\lambda_a}{\lambda_g}$.
Equating the two expressions for $\mu$, we get $\tan \theta = \frac{\lambda_a}{\lambda_g}$.
Rearranging this gives $\lambda_g = \frac{\lambda_a}{\tan \theta} = \lambda_a \cot \theta$.

(B) According to Rayleigh's criterion, the angular separation between the central maximum and the first minimum of a diffraction pattern for a circular aperture is given by $\theta = 1.22 \lambda / D$.
Here, $\lambda = 5000 \text{ Å} = 5000 \times 10^{-8} \text{ cm} = 5 \times 10^{-5} \text{ cm}$.
The diameter of the lens is $D = 5 \text{ cm}$.
The angular separation is $\theta = 1.22 \times (5 \times 10^{-5} \text{ cm}) / 5 \text{ cm} = 1.22 \times 10^{-5} \text{ radians}$.
The linear distance $x$ in the focal plane is given by $x = f \theta$, where $f = 25 \text{ cm}$.
$x = 25 \text{ cm} \times 1.22 \times 10^{-5} \text{ rad} = 30.5 \times 10^{-5} \text{ cm} = 3.05 \times 10^{-4} \text{ cm}$.

(C) In a single-slit diffraction experiment, the width of the central maximum $(w)$ is given by the formula:
$w = \frac{2\lambda D}{a}$
Given that the slit width $(a)$ is equal to the width of the central maximum $(w)$:
$a = w$
Substituting the expression for $w$:
$a = \frac{2\lambda D}{a}$
Rearranging the equation to solve for $D$:
$a^2 = 2\lambda D$
$D = \frac{a^2}{2\lambda}$
Therefore, the correct option is $C$.