MHT CET 2024 Physics Question Paper with Answer and Solution

(C) The angular width of the central maximum in a Fraunhofer diffraction pattern is given by the formula: $\theta = \frac{2 \lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the slit width.
From this relation,we can see that $\theta \propto \lambda$.
Let the initial wavelength be $\lambda_1 = 6000 Å$ and the initial angular width be $\theta_1$.
When the wavelength is changed to $\lambda_2$,the angular width becomes $\theta_2 = \theta_1 - 0.20 \theta_1 = 0.80 \theta_1$.
Using the proportionality $\frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1}$,we get:
$\frac{0.80 \theta_1}{\theta_1} = \frac{\lambda_2}{6000 Å}$
$0.80 = \frac{\lambda_2}{6000 Å}$
$\lambda_2 = 0.80 \times 6000 Å = 4800 Å$.

(B) The half-angular width of the central maxima for a single slit of width $d$ is given by $\theta = \frac{\lambda}{d}$.
For the first case: $\theta = \frac{\lambda}{d} \implies d = \frac{\lambda}{\theta}$.
For the second case: $q\theta = \frac{p\lambda}{d'} \implies d' = \frac{p\lambda}{q\theta}$.
The half-angular width of the $n^{th}$ secondary maxima is given by $\theta_n = \frac{(2n+1)\lambda}{2d}$.
For the first secondary maxima $(n=1)$,$\theta_{s1} = \frac{3\lambda}{2d}$ and $\theta_{s2} = \frac{3(p\lambda)}{2d'}$.
The ratio is $\frac{\theta_{s1}}{\theta_{s2}} = \frac{3\lambda / 2d}{3p\lambda / 2d'} = \frac{1}{d} \cdot \frac{d'}{p} = \frac{d'}{pd}$.
Substituting $d' = \frac{p\lambda}{q\theta}$ and $d = \frac{\lambda}{\theta}$,we get $\frac{\theta_{s1}}{\theta_{s2}} = \frac{p\lambda / q\theta}{p(\lambda / \theta)} = \frac{1}{q}$.
Thus,the ratio is $1:q$,which corresponds to the inverse of the given options if interpreted as $q:1$ or $1:q$. Given the standard form,the ratio of widths is $1:q$.

(C) For the formation of stable interference fringes,the two light sources must be coherent. Coherent sources are sources that emit light waves of the same frequency and maintain a constant phase difference over time.
When one slit is covered with a red filter and the other with a blue filter,the light passing through them has different frequencies and wavelengths.
Because the frequencies are different,the phase difference between the two waves at any point on the screen will change rapidly with time.
Consequently,the condition for a stable interference pattern is not satisfied,and no interference fringes will be observed on the screen.

(B) The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I_1}-\sqrt{I_2})^2}$
Given that $\frac{I_{\max }}{I_{\min }} = \frac{9}{1}$,we have:
$\frac{9}{1} = \frac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I_1}-\sqrt{I_2})^2}$
Taking the square root on both sides:
$\frac{3}{1} = \frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}$
Cross-multiplying gives:
$3(\sqrt{I_1}-\sqrt{I_2}) = 1(\sqrt{I_1}+\sqrt{I_2})$
$3\sqrt{I_1} - 3\sqrt{I_2} = \sqrt{I_1} + \sqrt{I_2}$
$2\sqrt{I_1} = 4\sqrt{I_2}$
$\sqrt{\frac{I_1}{I_2}} = \frac{4}{2} = 2$
Squaring both sides,we get:
$\frac{I_1}{I_2} = \frac{4}{1}$
Thus,the ratio of the intensities of the light sources is $4: 1$.

(C) The position of the $n^{\text{th}}$ bright fringe for wavelength $\lambda_1$ is given by $y_n = \frac{n \lambda_1 D}{d}$.
The position of the $m^{\text{th}}$ dark fringe for wavelength $\lambda_2$ is given by $y_m = \frac{(2m-1) \lambda_2 D}{2d}$.
Since the fringes coincide,we equate their positions:
$\frac{n \lambda_1 D}{d} = \frac{(2m-1) \lambda_2 D}{2d}$.
Canceling $D$ and $d$ from both sides,we get:
$n \lambda_1 = \frac{(2m-1) \lambda_2}{2}$.
Rearranging to find the ratio $\frac{\lambda_1}{\lambda_2}$:
$\frac{\lambda_1}{\lambda_2} = \frac{2m-1}{2n}$.

