Wave Nature and Interference of Light (Intensity) Questions in English

(C) The wave nature of light is demonstrated by phenomena that involve the superposition of waves and the interaction of waves with obstacles of comparable size to their wavelength.
$1$. Interference: This occurs when two or more light waves overlap,resulting in a new wave pattern.
$2$. Diffraction: This is the bending of light around the corners of an obstacle or through an aperture.
$3$. Polarization: This confirms the transverse wave nature of light.
Conversely,the photoelectric effect,Compton effect,and black body radiation are explained by the particle nature of light (photons).
Therefore,interference and diffraction are the correct phenomena explained by the wave nature of light.

(A) The optical path length of a wave traveling through a medium of refractive index $n$ and geometric path length $L$ is given by $OPL = n \times L$.
The optical path length for the first wave is $OPL_{1} = n_{1}L_{1}$.
The optical path length for the second wave is $OPL_{2} = n_{2}L_{2}$.
The optical path difference between the two waves is $\Delta p = |OPL_{1} - OPL_{2}| = |n_{1}L_{1} - n_{2}L_{2}|$.
The phase difference $\Delta \phi$ is related to the path difference $\Delta p$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta p$.
Substituting the path difference,we get $\Delta \phi = \frac{2 \pi}{\lambda} (n_{1}L_{1} - n_{2}L_{2})$.

(D) Let the intensity of each source be $I$. The resultant intensity is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) = 2I + 2I \cos(\Delta \phi) = 4I \cos^2(\Delta \phi / 2)$,where $\Delta \phi$ is the total phase difference.
$1$. For point $A$: The path difference $\Delta x = x_P - x_Q = 5\, m$. The phase difference due to path is $\Delta \phi_{path} = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{20} \times 5 = \frac{\pi}{2}$. Since $P$ is ahead of $Q$ by $90^{\circ}$ $(\pi/2)$,the wave from $Q$ travels a shorter distance to $A$. The total phase difference is $\Delta \phi_A = \Delta \phi_{source} - \Delta \phi_{path} = \frac{\pi}{2} - \frac{\pi}{2} = 0$. Thus,$I_A = 4I \cos^2(0) = 4I$.
$2$. For point $B$: The path difference $\Delta x = 0$ as $B$ is on the perpendicular bisector. The total phase difference is $\Delta \phi_B = \Delta \phi_{source} = \frac{\pi}{2}$. Thus,$I_B = 4I \cos^2(\pi/4) = 4I(1/2) = 2I$.
$3$. For point $C$: The path difference $\Delta x = x_Q - x_P = 5\, m$. The phase difference due to path is $\Delta \phi_{path} = \frac{\pi}{2}$. The total phase difference is $\Delta \phi_C = \Delta \phi_{source} + \Delta \phi_{path} = \frac{\pi}{2} + \frac{\pi}{2} = \pi$. Thus,$I_C = 4I \cos^2(\pi/2) = 0$.
Therefore,the ratio $I_A : I_B : I_C = 4I : 2I : 0 = 2 : 1 : 0$.

(B) In an interference pattern,the condition for constructive interference is that the path difference $\Delta x$ must be an integral multiple of the wavelength $\lambda$,i.e.,$\Delta x = n\lambda$ where $n = 0, 1, 2, \dots$.
The central maximum occurs at the point where the path difference between the two waves is zero $(\Delta x = 0)$.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting $\Delta x = 0$ into the equation,we get $\phi = \frac{2\pi}{\lambda} \times 0 = 0$.
Therefore,for the central maximum,the phase difference is $0$.

(D) The intensity of the resultant wave in an interference pattern is given by the formula: $I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} \cos \phi$.
For maxima,the phase difference $\phi = 2n\pi$,so $\cos \phi = 1$.
Thus,$I_{\max} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} = (\sqrt{I_{1}} + \sqrt{I_{2}})^2$.
Given $I_{1} = I_{2} = I_{0}$,we substitute these values into the formula:
$I_{\max} = (\sqrt{I_{0}} + \sqrt{I_{0}})^2 = (2\sqrt{I_{0}})^2 = 4I_{0}$.

