Young's Double Slit Experiment (YDSE) Questions in English

(C) The intensity of light in an interference pattern is given by $I_{res} = 4I_0 \cos^2(\frac{\phi}{2})$, where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.
At a path difference of $\Delta x = \lambda$, the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
The intensity is $I = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
Now, for a path difference of $\Delta x = \frac{\lambda}{4}$, the phase difference is $\phi' = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I'$ is $4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, we have $I' = 4I_0 (\frac{1}{\sqrt{2}})^2 = 4I_0 \cdot \frac{1}{2} = 2I_0$.
Since $I = 4I_0$, then $I_0 = \frac{I}{4}$.
Substituting this into the expression for $I'$, we get $I' = 2(\frac{I}{4}) = 0.5I$.

(A) The maximum intensity $I_{max}$ and minimum intensity $I_{min}$ in an interference pattern are given by $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Given that $I_{max} = 4 I_{min}$, we have $(\sqrt{I_1} + \sqrt{I_2})^2 = 4(\sqrt{I_1} - \sqrt{I_2})^2$.
Taking the square root on both sides: $\sqrt{I_1} + \sqrt{I_2} = 2(\sqrt{I_1} - \sqrt{I_2})$.
Rearranging the terms: $\sqrt{I_1} + \sqrt{I_2} = 2\sqrt{I_1} - 2\sqrt{I_2}$.
This simplifies to $3\sqrt{I_2} = \sqrt{I_1}$.
Squaring both sides gives $9I_2 = I_1$, so the ratio $\frac{I_1}{I_2} = 9/1$.
The ratio of the intensities of the two waves is $I_2/I_1 = 1/9$.

(C) The position of the $n^{th}$ order maximum is $x_n = n \frac{D \lambda}{d}$.
The position of the $n^{th}$ order minimum is $x'_n = (n - 0.5) \frac{D \lambda}{d}$.
The distance between the $3^{rd}$ order maximum $(n=3)$ and the $3^{rd}$ order minimum $(n=3)$ is:
$\Delta x = x_3 - x'_3 = 3 \frac{D \lambda}{d} - (3 - 0.5) \frac{D \lambda}{d} = 0.5 \frac{D \lambda}{d} = \frac{\beta}{2}$.
Given $\Delta x = 0.18 \,cm = 1.8 \times 10^{-3} \,m$, $D = 1.2 \,m$, and $d = 0.02 \,cm = 2 \times 10^{-4} \,m$.
$\frac{\beta}{2} = 1.8 \times 10^{-3} \,m \implies \beta = 3.6 \times 10^{-3} \,m$.
Since $\beta = \frac{D \lambda}{d}$, we have $\lambda = \frac{\beta d}{D}$.
$\lambda = \frac{(3.6 \times 10^{-3} \,m) \times (2 \times 10^{-4} \,m)}{1.2 \,m} = 6 \times 10^{-7} \,m = 600 \,nm$.

(A) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$.
Taking the logarithmic differentiation, we get the relative error formula: $\frac{\Delta \beta}{\beta} = \frac{\Delta D}{D} - \frac{\Delta d}{d}$.
Given that $D$ is increased by $0.5 \%$, so $\frac{\Delta D}{D} \times 100 = 0.5 \%$.
Given that $d$ is decreased by $0.3 \%$, so $\frac{\Delta d}{d} \times 100 = -0.3 \%$.
Substituting these values into the error formula:
$\frac{\Delta \beta}{\beta} \times 100 = 0.5 \% - (-0.3 \%) = 0.5 \% + 0.3 \% = 0.8 \%$.
Therefore, the fringe width increases by $0.8 \%$.

(C) In Young's double slit experiment, white light consists of wavelengths ranging from $4000 \ Å$ to $7000 \ Å$.
At the central position on the screen, the path difference for all wavelengths is zero.
Since the path difference is zero, all wavelengths interfere constructively at the center, resulting in a white fringe.
As we move away from the center, the path difference increases, causing different wavelengths to interfere constructively at different positions, resulting in coloured fringes.

(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 distance between the two slits.
When the entire arrangement is placed in a liquid of refractive index $n$, the wavelength of light changes to $\lambda' = \frac{\lambda}{n}$.
Since $D$ and $d$ remain unchanged, the new fringe width $\beta'$ becomes $\beta' = \frac{\lambda' D}{d} = \frac{(\lambda / n) D}{d} = \frac{1}{n} \left( \frac{\lambda D}{d} \right) = \frac{\beta}{n}$.
Therefore, the new fringe width is $\frac{\beta}{n}$.