(A) The optical path length of a ray in a medium is defined as the product of the refractive index of the medium and the geometric path length traveled by the ray.
Optical path of the first ray = $\mu_1 L_1$.
Optical path of the second ray = $\mu_2 L_2$.
The optical path difference between the two rays is $\Delta x = |\mu_1 L_1 - \mu_2 L_2|$.
The relationship between phase difference $(\Delta \phi)$ and optical path difference $(\Delta x)$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Substituting the value of path difference,the phase difference is $\Delta \phi = \frac{2 \pi}{\lambda} |\mu_1 L_1 - \mu_2 L_2|$.

(C) The resultant amplitude $A$ of two superposing waves with individual amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by: $A^2 = a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi$.
Since the waves are identical,$a_1 = a_2 = a$.
Substituting this into the equation: $A^2 = a^2 + a^2 + 2 a^2 \cos \phi = 2 a^2 (1 + \cos \phi)$.
Using the trigonometric identity $1 + \cos \phi = 2 \cos^2(\phi/2)$,we get: $A^2 = 2 a^2 (2 \cos^2(\phi/2)) = 4 a^2 \cos^2(\phi/2)$.
Since intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$,we have $I \propto \cos^2(\phi/2)$.

(B) The intensity of light in Young's double slit experiment is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual slit and $\phi$ is the phase difference.
Case $1$: Path difference $\Delta x = \lambda$.
The phase difference is $\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
Given intensity $I = x$,so $x = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
Thus,$4I_0 = x$.
Case $2$: Path difference $\Delta x = \frac{\lambda}{4}$.
The phase difference is $\phi' = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I'$ is $I' = 4I_0 \cos^2(\frac{\phi'}{2}) = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $I' = 4I_0 \times (\frac{1}{\sqrt{2}})^2 = 4I_0 \times \frac{1}{2} = 2I_0$.
Since $4I_0 = x$,then $2I_0 = \frac{x}{2}$.
Therefore,the intensity is $\frac{x}{2}$.

(C) According to Malus' law,the intensity of light transmitted through a polarizer is $I = I_{in} \cos^2 \theta$,where $\theta$ is the angle between the polarization direction of incident light and the pass axis of the polarizer.
$1$. When unpolarized light of intensity $I_0$ passes through the first polaroid $P_1$,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.
$2$. The light from $P_1$ is now polarized at an angle $0^{\circ}$ relative to $P_1$'s axis. $P_2$ is at $60^{\circ}$ to $P_1$. Thus,the intensity after $P_2$ is:
$I_2 = I_1 \cos^2(60^{\circ}) = \frac{I_0}{2} \times (0.5)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$.
$3$. The light from $P_2$ is polarized at $60^{\circ}$ relative to $P_1$. $P_3$ is at $90^{\circ}$ relative to $P_1$. The angle between the polarization of light from $P_2$ and the pass axis of $P_3$ is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
$4$. The intensity after $P_3$ is:
$I_3 = I_2 \cos^2(30^{\circ}) = \frac{I_0}{8} \times \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I_0}{8} \times \frac{3}{4} = \frac{3 I_0}{32}$.

(A) When light travels from a denser medium to a rarer medium, its speed increases $(v_2 > v_1)$.
Since the frequency $(f)$ of the wave remains constant, the wavelength $(\lambda = v/f)$ also increases.
The distance between successive wavefronts is equal to the wavelength.
Therefore, as the wavefronts move from a denser medium to a rarer medium, the width (or spacing) between successive wavefronts increases.
Thus, the correct option is $A$.

(A) Wavefront: The locus of all particles in a medium,vibrating in the same phase or constant phase,is called a wavefront.
The direction of propagation of light (ray of light) is always perpendicular to the wavefront.
Every point on the given wavefront acts as a source of a new disturbance called secondary wavelets,which travel in all directions with the velocity of light in the medium.
$A$ surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wavefront at that instant. This is called a secondary wavefront.