(D) Given the intensity ratio $\frac{I_1}{I_2} = 4$,we can write $I_1 = 4I_2$.
The ratio of maximum to minimum intensity in an interference pattern is given by:
$\frac{I_{\max}}{I_{\min}} = \left[ \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right]^2$
Substituting $I_1 = 4I_2$:
$\frac{I_{\max}}{I_{\min}} = \left[ \frac{\sqrt{4I_2} + \sqrt{I_2}}{\sqrt{4I_2} - \sqrt{I_2}} \right]^2 = \left[ \frac{2\sqrt{I_2} + \sqrt{I_2}}{2\sqrt{I_2} - \sqrt{I_2}} \right]^2 = \left( \frac{3}{1} \right)^2 = 9$.
Now,we need to calculate $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}$:
$\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}} = \frac{9 + 1}{9 - 1} = \frac{10}{8} = \frac{5}{4}$.
Comparing this with the given expression $\frac{5}{x}$,we get:
$\frac{5}{x} = \frac{5}{4} \implies x = 4$.

(B) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
At point $P$,the phase difference is $\phi_P = \frac{\pi}{2}$.
$I_P = I + 9I + 2\sqrt{I \times 9I} \cos \frac{\pi}{2} = 10I + 2(3I)(0) = 10I$.
At point $Q$,the phase difference is $\phi_Q = \pi$.
$I_Q = I + 9I + 2\sqrt{I \times 9I} \cos \pi = 10I + 2(3I)(-1) = 10I - 6I = 4I$.
The difference between the resultant intensities at $P$ and $Q$ is:
$I_P - I_Q = 10I - 4I = 6I$.

(B) Given the intensity ratio of two coherent sources is $\frac{I_1}{I_2} = \frac{1}{4}$,so $I_2 = 4I_1$.
The maximum intensity is given by $I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = I_1 + 4I_1 + 2\sqrt{I_1(4I_1)} = 5I_1 + 4I_1 = 9I_1$.
The minimum intensity is given by $I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = I_1 + 4I_1 - 2\sqrt{I_1(4I_1)} = 5I_1 - 4I_1 = I_1$.
Now,calculate the ratio $\frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}} = \frac{9I_1 + I_1}{9I_1 - I_1} = \frac{10I_1}{8I_1} = \frac{5}{4}$.
Comparing this with the given expression $\frac{2\alpha + 1}{\beta + 3} = \frac{5}{4}$,we have $2\alpha + 1 = 5 \implies 2\alpha = 4 \implies \alpha = 2$ and $\beta + 3 = 4 \implies \beta = 1$.
Therefore,$\frac{\alpha}{\beta} = \frac{2}{1} = 2$.

(D) 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 $A$,$\phi_A = \pi/2$. Thus,$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/2) = 5I + 4I(0) = 5I$.
At point $B$,$\phi_B = \pi/3$. Thus,$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/3) = 5I + 4I(1/2) = 5I + 2I = 7I$.
The difference between the resultant intensities is $I_B - I_A = 7I - 5I = 2I$.
Comparing this with $xI$,we get $x = 2$.

(A) The resultant intensity $I_R$ of two interfering beams is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
At point $A$,the phase difference $\phi = 0$. Therefore,the intensity is maximum:
$I_A = I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{9I} + \sqrt{4I})^2 = (3\sqrt{I} + 2\sqrt{I})^2 = (5\sqrt{I})^2 = 25I$.
At point $B$,the phase difference $\phi = \pi$. Therefore,the intensity is minimum:
$I_B = I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{9I} - \sqrt{4I})^2 = (3\sqrt{I} - 2\sqrt{I})^2 = (\sqrt{I})^2 = I$.
The difference between the resultant intensities at points $A$ and $B$ is:
$I_A - I_B = 25I - I = 24I$.