(A) The condition for the $n$-th bright fringe is given by $y = n \frac{\lambda D}{d}$.
For the bright fringes of both wavelengths $\lambda_1 = 650 \ nm$ and $\lambda_2 = 550 \ nm$ to coincide, their positions must be equal: $y_1 = y_2$.
$n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$.
$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$.
Since we want the least distance, we take the smallest integers $n_1 = 11$ and $n_2 = 13$.
Substituting $n_1 = 11$ into the position formula:
$y = 11 \times \frac{650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$.
$y = 11 \times 325 \times 1.2 \times 10^{-6} = 4290 \times 10^{-6} \ m = 429 \times 10^{-5} \ m$.

(B) Given: Slit separation $d = 2 \ mm = 2 \times 10^{-3} \ m$, Screen distance $D = 10 \ m$, Wavelength $\lambda = 6000 \ \mathring{A} = 6 \times 10^{-7} \ m$.
In $Young's$ double slit experiment, the intensity at any point on the screen is given by $I = 4 I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference.
The path difference $\Delta x$ at a position $y$ on the screen is $\Delta x = \frac{yd}{D}$.
For a point in front of one of the slits, $y = \frac{d}{2}$.
Thus, $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.
Substituting the values: $\Delta x = \frac{(2 \times 10^{-3})^2}{2 \times 10} = \frac{4 \times 10^{-6}}{20} = 2 \times 10^{-7} \ m$.
The phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{6 \times 10^{-7}} \times 2 \times 10^{-7} = \frac{2\pi}{3}$.
The intensity $I = I_{max} \cos^2(\frac{\phi}{2}) = (4 I_0) \cos^2(\frac{2\pi/3}{2}) = 4 I_0 \cos^2(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, we have $I = 4 I_0 (\frac{1}{2})^2 = 4 I_0 \times \frac{1}{4} = I_0$.

(C) The angular fringe width $\theta$ in a Young's double slit experiment is given by the formula $\theta = \frac{\lambda}{d}$, where $\lambda$ is the wavelength of the light used and $d$ is the distance between the two slits.
Statement $I$: The angular fringe width depends only on the wavelength $\lambda$ and the slit separation $d$. It is independent of the distance $D$ between the screen and the slits. Therefore, moving the screen away does not change the angular separation. Thus, Statement $I$ is false.
Statement $II$: Since $\theta = \frac{\lambda}{d}$, the angular fringe width is directly proportional to the wavelength $\lambda$. If the source is replaced by one with a higher wavelength, $\theta$ will increase. Thus, Statement $II$ is true.
Conclusion: Statement $I$ is false and Statement $II$ is true.

(C) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{D\lambda}{d}$.
Given that the fringe widths are equal, we have $\beta_1 = \beta_2$.
Therefore, $\frac{D_1 \lambda_1}{d_1} = \frac{D_2 \lambda_2}{d_2}$.
Rearranging the terms to find the ratio $\frac{D_1}{D_2}$, we get $\frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2}$.
Given the ratios of slit separations $\frac{d_1}{d_2} = 2$ and the ratio of wavelengths $\frac{\lambda_1}{\lambda_2} = \frac{1}{2}$, which implies $\frac{\lambda_2}{\lambda_1} = 2$.
Substituting these values, we get $\frac{D_1}{D_2} = 2 \times 2 = 4$.

(B) Given: Slit separation $d = 0.54$ mm $= 0.54 \times 10^{-3}$ m.
Distance of screen $D = 1.8$ m.
Distance of the $n^{th}$ bright fringe from the central fringe is given by $y_n = \frac{n\lambda D}{d}$.
For the $6^{th}$ bright fringe, $n = 6$ and $y_6 = 1.2$ cm $= 1.2 \times 10^{-2}$ m.
Rearranging the formula for wavelength $\lambda$: $\lambda = \frac{y_n d}{n D}$.
Substituting the values: $\lambda = \frac{1.2 \times 10^{-2} \times 0.54 \times 10^{-3}}{6 \times 1.8}$.
$\lambda = \frac{0.648 \times 10^{-5}}{10.8} = 0.06 \times 10^{-5}$ m.
$\lambda = 600 \times 10^{-9}$ m $= 600$ nm.