(B) Two coherent sources of intensities $I_1$ and $I_2$ produce maximum intensity $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and minimum intensity $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$ in an interference pattern.
The ratio is given by $\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$.
Given $I_1 = 2I_2$,we substitute this into the ratio:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{2I_2} + \sqrt{I_2})^2}{(\sqrt{2I_2} - \sqrt{I_2})^2} = \frac{(\sqrt{I_2}(\sqrt{2} + 1))^2}{(\sqrt{I_2}(\sqrt{2} - 1))^2} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2}$.
Expanding the squares: $\frac{2 + 1 + 2\sqrt{2}}{2 + 1 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$.
Rationalizing the denominator: $\frac{(3 + 2\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} = \frac{9 + 8 + 12\sqrt{2}}{9 - 8} = 17 + 12\sqrt{2} \approx 33.97 \approx 34$.
Thus,the ratio is $34:1$.

(B) The intensity in Young's double slit experiment is given by $I = I_{max} \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase difference.
Given that at path difference $\Delta x = \lambda$,the intensity is $K$. Since $\Delta x = \lambda$ corresponds to a phase difference $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$,we have $I = I_{max} \cos^2(\pi) = I_{max} = K$.
Now,for a path difference $\Delta x = \frac{\lambda}{6}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}$.
The intensity at this point is $I = K \cos^2 \left( \frac{\pi/3}{2} \right) = K \cos^2 \left( \frac{\pi}{6} \right)$.
Substituting $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,we get $I = K \left( \frac{\sqrt{3}}{2} \right)^2 = K \left( \frac{3}{4} \right) = \frac{3K}{4}$.

(C) The position of the $n^{th}$ minimum in Young's double slit experiment is given by $y_n' = \frac{(2n-1) \lambda D}{2d}$.
Since the second minimum $(n=2)$ is observed exactly in front of one slit,its distance from the central axis is $y_2 = \frac{d}{2}$.
Substituting $n=2$ into the formula: $\frac{d}{2} = \frac{(2(2)-1) \lambda D}{2d}$.
Simplifying the equation: $\frac{d}{2} = \frac{3 \lambda D}{2d}$.
Canceling the $2$ from the denominators: $d = \frac{3 \lambda D}{d}$.
Solving for $\lambda$: $\lambda = \frac{d^2}{3D}$.

(A) Given that $25 \%$ of the total intensity of incident light is reflected from the upper surface. This implies that if the intensity of incident light is $I_0$,the intensity of the first reflected ray is $I_1 = 0.25 I_0 = \frac{I_0}{4}$.
The intensity of light reaching the lower surface of the plate is $I_0 - 0.25 I_0 = 0.75 I_0 = \frac{3}{4} I_0$.
Since $50 \%$ of this intensity is reflected from the lower surface,the intensity of the second reflected ray is $I_2 = 0.50 \times \frac{3}{4} I_0 = \frac{3}{8} I_0$.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}$
Substituting the values of $I_1$ and $I_2$:
$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{I_0}{4}}+\sqrt{\frac{3 I_0}{8}}\right)^2}{\left(\sqrt{\frac{I_0}{4}}-\sqrt{\frac{3 I_0}{8}}\right)^2} = \left(\frac{\frac{1}{2} \sqrt{I_0} + \sqrt{\frac{3}{8}} \sqrt{I_0}}{\frac{1}{2} \sqrt{I_0} - \sqrt{\frac{3}{8}} \sqrt{I_0}}\right)^2 = \left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)^2$

(A) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $I_{max} = K$ and $\phi$ 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 $\Delta x = \frac{\lambda}{6}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
Substituting this into the intensity formula: $I = K \cos^2(\frac{\pi/3}{2}) = K \cos^2(\frac{\pi}{6})$.
Since $\cos(\frac{\pi}{6}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $I = K (\frac{\sqrt{3}}{2})^2 = K (\frac{3}{4}) = \frac{3K}{4}$.