(B) The path difference $\Delta x$ at any point on the screen is given by $\Delta x = d \cos \theta$,where $d = 2 \lambda$ is the distance between the sources and $\theta$ is the angle with the line joining the sources.
For maxima,the path difference must be an integer multiple of $\lambda$,i.e.,$\Delta x = n \lambda$.
Since the screen is perpendicular to the line joining the sources,the range of $\theta$ is from $0$ to $90^\circ$ (or $0$ to $\pi/2$ radians).
Thus,$\Delta x = 2 \lambda \cos \theta$.
As $\theta$ varies from $0$ to $90^\circ$,$\cos \theta$ varies from $1$ to $0$.
Therefore,the path difference $\Delta x$ varies from $2 \lambda$ to $0$.
The possible values for $n$ in $\Delta x = n \lambda$ are $n = 0, 1, 2$.
For $n = 2$,$\Delta x = 2 \lambda$ (at $\theta = 0$,the point on the screen closest to the sources).
For $n = 1$,$\Delta x = \lambda$ (at $\cos \theta = 0.5$,i.e.,$\theta = 60^\circ$).
For $n = 0$,$\Delta x = 0$ (at $\theta = 90^\circ$,i.e.,at infinity).
Since the screen is symmetric on both sides of the line joining the sources,we have:
- At $\theta = 0$: $1$ central maxima $(n=2)$.
- For $\theta > 0$: $2$ maximas for $n=1$ (one on each side) and $2$ maximas for $n=0$ (one on each side at infinity).
However,the question typically asks for finite points on the screen. At $\theta = 90^\circ$,the path difference is $0$,which corresponds to the point at infinity. Excluding the points at infinity,we have $n=2$ (one point) and $n=1$ (two points),totaling $3$ maximas.

(C) The optical path length of a wave traveling through a medium of thickness $t$ and refractive index $\mu$ is given by $\Delta = \mu t$.
When two waves travel through media with different refractive indices $\mu_1$ and $\mu_2$ but the same thickness $t$,their optical paths are $\mu_1 t$ and $\mu_2 t$ respectively.
The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the relation $\Delta \phi = \frac{2 \pi}{\lambda_0} \Delta x$,where $\lambda_0$ is the wavelength in vacuum.
Since the refractive index $\mu = \frac{\lambda_0}{\lambda_m}$ (where $\lambda_m$ is the wavelength in the medium),the wavelength in the medium is $\lambda_m = \frac{\lambda_0}{\mu}$.
Because the wavelengths in the media are different due to different refractive indices,the optical paths differ,leading to a change in phase difference.
Thus,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

(B) Let the intensities of the two waves be $I_A = I_0$ and $I_B = 9I_0$.
For incoherent waves,the resultant intensity is the sum of individual intensities:
$I_1 = I_A + I_B = I_0 + 9I_0 = 10I_0$.
For coherent waves,the resultant intensity is given by $I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cos \phi$,where $\phi = 60^{\circ}$.
$I_2 = I_0 + 9I_0 + 2\sqrt{I_0 \cdot 9I_0} \cos 60^{\circ}$.
$I_2 = 10I_0 + 2(3I_0) \cdot \frac{1}{2} = 10I_0 + 3I_0 = 13I_0$.
Given $\frac{I_1}{I_2} = \frac{10}{x}$,we have $\frac{10I_0}{13I_0} = \frac{10}{x}$.
Therefore,$x = 13$.

(D) The maximum intensity $I_{\max}$ for two interfering beams with intensities $I_1$ and $I_2$ is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Given $I_1 = I$ and $I_2 = 4I$,we have $I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.
The minimum intensity $I_{\min}$ is given by $I_{\min} = (\sqrt{I_2} - \sqrt{I_1})^2$.
Substituting the values,$I_{\min} = (\sqrt{4I} - \sqrt{I})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$.
The difference between maximum and minimum intensity is $I_{\max} - I_{\min} = 9I - I = 8I$.
Comparing this with $xI$,we get $x = 8$.