(B) Given: Slit separation $d = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}$.
Distance to screen $D = 2.0 \text{ m}$.
Order of fringe $n = 3$.
Distance of $n^{th}$ bright fringe $y_n = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}$.
The formula for the position of the $n^{th}$ bright fringe is $y_n = \frac{n \lambda D}{d}$.
Rearranging for wavelength $\lambda$: $\lambda = \frac{y_n d}{n D}$.
Substituting the values: $\lambda = \frac{1.5 \times 10^{-2} \times 2 \times 10^{-4}}{3 \times 2.0}$.
$\lambda = \frac{3 \times 10^{-6}}{6} = 0.5 \times 10^{-6} \text{ m}$.
Converting to $ \mathring{A}$s: $\lambda = 0.5 \times 10^{-6} \times 10^{10} \mathring{A} = 5000 \mathring{A}$.
Thus, the wavelength of light used is $5000 \mathring{A}$.

(B) The fringe width $\beta$ in a medium of refractive index $\mu$ is given by the formula $\beta' = \frac{\beta}{\mu}$.
Given the initial fringe width $\beta = 2.4 \text{ } \mu\text{m}$ and the refractive index of the new medium $\mu = 1.2$.
Substituting these values into the formula, we get $\beta' = \frac{2.4}{1.2} = 2 \text{ } \mu\text{m}$.
Therefore, the new fringe width is $2 \text{ } \mu\text{m}$.

(C) Let $I_s$ be the intensity due to an individual slit.
The resultant intensity $I_R$ is given by the formula: $I_R = I_s + I_s + 2 \sqrt{I_s I_s} \cos \phi = 2I_s + 2I_s \cos \phi = 2I_s(1 + \cos \phi) = 4I_s \cos^2(\phi/2)$.
Given that the resultant intensity $I_R$ at point $P$ is equal to the intensity due to an individual slit $(I_s)$, we have:
$4I_s \cos^2(\phi/2) = I_s$
$\cos^2(\phi/2) = 1/4$
$\cos(\phi/2) = 1/2$
This implies $\phi/2 = \pi/3$, so the phase difference $\phi = 2\pi/3$.
The path difference $\Delta x$ is related to the phase difference $\phi$ by the formula $\Delta x = (\lambda / 2\pi) \phi$.
Substituting the given values: $\Delta x = (6000 \mathring{A} / 2\pi) \times (2\pi/3) = 2000 \mathring{A}$.

(C) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference.
Given $I = \frac{3}{4} I_{max}$, we have $\frac{3}{4} I_{max} = I_{max} \cos^2(\frac{\phi}{2})$.
This simplifies to $\cos^2(\frac{\phi}{2}) = \frac{3}{4}$, which implies $\cos(\frac{\phi}{2}) = \frac{\sqrt{3}}{2}$.
Therefore, $\frac{\phi}{2} = \frac{\pi}{6}$, which gives the phase difference $\phi = \frac{\pi}{3}$.
The relationship between path difference $\Delta x$ and phase difference $\phi$ is $\Delta x = \frac{\lambda}{2\pi} \phi$.
Substituting $\phi = \frac{\pi}{3}$, we get $\Delta x = \frac{\lambda}{2\pi} \cdot \frac{\pi}{3} = \frac{\lambda}{6}$.
Comparing this with $\frac{\lambda}{x}$, we find $x = 6$.

(C) The intensity of light 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 $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x = \frac{\lambda}{3}$, the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
The intensity at this point is $I_1 = I_{max} \cos^2(\frac{2\pi/3}{2}) = I_{max} \cos^2(\frac{\pi}{3}) = I_{max} (\frac{1}{2})^2 = \frac{I_{max}}{4}$.
Given $I_1 = K$, we have $K = \frac{I_{max}}{4}$, which implies $I_{max} = 4K$.
For path difference $\Delta x = \lambda$, the phase difference is $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
The intensity at this point is $I_2 = I_{max} \cos^2(\frac{2\pi}{2}) = I_{max} \cos^2(\pi) = I_{max} (-1)^2 = I_{max}$.
Since $I_{max} = 4K$, the intensity at path difference $\lambda$ is $4K$.

(A) 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 point exactly in front of one slit, $y = d/2$.
Therefore, the path difference is $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.
For a minimum (destructive interference) to occur, the path difference must be an odd multiple of $\lambda/2$, i.e.,$\Delta x = (2n-1)\frac{\lambda}{2}$ where $n = 1, 2, 3, \dots$.
Equating the two expressions: $\frac{d^2}{2D} = (2n-1)\frac{\lambda}{2}$.
Solving for $\lambda$: $\lambda = \frac{d^2}{D(2n-1)}$.
Thus, $\lambda \propto \frac{1}{2n-1}$.
For $n=1, 2, 3, \dots$, the values of $(2n-1)$ are $1, 3, 5, \dots$.
Therefore, the possible wavelengths are proportional to $\frac{d^2}{D}, \frac{d^2}{3D}, \frac{d^2}{5D}, \dots$.

