MHT CET 2025 Physics Question Paper with Answer and Solution

(A) The position of the $n^{th}$ order maxima is given by $x_n = x_0 + n \beta$,where $x_0$ is the position of the zero-order maxima and $\beta = \frac{\lambda D}{d}$ is the fringe width.
Given for $\lambda_1 = 6000 \ Å$:
$x_1 = x_0 + 1 \beta_1 = 14.50 \ mm$
$x_{10} = x_0 + 10 \beta_1 = 16.75 \ mm$
Subtracting the two equations: $9 \beta_1 = 16.75 - 14.50 = 2.25 \ mm$,so $\beta_1 = 0.25 \ mm$.
Then $x_0 = 14.50 - 0.25 = 14.25 \ mm$.
Now,for $\lambda_2 = 5500 \ Å$,the new fringe width $\beta_2 = \beta_1 \times (\frac{\lambda_2}{\lambda_1}) = 0.25 \times (\frac{5500}{6000}) = 0.25 \times (\frac{11}{12}) \approx 0.229 \ mm$.
The zero-order maxima $x_0$ remains at the same position relative to the central axis,but the problem implies a shift relative to the reference point. Since $x_0$ is the central fringe,its position relative to the slits remains $14.25 \ mm$.
The new position of the $10^{th}$ order maxima is $x'_{10} = x_0 + 10 \beta_2 = 14.25 + 10 \times (0.25 \times \frac{11}{12}) = 14.25 + 2.29 = 16.54 \ mm \approx 16.55 \ mm$.
Thus,the positions are $14.25 \ mm$ and $16.55 \ mm$.

(D) The formula for fringe width $(\beta)$ in Young's double slit 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 slits.
Given values are: $\lambda = 6000 \, \text{\AA} = 6000 \times 10^{-8} \, \text{cm} = 6 \times 10^{-5} \, \text{cm}$, $D = 40 \, \text{cm}$, and $\beta = 0.012 \, \text{cm}$.
Rearranging the formula to solve for $d$: $d = \frac{\lambda D}{\beta}$.
Substituting the values: $d = \frac{(6 \times 10^{-5} \, \text{cm}) \times (40 \, \text{cm})}{0.012 \, \text{cm}}$.
$d = \frac{240 \times 10^{-5}}{0.012} \, \text{cm} = \frac{2.4 \times 10^{-3}}{1.2 \times 10^{-2}} \, \text{cm} = 2 \times 10^{-1} \, \text{cm} = 0.2 \, \text{cm}$.
Therefore, the distance between the two slits is $0.2 \, \text{cm}$.

(A) The intensity at any point on the screen in Young's double slit experiment is given by $I = I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given path difference $\Delta x = \frac{\lambda}{4}$,so $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
Now,substitute $\phi$ into the intensity formula: $I = I_0 \cos^2(\frac{\pi/2}{2}) = I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $I = I_0 (\frac{1}{\sqrt{2}})^2 = I_0 \times \frac{1}{2}$.
Therefore,the ratio $\frac{I_0}{I} = \frac{I_0}{I_0/2} = 2:1$.

(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 the formula $\phi = (2\pi / \lambda) \Delta x$.
For path difference $\Delta x = \lambda$,the phase difference is $\phi_1 = (2\pi / \lambda) \times \lambda = 2\pi$.
The intensity at this point is $I = I_{max} \cos^2(2\pi / 2) = I_{max} \cos^2(\pi) = I_{max} (-1)^2 = I_{max}$. Thus,$I_{max} = I$.
For path difference $\Delta x = \lambda / 6$,the phase difference is $\phi_2 = (2\pi / \lambda) \times (\lambda / 6) = \pi / 3$.
The intensity at this point is $I' = I_{max} \cos^2(\phi_2 / 2) = I \cos^2((\pi / 3) / 2) = I \cos^2(\pi / 6)$.
Given $\cos(\pi / 6) = \sqrt{3} / 2$,we have $I' = I \times (\sqrt{3} / 2)^2 = I \times (3 / 4) = 3I / 4$.

(D) The fringe width $\beta$ in a biprism experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the source and the screen,and $d$ is the distance between the two virtual sources.
Since $D$ and $d$ remain constant,the fringe width is directly proportional to the wavelength: $\beta \propto \lambda$.
Given $\lambda_1 = 5000 \ Å$ and $\lambda_2 = 6400 \ Å$.
The ratio of the new fringe width $\beta_2$ to the initial fringe width $\beta_1$ is $\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} = \frac{6400}{5000} = 1.28$.
This means $\beta_2 = 1.28 \beta_1$.
The percentage increase in fringe width is given by $\frac{\beta_2 - \beta_1}{\beta_1} \times 100\% = (1.28 - 1) \times 100\% = 0.28 \times 100\% = 28\%$.
Therefore,the fringe width increases by $28\%$.

(B) The intensity $I$ at any point on the screen in Young's double slit experiment is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Phase difference $\phi = \frac{2\pi}{\lambda} \times (\text{path difference} \Delta x)$.
For point $P$,path difference $\Delta x_P = 0$,so $\phi_P = 0$. Intensity $I_P = I_{max} \cos^2(0) = I_{max}$.
For point $Q$,path difference $\Delta x_Q = \frac{\lambda}{4}$,so $\phi_Q = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} = 90^{\circ}$.
Intensity $I_Q = I_{max} \cos^2(\frac{90^{\circ}}{2}) = I_{max} \cos^2(45^{\circ}) = I_{max} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_{max}}{2}$.
The ratio of intensities is $\frac{I_P}{I_Q} = \frac{I_{max}}{I_{max}/2} = 2: 1$.