(D) The path difference at a point $P$ on the circumference,where the line joining the center to $P$ makes an angle $\theta$ with the line joining the sources,is given by $\Delta x = d \cos \theta$.
At point $P_1$,$\theta = 90^{\circ}$,so $\Delta x = d \cos 90^{\circ} = 0$. This corresponds to the central maximum.
At point $P_2$,$\theta = 0^{\circ}$,so $\Delta x = d \cos 0^{\circ} = d = 1.8 \ mm$.
The order of the fringe $n$ at $P_2$ is $n = \frac{d}{\lambda} = \frac{1.8 \times 10^{-3} \ m}{600 \times 10^{-9} \ m} = 3000$.
Since $n = 3000$ is an integer,a bright spot (maxima) is formed at $P_2$. Thus,option $[A]$ is incorrect and option $[B]$ is correct.
The number of fringes between $P_1$ $(\theta = 90^{\circ}, n=0)$ and $P_2$ $(\theta = 0^{\circ}, n=3000)$ is $3000$. Thus,option $[C]$ is correct.
For maxima,$d \cos \theta = n \lambda$. Differentiating,$-d \sin \theta \ d\theta = \lambda \ dn$,so the angular fringe width is $\Delta \theta \approx |d\theta| = \frac{\lambda}{d \sin \theta}$.
As we move from $P_1$ $(\theta = 90^{\circ})$ to $P_2$ $(\theta = 0^{\circ})$,$\sin \theta$ decreases,so $\Delta \theta$ increases. Thus,option $[D]$ is incorrect.
Therefore,the correct options are $[B]$ and $[C]$.

(C) The maximum intensity in an interference pattern is given by $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Substituting the given values,$I_{\max} = (\sqrt{4I} + \sqrt{9I})^2 = (2\sqrt{I} + 3\sqrt{I})^2 = (5\sqrt{I})^2 = 25I$.
The minimum intensity in an interference pattern is given by $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Substituting the given values,$I_{\min} = (\sqrt{4I} - \sqrt{9I})^2 = (2\sqrt{I} - 3\sqrt{I})^2 = (-\sqrt{I})^2 = I$.
The difference between the maximum and minimum intensities is $I_{\max} - I_{\min} = 25I - I = 24I$.
Comparing this with $xI$,we get $x = 24$.

(D) Given the ratio of intensities $I_1 : I_2 = 1 : 9$. Let $I_1 = I$ and $I_2 = 9I$.
The formula for the ratio of maximum to minimum intensity in an interference pattern is given by:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I} + \sqrt{9I})^2}{(\sqrt{I} - \sqrt{9I})^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I} + 3\sqrt{I})^2}{(\sqrt{I} - 3\sqrt{I})^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(4\sqrt{I})^2}{(-2\sqrt{I})^2} = \frac{16I}{4I} = \frac{4}{1}$
Thus,the ratio is $4 : 1$.

(C) The resultant amplitude $A$ of two waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by the formula: $A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}$.
Here,$A_1 = E_0$,$A_2 = E_0$,and $\phi = \frac{\pi}{3}$.
Substituting these values into the formula:
$A = \sqrt{E_0^2 + E_0^2 + 2(E_0)(E_0) \cos(\frac{\pi}{3})}$
Since $\cos(\frac{\pi}{3}) = 0.5$,we have:
$A = \sqrt{E_0^2 + E_0^2 + 2E_0^2(0.5)}$
$A = \sqrt{2E_0^2 + E_0^2} = \sqrt{3E_0^2}$
$A = \sqrt{3} E_0 \approx 1.732 E_0$.

