MHT CET 2021 Physics Question Paper with Answer and Solution

(D) The dark band formed between the $4^{\text{th}}$ and $5^{\text{th}}$ bright band is the $5^{\text{th}}$ dark band.
For the $n^{\text{th}}$ dark band,the path difference $\Delta x$ is given by $\Delta x = (n - 0.5) \lambda$.
Here,$n = 5$ and $\lambda = 6000 \text{ Å} = 6000 \times 10^{-8} \text{ cm} = 6 \times 10^{-5} \text{ cm}$.
Substituting the values:
$\Delta x = (5 - 0.5) \times 6 \times 10^{-5} \text{ cm}$
$\Delta x = 4.5 \times 6 \times 10^{-5} \text{ cm} = 27 \times 10^{-5} \text{ cm} = 2.7 \times 10^{-4} \text{ cm}$.

(D) The intensity of light in an interference pattern is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For destructive interference, the phase difference $\phi = (2n+1)\pi$, which leads to $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
If the amplitudes are different, then $I_1 \neq I_2$, which implies $\sqrt{I_1} \neq \sqrt{I_2}$.
Therefore, $I_{min} \neq 0$.
This means that in the region of destructive interference, the waves do not cancel each other out completely, and there remains some residual intensity of light.

(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:
$W = \frac{(6300 \times 10^{-10} / 1.33) \times 1.33}{10^{-3}}$
$W = \frac{6300 \times 10^{-10}}{10^{-3}} = 6300 \times 10^{-7} \,m = 6.3 \times 10^{-4} \,m$.

(D) In Young's double slit experiment,the condition for constructive interference (maxima) is given by the path difference $\Delta x = n \lambda$,where $n = 0, 1, 2, ...$ and $\lambda$ is the wavelength of light.
For the central maximum,$n = 0$,and for the first maximum,$n = 1$.
Given the wavelength $\lambda = 6300 Å$,the path difference for the first maximum is $\Delta x = 1 \times 6300 Å = 6300 Å$.

(D) Let $I'$ be the intensity of each individual wave. The resultant intensity $I$ is given by the formula:
$I = 4I' \cos^2 \left( \frac{\phi}{2} \right)$
where $\phi$ is the phase difference.
The maximum intensity $I_0$ occurs when $\cos^2 \left( \frac{\phi}{2} \right) = 1$,so $I_0 = 4I'$.
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:
$I = 4I' \cos^2 \left( \frac{\pi/2}{2} \right) = 4I' \cos^2 \left( \frac{\pi}{4} \right) = 4I' \times \left( \frac{1}{\sqrt{2}} \right)^2 = 4I' \times \frac{1}{2} = 2I'$.
Therefore,the ratio is:
$\frac{I}{I_0} = \frac{2I'}{4I'} = \frac{1}{2}$.

(B) The position of the $n^{\text{th}}$ bright fringe (maximum) in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
For the first case,the $n^{\text{th}}$ maximum with wavelength $\lambda_1$ is at distance $y_1 = \frac{n \lambda_1 D}{d}$.
For the second case,the $(n/2)^{\text{th}}$ maximum with wavelength $\lambda_2$ is at distance $y_2 = \frac{(n/2) \lambda_2 D}{d} = \frac{n \lambda_2 D}{2d}$.
Now,taking the ratio of $y_1$ and $y_2$:
$\frac{y_1}{y_2} = \frac{(n \lambda_1 D / d)}{(n \lambda_2 D / 2d)} = \frac{n \lambda_1 D}{d} \times \frac{2d}{n \lambda_2 D} = \frac{2 \lambda_1}{\lambda_2}$.
Therefore,the ratio is $\frac{2 \lambda_1}{\lambda_2}$.