(D) The fringe width is given by $W = \frac{\lambda D}{d} = X$ (given).
The position of the $n^{\text{th}}$ bright fringe from the central maximum is $y_n = n \frac{\lambda D}{d} = nX$.
For the $4^{\text{th}}$ bright fringe,$y_4 = 4X$.
The position of the $n^{\text{th}}$ dark fringe from the central maximum is $y'_n = (2n - 1) \frac{\lambda D}{2d} = (2n - 1) \frac{X}{2}$.
For the $6^{\text{th}}$ dark fringe,$y'_6 = (2(6) - 1) \frac{X}{2} = \frac{11X}{2} = 5.5X$.
Since the fringes are on opposite sides of the central bright band,the total distance is the sum of the magnitudes of their positions:
Total distance $= y_4 + y'_6 = 4X + 5.5X = 9.5X$.

(C) The intensity $I$ in Young's double slit experiment is given by the formula $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.
Given that the intensity at a path difference of $\lambda$ is $\beta$,we know that for a path difference of $\lambda$,the phase difference $\phi = (2\pi/\lambda) \times \lambda = 2\pi$.
Thus,$\beta = I_{max} \cos^2(2\pi/2) = I_{max} \cos^2(\pi) = I_{max}(1)^2 = I_{max}$.
Now,for a path difference of $\Delta x = \lambda/3$,the phase difference is $\phi = (2\pi/\lambda) \times (\lambda/3) = 2\pi/3$.
The intensity at this point is $I = I_{max} \cos^2(\phi/2) = \beta \cos^2((2\pi/3)/2) = \beta \cos^2(\pi/3)$.
Since $\cos(\pi/3) = 1/2$,we have $I = \beta (1/2)^2 = \beta/4$.

(A) The position of the $n^{\text{th}}$ maximum in Young's double-slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
For the $n^{\text{th}}$ maximum with wavelength $\lambda_1$,the distance is $y_1 = \frac{n \lambda_1 D}{d} \quad (i)$.
For the $(\frac{n}{3})^{\text{th}}$ maximum with wavelength $\lambda_2$,the distance is $y_2 = \frac{(\frac{n}{3}) \lambda_2 D}{d} \quad (ii)$.
Taking the ratio of $(i)$ and $(ii)$:
$\frac{y_1}{y_2} = \frac{\frac{n \lambda_1 D}{d}}{\frac{n \lambda_2 D}{3d}} = \frac{n \lambda_1 D}{d} \times \frac{3d}{n \lambda_2 D} = \frac{3 \lambda_1}{\lambda_2}$.
Thus,the ratio $\frac{y_1}{y_2}$ is $\frac{3 \lambda_1}{\lambda_2}$.

(C) For the $n^{\text{th}}$ bright fringe,the position is given by $y_n = \frac{n \lambda D}{d}$.
For the $10^{\text{th}}$ bright fringe $(n=10)$:
$y_{10} = \frac{10 \lambda D}{d} = \frac{10 \times \lambda \times 1.2}{0.6 \times 10^{-3}} = (20 \times 10^3) \lambda \quad \dots(i)$
For the $n^{\text{th}}$ dark fringe,the position is given by $y'_n = \frac{(2n-1) \lambda D}{2d}$.
For the $3^{\text{rd}}$ dark fringe $(n=3)$:
$y'_3 = \frac{(2 \times 3 - 1) \lambda D}{2d} = \frac{5 \lambda D}{2d} = \frac{5 \times \lambda \times 1.2}{2 \times 0.6 \times 10^{-3}} = (5 \times 10^3) \lambda \quad \dots(ii)$
Given that the distance between them is $8.85 \ mm$:
$y_{10} - y'_3 = 8.85 \times 10^{-3} \ m$
$(20 \times 10^3) \lambda - (5 \times 10^3) \lambda = 8.85 \times 10^{-3}$
$(15 \times 10^3) \lambda = 8.85 \times 10^{-3}$
$\lambda = \frac{8.85 \times 10^{-3}}{15 \times 10^3} = 0.59 \times 10^{-6} \ m = 5.9 \times 10^{-7} \ m$
$\lambda = 5900 \ Å$.