(C) The resultant intensity $I_R$ in an interference pattern is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.
Given $I_1 = I$ and $I_2 = 9I$.
At point $P$,the phase difference $\phi_P = \pi / 2$. Thus,$I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi / 2) = 10I + 6I(0) = 10I$.
At point $Q$,the phase difference $\phi_Q = \pi$. Thus,$I_Q = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi) = 10I + 6I(-1) = 10I - 6I = 4I$.
The difference between the resultant intensities at points $P$ and $Q$ is $|I_P - I_Q| = |10I - 4I| = 6I$.

(A) In the phenomenon of interference, the intensity of light is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$, where $I_1$ and $I_2$ are the intensities of the two waves.
Since $I \propto A^2$, where $A$ is the amplitude, the intensities at the points of destructive interference are given by $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (A_1 - A_2)^2$.
If the amplitudes $A_1$ and $A_2$ are different, then $(A_1 - A_2)^2 \neq 0$.
Therefore, the intensity at the region of destructive interference is not zero, meaning there is some residual intensity of light.

(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 $I \propto \cos^2(\phi/2)$.

(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \Delta x$.
For the first point,$\Delta x_1 = 0$,so $\Delta \phi_1 = 0$. The intensity is $I_1 = I_{max} \cos^2(\frac{\Delta \phi_1}{2}) = I_{max} \cos^2(0) = I_{max}$.
For the second point,$\Delta x_2 = \frac{\lambda}{2}$,so $\Delta \phi_2 = \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{2} = \pi$. The intensity is $I_2 = I_{max} \cos^2(\frac{\pi}{2}) = 0$.
The ratio of intensities is $\frac{I_1}{I_2} = \frac{I_{max}}{0} = \infty$.
Thus,the ratio is $\infty: 1$.

(C) The total phase difference $\Delta \phi$ at point $P$ is given by the sum of the initial phase difference between the sources and the phase difference due to the path difference.
Let $\phi_1$ be the initial phase difference between the sources $A$ and $B$. Given $\phi_1 = 54^{\circ}$.
Converting to radians: $\phi_1 = 54 \times \frac{\pi}{180} = 0.3 \pi \text{ rad}$.
Let $\phi_2$ be the phase difference due to the path difference $\Delta x = PB - PA = 2.5 \lambda$.
The formula for phase difference due to path difference is $\phi_2 = \frac{2 \pi}{\lambda} \times \Delta x$.
Substituting the values: $\phi_2 = \frac{2 \pi}{\lambda} \times 2.5 \lambda = 5 \pi \text{ rad}$.
Since source $A$ is ahead in phase, the total phase difference is $\Delta \phi = \phi_2 + \phi_1 = 5 \pi + 0.3 \pi = 5.3 \pi \text{ rad}$.

(C) The correct option is $C$.
Statement $A$ is false because when the crest of one wave coincides with the crest of another wave,the amplitudes add up,resulting in constructive interference,not destructive interference.
Statement $B$ is true because,by definition,coherent sources are sources that emit light waves of the same frequency and maintain a constant phase difference over time.

(A) In interference,the fringes are equally bright and equally spaced,and the intensity of the bright fringe is four times the intensity of each incident wave.
Resultant intensity at a point is given by $I = I_1 + I_2 + 2\sqrt{I_1}\sqrt{I_2}\cos\delta$,where $I_1$ and $I_2$ are the intensities of the waves from the two sources.
For constructive interference,the crest of one wave coincides with the crest of another wave,or the trough coincides with the trough. Statement $C$ is incorrect because the condition described (crest coinciding with trough) leads to destructive interference.
For constructive interference at the central maxima,the phase difference $\delta = 0$.
If $I_1 = I_2 = I$,then $I_{\max} = I + I + 2\sqrt{I}\sqrt{I}\cos(0) = 4I$.
Thus,statements $A$ and $B$ are correct.