(C) The 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 separation between the slits.
Initially, $\beta_1 = \frac{\lambda D_1}{d_1}$, where $D_1 = 75 \,cm$.
When the slit separation is doubled, $d_2 = 2d_1$. To keep the fringe width constant $(\beta_2 = \beta_1)$, we must have $\frac{\lambda D_2}{d_2} = \frac{\lambda D_1}{d_1}$.
Substituting $d_2 = 2d_1$, we get $\frac{D_2}{2d_1} = \frac{D_1}{d_1}$, which implies $D_2 = 2D_1$.
$D_2 = 2 \times 75 \,cm = 150 \,cm$.
The screen must be moved from $75 \,cm$ to $150 \,cm$, which means it should be moved $150 \,cm - 75 \,cm = 75 \,cm$ away from the slits.

(A) The position of the $n^{\text{th}}$ bright fringe (maximum) in Young's double slit experiment is given by $Y_n = \frac{n \lambda D}{d}$,where $D$ is the distance between the slits and the screen,and $d$ is the distance between the slits.
For the $10^{\text{th}}$ maximum with wavelength $\lambda_1$,the distance is $Y_1 = \frac{10 \lambda_1 D}{d}$.
For the $5^{\text{th}}$ maximum with wavelength $\lambda_2$,the distance is $Y_2 = \frac{5 \lambda_2 D}{d}$.
Taking the ratio of $Y_1$ to $Y_2$:
$\frac{Y_1}{Y_2} = \frac{10 \lambda_1 D / d}{5 \lambda_2 D / d} = \frac{10 \lambda_1}{5 \lambda_2} = \frac{2 \lambda_1}{\lambda_2}$.

(A) The intensity $I$ in Young's double slit experiment is given by the formula: $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual slit and $\phi$ is the phase difference.
Maximum intensity $K$ occurs when $\cos^2(\frac{\phi}{2}) = 1$,so $K = 4I_0$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the path difference $\Delta x = \frac{\lambda}{3}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
Substituting this into the intensity formula: $I = K \cos^2(\frac{2\pi/3}{2}) = K \cos^2(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I = K (\frac{1}{2})^2 = \frac{K}{4}$.

(B) In Young's double-slit experiment,the path difference $\Delta x$ at a point exactly in front of one slit is given by $\Delta x = \frac{d^2}{2D}$.
For a minimum (destructive interference) to occur at this point,the path difference must be an odd multiple of $\frac{\lambda}{2}$,i.e.,$\Delta x = (2n-1)\frac{\lambda}{2}$ where $n = 1, 2, 3, \ldots$.
Equating the two expressions: $\frac{d^2}{2D} = (2n-1)\frac{\lambda}{2}$.
This simplifies to $\lambda = \frac{d^2}{(2n-1)D}$.
For $n=1, 2, 3, \ldots$,the values of $(2n-1)$ are $1, 3, 5, \ldots$.
Thus,$\lambda$ is proportional to $\frac{1}{D}, \frac{1}{3D}, \frac{1}{5D}, \ldots$.
Therefore,the possible wavelengths are inversely proportional to $D, 3D, 5D, \ldots$.

(D) In Young's double-slit experiment,the formation of interference fringes requires the superposition of coherent light waves originating from two separate slits.
When one of the slits is covered with black opaque paper,light can only pass through the remaining single slit.
Since there is no longer a second source to interfere with the light from the first,the condition for interference is lost.
Consequently,no interference pattern (fringes) will be observed on the screen.

(B) The formula for fringe width in $YDSE$ is given by $W = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance between the slits and the screen, and $d$ is the separation between the slits.
Let the original fringe width be $W_1 = \frac{\lambda D_1}{d_1} = W$.
According to the problem, the new distance $D_2 = D_1 + 0.25 D_1 = 1.25 D_1$ and the new separation $d_2 = \frac{d_1}{2}$.
The new fringe width $W_2$ is given by $W_2 = \frac{\lambda D_2}{d_2}$.
Substituting the values, we get $W_2 = \frac{\lambda (1.25 D_1)}{(d_1 / 2)} = 1.25 \times 2 \times \frac{\lambda D_1}{d_1}$.
Since $W = \frac{\lambda D_1}{d_1}$, we have $W_2 = 2.5 W$.