(B) Let the point $P$ be exactly in front of slit $S_1$. The distance $S_1P = D$ and $S_2P = \sqrt{D^2 + d^2}$.
Using the binomial expansion for $S_2P$:
$S_2P = D(1 + \frac{d^2}{D^2})^{1/2} \approx D(1 + \frac{d^2}{2D^2}) = D + \frac{d^2}{2D}$.
The path difference $\Delta x$ at point $P$ is:
$\Delta x = S_2P - S_1P = (D + \frac{d^2}{2D}) - D = \frac{d^2}{2D}$.
For the $n^{\text{th}}$ dark fringe,the condition for path difference is:
$\Delta x = (2n - 1) \frac{\lambda}{2}$.
Given $\lambda = \frac{d^2}{3D}$,we substitute this into the condition:
$\frac{d^2}{2D} = (2n - 1) \frac{d^2}{6D}$.
Dividing both sides by $\frac{d^2}{D}$:
$\frac{1}{2} = \frac{2n - 1}{6}$.
$3 = 2n - 1 \Rightarrow 2n = 4 \Rightarrow n = 2$.

(C) The distance of the $1^{st}$ minima from the central maxima is given by $y_{1d} = \frac{\lambda D}{a}$.
Here,$\lambda = 5500 \ Å = 5500 \times 10^{-10} \ m$,$D = 2 \ m$,and $a = 0.5 \ mm = 0.5 \times 10^{-3} \ m$.
The distance between two minima on either side of the central maxima is $2 y_{1d} = \frac{2 \lambda D}{a}$.
Substituting the values: $2 y_{1d} = \frac{2 \times 5500 \times 10^{-10} \times 2}{0.5 \times 10^{-3}} = \frac{22000 \times 10^{-10}}{0.5 \times 10^{-3}} = 44000 \times 10^{-7} \ m = 4.4 \times 10^{-3} \ m = 4.4 \ mm$.

(C) In Young's double slit experiment,the fringe width $W$ is given by the formula $W = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
Since the wavelength of violet light $(\lambda_{violet} \approx 400 \ nm)$ is significantly smaller than the wavelength of sodium light $(\lambda_{sodium} \approx 589 \ nm)$,the fringe width $W$ will decrease when violet light is used instead of sodium light.

(C) In Young's double-slit experiment,dark fringes are formed due to destructive interference.
For destructive interference,the path difference $\Delta x$ must be an odd multiple of half-wavelength,i.e.,$\Delta x = (2n-1) \frac{\lambda}{2}$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the value of $\Delta x$: $\Delta \phi = \frac{2\pi}{\lambda} \times (2n-1) \frac{\lambda}{2} = (2n-1) \pi$.
Thus,the phase difference for a dark fringe is $(2n-1) \pi$.

(B) Let the intensities of the two individual waves be $I_0$. The resultant intensity $I$ is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi) = 4I_0 \cos^2(\phi/2)$.
- The minimum intensity $I_{min}$ occurs when $\phi = \pi$,so $I_{min} = 4I_0 \cos^2(\pi/2) = 0$.
- The path difference $\Delta x = \lambda/4$ corresponds to a phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
- The intensity $I_1$ at this point is $I_1 = 4I_0 \cos^2(\pi/4) = 4I_0 (1/\sqrt{2})^2 = 2I_0$.
- The ratio $I_{min} / I_1 = 0 / 2I_0 = 0$.

(D) Let the amplitude of light from the first slit be $a_1 = a$ and from the second slit be $a_2 = a$. The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.
Initially,for equal widths,the amplitudes are equal $(a_1 = a_2 = a)$. The minimum intensity is $I_{\min} \propto (a - a)^2 = 0$,and the maximum intensity is $I_{\max} \propto (a + a)^2 = 4a^2$.
When one slit is made twice as wide,the amplitude of light from that slit becomes $a_1 = 2a$,while the other remains $a_2 = a$.
The new minimum intensity is $I_{\min}' \propto (2a - a)^2 = a^2$. Since $a^2 > 0$,the intensity of the minima increases.
The new maximum intensity is $I_{\max}' \propto (2a + a)^2 = 9a^2$. Since $9a^2 > 4a^2$,the intensity of the maxima also increases.
Therefore,both the intensity of the maxima and the minima increase.