(D) In an interference pattern produced by two coherent sources,the condition for constructive interference (maximum intensity) is given by the path difference $\Delta x = n \lambda$,where $n$ is the order of the maximum $(n = 0, 1, 2, 3, \dots)$.
For the $10^{\text{th}}$ order maximum,we substitute $n = 10$ into the formula.
Therefore,the path difference $\Delta x = 10 \lambda$.

(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.

(A) The intensity $I$ of a wave is proportional to the square of its amplitude $a$,i.e.,$I \propto a^{2}$.
This implies that the amplitude $a$ is proportional to the square root of the intensity,i.e.,$a \propto \sqrt{I}$.
Let the amplitudes of the two coherent sources be $a_{1}$ and $a_{2}$,where $a_{1} \propto \sqrt{I_{1}}$ and $a_{2} \propto \sqrt{I_{2}}$.
In an interference pattern,the maximum intensity $(I_{\max})$ occurs at points of constructive interference,where the phase difference is an even multiple of $\pi$ $(2n\pi)$.
At these points,the resultant amplitude is the sum of the individual amplitudes: $A_{\max} = a_{1} + a_{2}$.
Since $I_{\max} \propto A_{\max}^{2}$,we have $I_{\max} \propto (a_{1} + a_{2})^{2}$.
Substituting the expressions for $a_{1}$ and $a_{2}$,we get $I_{\max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$.

(D) The intensity of the resultant wave in interference is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference and $I_0$ is the intensity of each individual source.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x = 0$,the phase difference $\phi = 0$. Thus,$I_1 = 4I_0 \cos^2(0) = 4I_0$.
For path difference $\Delta x = \lambda/2$,the phase difference $\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{2} = \pi$. Thus,$I_2 = 4I_0 \cos^2(\pi/2) = 4I_0 \cdot 0 = 0$.
The ratio of intensities is $\frac{I_1}{I_2} = \frac{4I_0}{0} = \infty$.
Therefore,the ratio is $\infty:1$.

(C) In an interference pattern, the intensity $I$ is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For destructive interference, the phase difference $\phi = (2n+1)\pi$, so $\cos \phi = -1$.
The minimum intensity is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
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, there is some residual intensity of light instead of complete darkness.

(D) For destructive interference to occur,the two light waves must arrive at a point with a phase difference that results in the cancellation of their amplitudes.
This occurs when the waves are in anti-phase,meaning one wave is at a crest while the other is at a trough.
The condition for destructive interference is given by the phase difference $\Delta \phi = (2n + 1)\pi$,where $n = 0, 1, 2, \ldots$.
Substituting values for $n$,we get $\Delta \phi = \pi, 3\pi, 5\pi, \ldots$.

(C) In the phenomenon of interference,the energy is redistributed in the medium.
At points of constructive interference,the intensity (and thus energy) is maximum,while at points of destructive interference,the intensity is minimum.
Since the total energy remains constant throughout the medium,the phenomenon of interference is based on the principle of conservation of energy.

(C) In an interference experiment,the resultant intensity $I$ is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference between the two interfering waves.
The intensity is minimum when $\cos \phi = -1$.
This occurs when the phase difference $\phi$ is an odd multiple of $\pi$.
Therefore,the condition for minimum intensity is $\phi = (2n-1) \pi$,where $n = 1, 2, 3, \ldots$.

(D) The optical path length is defined as the product of the refractive index of the medium and the geometric distance traveled by the light.
In vacuum (or air),the optical path for a distance $t$ is $t_{opt} = 1 \times t = t$.
When a mica film of thickness $t$ and refractive index $n$ is introduced,the light travels a distance $t$ through the film.
The optical path through the film is $t'_{opt} = n \times t = nt$.
The change in optical path is $\Delta = t'_{opt} - t_{opt} = nt - t = (n-1)t$.
Since $n > 1$,the optical path increases by $(n-1)t$.