(B) The position of the $n^{\text{th}}$ dark fringe from the central maximum is given by $y_n = (n - 0.5) \frac{\lambda D}{d}$.
For the $4^{\text{th}}$ dark band,$n = 4$,so $y_4 = (4 - 0.5) \frac{\lambda D}{d} = 3.5 \frac{\lambda D}{d}$.
Since the dark band is formed opposite to one of the slits,its distance from the central axis is $y = \frac{d}{2}$.
Equating the two expressions: $\frac{d}{2} = 3.5 \frac{\lambda D}{d}$.
Rearranging for $\lambda$: $\lambda = \frac{d^2}{2 \times 3.5 D} = \frac{d^2}{7 D}$.

(C) In Young's double slit experiment,the position of the $n^{\text{th}}$ dark fringe from the central bright fringe is given by the formula:
$y_n = (n - \frac{1}{2}) \beta$
where $n = 1, 2, 3, \dots$ represents the order of the dark fringe.
Thus,the distance is $(n - 0.5) \beta$.

(D) The resultant intensity $I_R$ in an interference pattern is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For maximum intensity $(I_{\max})$,the phase difference $\phi = 0$,so $I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Given $I_1 = I$ and $I_2 = 4I$:
$I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.
For minimum intensity $(I_{\min})$,the phase difference $\phi = \pi$,so $I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = (\sqrt{I_1} - \sqrt{I_2})^2$.
$I_{\min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$.
Thus,the maximum and minimum intensities are $9I$ and $I$ respectively.

(D) The path difference is given as $\Delta x = \frac{57}{2} \lambda = 28.5 \lambda$.
For constructive interference (bright band),the path difference must be an integral multiple of $\lambda$,i.e.,$\Delta x = n \lambda$ where $n = 1, 2, 3, \dots$. Since $28.5 \lambda$ is not an integer multiple,it is not a bright band.
For destructive interference (dark band),the path difference is given by $\Delta x = (n - \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$.
Equating the two: $28.5 \lambda = (n - 0.5) \lambda$.
Solving for $n$: $n - 0.5 = 28.5$,which gives $n = 29$.
Therefore,the point corresponds to the $29^{\text{th}}$ dark band.

(A) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Given $I_1 = I$ and $I_2 = 4I$.
At point $A$,the phase difference $\phi_A = \pi / 2$. Thus,$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi / 2) = 5I + 4I(0) = 5I$.
At point $B$,the phase difference $\phi_B = \pi$. Thus,$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 4I(-1) = 5I - 4I = I$.
The difference between the resultant intensities at $A$ and $B$ is $I_A - I_B = 5I - I = 4I$.

(D) In Young's double slit experiment $(YDSE)$,the path difference is given as $\Delta x = \frac{\lambda}{6}$.
Assuming the intensities of the two slits are equal,let $I_1 = I_2 = I_s$.
The phase difference $\Delta \phi$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x = \frac{2 \pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
The resultant intensity $I$ at any point is given by $I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\Delta \phi)$.
Substituting the values: $I = I_s + I_s + 2 \sqrt{I_s I_s} \cos(\frac{\pi}{3}) = 2 I_s + 2 I_s (\frac{1}{2}) = 2 I_s + I_s = 3 I_s$.
The maximum intensity $I_0$ occurs when $\cos(\Delta \phi) = 1$,so $I_0 = I_1 + I_2 + 2 \sqrt{I_1 I_2} = (\sqrt{I_s} + \sqrt{I_s})^2 = (2 \sqrt{I_s})^2 = 4 I_s$.
Therefore,the ratio $\frac{I}{I_0} = \frac{3 I_s}{4 I_s} = \frac{3}{4}$.