(B) When a thin plate of refractive index $n$ is introduced in front of one of the slits in Young's double-slit experiment,it adds an extra optical path (and hence a phase difference) to the light coming from that slit.
This shifts the entire interference pattern on the screen but does not change the fringe spacing or make the pattern disappear.
Fringe width $\beta = \frac{\lambda D}{d}$ does not change because it depends on the slit separation $d$,the wavelength $\lambda$,and the screen distance $D$,not on a uniform phase shift introduced in one path.
The net effect of introducing a uniform optical path difference in one slit is a lateral shift of the central maximum (and all fringes) toward the slit with the plate.
The shift is given by $\Delta x = \frac{D}{d}(n-1)t$.
Hence,the correct statement is: $(B)$ The central maximum will shift towards the side on which the plate is introduced.

(C) For any point in the interference pattern,the intensity $I$ is given by $I = I_{\max} \cos^2 \left(\frac{\phi}{2}\right)$.
Given that $I = \frac{I_{\max}}{4}$,we have $\frac{I_{\max}}{4} = I_{\max} \cos^2 \left(\frac{\phi}{2}\right)$.
This simplifies to $\cos^2 \left(\frac{\phi}{2}\right) = \frac{1}{4}$,which implies $\cos \left(\frac{\phi}{2}\right) = \frac{1}{2}$.
Thus,$\frac{\phi}{2} = 60^{\circ} = \frac{\pi}{3}$,so the phase difference $\phi = \frac{2\pi}{3}$.
The phase difference is related to the path difference $\Delta x$ by $\phi = \left(\frac{2\pi}{\lambda}\right) \Delta x$.
Substituting $\Delta x = d \sin \theta$,we get $\frac{2\pi}{3} = \left(\frac{2\pi}{\lambda}\right) d \sin \theta$.
Solving for $\sin \theta$,we find $\sin \theta = \frac{\lambda}{3d}$,which means $\theta = \sin^{-1} \left(\frac{\lambda}{3d}\right)$.

(C) The formula for fringe width $(\beta)$ in a Young's double slit experiment is given by $\beta = \frac{D \lambda}{d}$,where $D$ is the distance between the slits and the screen,$\lambda$ is the wavelength of light,and $d$ is the distance between the slits.
Given that the new distance between slits $d' = 10d$ and the new distance from the screen $D' = \frac{D}{2}$.
The new fringe width $\beta'$ is given by $\beta' = \frac{D' \lambda}{d'} = \frac{(\frac{D}{2}) \lambda}{10d} = \frac{D \lambda}{20d}$.
Since $\beta = \frac{D \lambda}{d}$,we have $\beta' = \frac{\beta}{20}$.
Thus,the fringe width becomes $\frac{1}{20}$ times the original fringe width.

(A) The resultant intensity $I'$ of two interfering waves with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by the formula:
$I' = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
Given $I_1 = I$,$I_2 = 4I$,and $\phi = \pi$:
$I' = I + 4I + 2\sqrt{I \times 4I} \cos \pi$
Since $\cos \pi = -1$:
$I' = 5I + 2\sqrt{4I^2} (-1)$
$I' = 5I + 2(2I)(-1)$
$I' = 5I - 4I = I$

(D) The electric field $\vec{E}$ due to a point charge $Q$ is always directed radially outward.
Since the path from $A$ to $B$ is an arc of a circle centered at $Q$,the displacement vector $d\vec{s}$ at any point on the path is tangent to the circle.
The radial direction (direction of $\vec{E}$) is always perpendicular to the tangent of the circle at any point.
Therefore,the force $\vec{F} = q\vec{E}$ is always perpendicular to the displacement $d\vec{s}$.
The work done $W$ is given by the integral $W = \int_A^B \vec{F} \cdot d\vec{s}$.
Since $\vec{F} \perp d\vec{s}$,the dot product $\vec{F} \cdot d\vec{s} = F \cdot ds \cos 90^{\circ} = 0$.
Thus,the total work done $W = 0$.