(D) Let the amplitudes of the two waves be $A_1 = 5x$ and $A_2 = x$.
The maximum intensity $I_{max}$ is proportional to the square of the sum of amplitudes: $I_{max} \propto (A_1 + A_2)^2 = (5x + x)^2 = (6x)^2 = 36x^2$.
The minimum intensity $I_{min}$ is proportional to the square of the difference of amplitudes: $I_{min} \propto (A_1 - A_2)^2 = (5x - x)^2 = (4x)^2 = 16x^2$.
The ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{36x^2}{16x^2} = \frac{36}{16} = \frac{9}{4}$.

(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.

(C) Given the ratio of intensities of two waves is $\frac{I_1}{I_2} = \frac{9}{4}$.
Since intensity $I \propto a^2$,the ratio of amplitudes is $\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The maximum intensity is $I_{max} = (a_1 + a_2)^2$ and the minimum intensity is $I_{min} = (a_1 - a_2)^2$.
The ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.
Substituting the values,$\frac{I_{max}}{I_{min}} = \left( \frac{3 + 2}{3 - 2} \right)^2 = \left( \frac{5}{1} \right)^2 = \frac{25}{1}$.
Thus,the ratio is $25: 1$.

(C) The optical path length in a medium of refractive index $\mu$ for a physical path length $L$ is defined as $\Delta = \mu L$.
For the first ray,the optical path is $\Delta_1 = \mu_1 L_1$.
For the second ray,the optical path is $\Delta_2 = \mu_2 L_2$.
The path difference between the two rays is $\Delta x = |\Delta_1 - \Delta_2| = |\mu_1 L_1 - \mu_2 L_2|$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2 \pi}{\lambda} \Delta x$.
Therefore,the phase difference is $\frac{2 \pi}{\lambda} |\mu_1 L_1 - \mu_2 L_2|$.
Comparing this with the given options,the correct expression is $\frac{2 \pi}{\lambda} [\mu_1 L_1 - \mu_2 L_2]$.

(B) Let the intensities of the two waves be $I_1$ and $I_2$.
Given: $I_1/I_2 = 49/16$.
Since intensity $I \propto A^2$,the ratio of amplitudes is $\sqrt{I_1}/\sqrt{I_2} = A_1/A_2 = \sqrt{49}/\sqrt{16} = 7/4$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$.
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \left(\frac{7 + 4}{7 - 4}\right)^2 = \left(\frac{11}{3}\right)^2 = \frac{121}{9}$.
Thus,the ratio is $121:9$.

(B) The condition for destructive interference is that the path difference $\Delta x$ must be an odd multiple of $\frac{\lambda}{2}$.
Given path difference $\Delta x = 31.5 \lambda = \frac{63}{2} \lambda = 63 \left( \frac{\lambda}{2} \right)$.
Since $63$ is an odd integer,the path difference is an odd multiple of $\frac{\lambda}{2}$.
Therefore,the point corresponds to destructive interference,which results in a dark point.

(B) The intensity of a wave is proportional to the square of its amplitude,$I \propto a^2$,so $I = ka^2$. Let $I_1 = ka_1^2$ and $I_2 = ka_2^2$.
The maximum intensity during interference is given by $I_{\max} = k(a_1 + a_2)^2 = k(a_1^2 + a_2^2 + 2a_1a_2)$.
The minimum intensity during interference is given by $I_{\min} = k(a_1 - a_2)^2 = k(a_1^2 + a_2^2 - 2a_1a_2)$.
The sum of the maximum and minimum intensities is $I_{\max} + I_{\min} = k(a_1^2 + a_2^2 + 2a_1a_2) + k(a_1^2 + a_2^2 - 2a_1a_2)$.
$I_{\max} + I_{\min} = k(2a_1^2 + 2a_2^2) = 2(ka_1^2 + ka_2^2)$.
Substituting the intensities back,we get $I_{\max} + I_{\min} = 2(I_1 + I_2)$.