(A) The position of the $n^{th}$ order bright fringe in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
Given: $d = 2 \ mm = 2 \times 10^{-3} \ m$,$D = 1 \ m$,$n = 3$.
For wavelength $\lambda_1$,the position of the $3^{rd}$ bright fringe is $y_1 = \frac{3 \lambda_1 D}{d}$.
For wavelength $\lambda_2$,the position of the $3^{rd}$ bright fringe is $y_2 = \frac{3 \lambda_2 D}{d}$.
Given $\lambda_2 = 1.5 \lambda_1$,so $y_2 = \frac{3(1.5 \lambda_1) D}{d} = \frac{4.5 \lambda_1 D}{d}$.
The separation between the fringes is $\Delta y = |y_2 - y_1| = \frac{4.5 \lambda_1 D}{d} - \frac{3 \lambda_1 D}{d} = \frac{1.5 \lambda_1 D}{d}$.
Substituting the values: $\Delta y = \frac{1.5 \times \lambda_1 \times 1}{2 \times 10^{-3}} = 0.75 \times 10^3 \lambda_1 \ m$.

(C) In an interference pattern,the distance of the $n^{\text{th}}$ bright band from the central bright band is given by $y_n = n \beta$,where $\beta$ is the fringe width.
The distance of the $m^{\text{th}}$ dark band from the central bright band is given by $y_m = (m - 1/2) \beta$.
The ratio of the distance of the $n^{\text{th}}$ bright band to the $m^{\text{th}}$ dark band is:
$\text{Ratio} = \frac{y_n}{y_m} = \frac{n \beta}{(m - 1/2) \beta} = \frac{n}{m - 1/2}$.
Therefore,the ratio is $n : (m - 1/2)$.

(B) In Young's double slit experiment,the path difference $\Delta x$ at a point $y$ on the screen is given by $\Delta x = \frac{yd}{D}$.
For a minimum (destructive interference),the path difference is given by $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$.
The second minimum corresponds to $n = 2$,so $\Delta x = (2(2) - 1) \frac{\lambda}{2} = \frac{3\lambda}{2}$.
The point is exactly in front of one slit,so the distance from the center is $y = \frac{d}{2}$.
Equating the two expressions for path difference: $\frac{(\frac{d}{2})d}{D} = \frac{3\lambda}{2}$.
$\frac{d^2}{2D} = \frac{3\lambda}{2}$.
Solving for $\lambda$,we get $\lambda = \frac{d^2}{3D}$.

(D) In Young's double-slit experiment,the fringe width $(X)$ is given by the formula:
$X = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light,$D$ is the distance of the screen from the slits,and $d$ is the distance between the slits.
Comparing this equation with the equation of a straight line $y = mx + c$,where $y = X$ and $x = D$,we get:
$X = (\frac{\lambda}{d}) D$
The slope of the graph is given by $m = \frac{\lambda}{d}$.
Therefore,the wavelength $\lambda$ can be calculated as:
$\lambda = \text{slope} \times d$.

(A) The distance of the $n^{th}$ dark fringe from the central fringe is given by $y_n = (n - 1/2) \beta$,where $\beta$ is the fringe width.
For the second dark fringe $(n = 2)$:
$y_2 = (2 - 1/2) \beta = 1.5 \beta = 3 \ mm$.
Therefore,$\beta = 3 / 1.5 = 2 \ mm$.
The distance of the $n^{th}$ bright fringe from the central fringe is given by $y_n = n \beta$.
For the sixth bright fringe $(n = 6)$:
$y_6 = 6 \times \beta = 6 \times 2 \ mm = 12 \ mm$.

(D) 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 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 $\beta'$:
$\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{d/2} = 4 \left( \frac{\lambda D}{d} \right) = 4\beta$.
Therefore,the fringe width increases four times.

(D) The intensity $I$ at any point on the screen is given by the formula $I = I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the maximum intensity and $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For the first case,$\Delta x = \frac{\lambda}{4}$,so $\phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity is $I_1 = I_0 \cos^2(\frac{\pi/2}{2}) = I_0 \cos^2(\frac{\pi}{4}) = I_0 (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{2}$.
Given $I_1 = \frac{K}{4}$,we have $\frac{I_0}{2} = \frac{K}{4}$,which implies $I_0 = \frac{K}{2}$.
For the second case,$\Delta x = \lambda$,so $\phi_2 = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
The intensity is $I_2 = I_0 \cos^2(\frac{2\pi}{2}) = I_0 \cos^2(\pi) = I_0 (-1)^2 = I_0$.
Substituting $I_0 = \frac{K}{2}$,we get $I_2 = \frac{K}{2}$.

(B) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the actual wavelength,$v$ is the velocity of the star,and $c$ is the speed of light.
Given that the apparent wavelength is $0.02 \%$ more than the actual wavelength,we have $\frac{\Delta \lambda}{\lambda} = 0.02 \% = \frac{0.02}{100} = 2 \times 10^{-4}$.
Substituting the values into the formula: $2 \times 10^{-4} = \frac{v}{3 \times 10^8 \ m/s}$.
Solving for $v$: $v = (2 \times 10^{-4}) \times (3 \times 10^8 \ m/s) = 6 \times 10^4 \ m/s$.
Converting to $km/s$: $v = \frac{6 \times 10^4}{10^3} \ km/s = 60 \ km/s$.