(B) Let the intensity of each individual wave be $I$.
When two identical waves with phase difference $\phi$ superpose,the resultant intensity $I_R$ is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
Since the waves are identical,$I_1 = I_2 = I$.
Substituting these values:
$I_R = I + I + 2\sqrt{I \cdot I} \cos \phi$
$I_R = 2I + 2I \cos \phi$
$I_R = 2I(1 + \cos \phi)$
Using the trigonometric identity $1 + \cos \phi = 2 \cos^2 \frac{\phi}{2}$:
$I_R = 2I(2 \cos^2 \frac{\phi}{2}) = 4I \cos^2 \frac{\phi}{2}$
Therefore,the resultant intensity is proportional to $\cos^2 \frac{\phi}{2}$.

(C) The resultant intensity $I_R$ for two interfering waves 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 $I_2 = 9I$.
At point $P$,the phase difference $\phi_P = \frac{\pi}{2}$.
$I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos(\frac{\pi}{2}) = 10I + 6I(0) = 10I$.
At point $Q$,the phase difference $\phi_Q = \pi$.
$I_Q = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi) = 10I + 6I(-1) = 10I - 6I = 4I$.
The difference between the resultant intensities at points $P$ and $Q$ is $\Delta I = I_P - I_Q = 10I - 4I = 6I$.

(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 resultant intensity of two periodic waves is given by the formula:
$I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\delta$
where $\delta$ is the phase difference between the waves.
For maximum intensity,$\cos\delta = 1$,so:
$I_{\max} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$
For minimum intensity,$\cos\delta = -1$,so:
$I_{\min} = I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$
The sum of the maximum and minimum intensities is:
$I_{\max} + I_{\min} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2} + (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$
Expanding the squares:
$I_{\max} + I_{\min} = (I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}) + (I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}})$
$I_{\max} + I_{\min} = 2(I_{1} + I_{2})$

(B) The number of waves $N$ in a given length $L$ is calculated by the formula: $N = \frac{L}{\lambda}$.
Given:
Length $L = 1 \text{ mm} = 10^{-3} \text{ m}$.
Wavelength $\lambda = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} = 4 \times 10^{-7} \text{ m}$.
Substituting the values:
$N = \frac{10^{-3}}{4 \times 10^{-7}}$
$N = \frac{1}{4} \times 10^{-3 - (-7)}$
$N = 0.25 \times 10^4$
$N = 2500$.
Therefore, the number of waves is $2500$.

(C) The resultant intensity $I$ of two superposing waves is given by the formula: $I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} \cos \phi$.
For non-coherent sources,the phase difference $\phi$ changes randomly with time.
Therefore,the average value of the interference term $\langle 2\sqrt{I_{1}I_{2}} \cos \phi \rangle$ over time is $0$ because the average value of $\cos \phi$ is $0$.
Given that $I_{1} = I_{2} = I_{0}$,the resultant average intensity is:
$I_{avg} = I_{1} + I_{2} = I_{0} + I_{0} = 2I_{0}$.

(A) When a light wave travels from a rarer medium to a non-reflecting and non-absorbing medium,the total energy it carries remains the same.
Energy in a light wave is proportional to the frequency of the wave and the number of photons present.
Since the frequency of light remains constant when it changes mediums and the medium is non-absorbing (no energy loss) and non-reflecting (no energy diverted),the total energy is conserved.

(D) The resultant amplitude $R$ of two interfering waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$.
Given $A_1 = 3 A$,$A_2 = 2 A$,and $\phi = 60^{\circ}$.
The intensity $I$ is proportional to the square of the resultant amplitude,$I \propto R^2$.
$R^2 = (3 A)^2 + (2 A)^2 + 2(3 A)(2 A) \cos(60^{\circ})$.
$R^2 = 9 A^2 + 4 A^2 + 12 A^2 \times (0.5)$.
$R^2 = 13 A^2 + 6 A^2 = 19 A^2$.
Therefore,the intensity is proportional to $19 A^2$.