(B) The angular width of a fringe in Young's double slit experiment is given by the formula $\theta = \frac{\lambda}{d}$, where $\lambda$ is the wavelength of light and $d$ is the slit separation.
When the apparatus is immersed in a medium of refractive index $\mu$, the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Therefore, the new angular width $\theta'$ is given by $\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}$.
Given $\theta = 0.2^\circ$ and $\mu = 4/3$.
Substituting the values, we get $\theta' = \frac{0.2}{4/3} = 0.2 \times \frac{3}{4} = 0.15^\circ$.
Thus, the new angular width of the fringe is $0.15^\circ$.

(B) The path difference introduced by the slabs is given by $\Delta x = (\mu_2 - 1)t - (\mu_1 - 1)t = (\mu_2 - \mu_1)t$.
Substituting the given values: $\Delta x = (1.55 - 1.51) \times 0.1 \text{ mm} = 0.04 \times 0.1 \text{ mm} = 0.004 \text{ mm} = 4 \times 10^{-6} \text{ m}$.
The shift in the central fringe in terms of fringe width $\beta$ is given by $n = \frac{\Delta x}{\lambda}$.
Given $\lambda = 4000 \text{ Å} = 4000 \times 10^{-10} \text{ m} = 4 \times 10^{-7} \text{ m}$.
Therefore, $n = \frac{4 \times 10^{-6}}{4 \times 10^{-7}} = 10$.
Thus, the central bright fringe shifts by $10$ fringes.

(D) In Young's double slit experiment, the path difference $\Delta x$ at any point $P$ is given by $S_2P - S_1P = d \sin \theta$. From the geometry of the figure, $S_2P - S_1P = S_2B = \Delta x$.
For a bright fringe, the path difference is $\Delta x = n\lambda$, where $n$ is the order of the fringe.
Point $Q$ is the $2^{nd}$ bright fringe on one side of the central maximum $O$, so its position is $y_Q = 2\beta$, where $\beta$ is the fringe width.
Point $P$ is the $11^{th}$ bright fringe on the other side of $Q$. The distance of $P$ from $O$ is $y_P = (11 - 2)\beta = 9\beta$ on the opposite side of $O$.
Thus, the path difference at point $P$ is $\Delta x = 9\lambda$.
Given $\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$.
Therefore, $S_2B = 9 \times 6 \times 10^{-7} \text{ m} = 54 \times 10^{-7} \text{ m} = 5.4 \times 10^{-6} \text{ m}$.

(B) The fringe width $\beta$ in a medium of refractive index $n$ is given by the formula: $\beta = \frac{\lambda_n D}{d}$, where $\lambda_n = \frac{\lambda_0}{n}$.
Given:
$\lambda_0 = 6300 \text{ Å} = 6300 \times 10^{-10} \text{ m}$
$n = 1.33$
$D = 1.33 \text{ m}$
$d = 1 \text{ mm} = 10^{-3} \text{ m}$
Substituting the values:
$\lambda_n = \frac{6300 \times 10^{-10}}{1.33} \text{ m}$
$\beta = \frac{(6300 \times 10^{-10} / 1.33) \times 1.33}{10^{-3}}$
$\beta = \frac{6300 \times 10^{-10}}{10^{-3}} = 6300 \times 10^{-7} \text{ m} = 6.3 \times 10^{-4} \text{ m}$.

(D) 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:
Slit separation $d = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}$.
Distance to screen $D = 2 \text{ m}$.
Order $n = 5$.
Wavelengths $\lambda_1 = 480 \text{ nm} = 480 \times 10^{-9} \text{ m}$ and $\lambda_2 = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$.
The position of the $5^{th}$ bright fringe for $\lambda_1$ is $y_1 = \frac{5 \times 480 \times 10^{-9} \times 2}{3 \times 10^{-3}} = 1600 \times 10^{-6} \text{ m} = 1.6 \times 10^{-3} \text{ m}$.
The position of the $5^{th}$ bright fringe for $\lambda_2$ is $y_2 = \frac{5 \times 600 \times 10^{-9} \times 2}{3 \times 10^{-3}} = 2000 \times 10^{-6} \text{ m} = 2.0 \times 10^{-3} \text{ m}$.
The separation between the two fringes is $\Delta y = |y_2 - y_1| = (2.0 - 1.6) \times 10^{-3} \text{ m} = 0.4 \times 10^{-3} \text{ m} = 4 \times 10^{-4} \text{ m}$.

(B) The intensity $I$ at any point in Young's double slit experiment is given by $I = 4I_0 \cos^2(\phi/2)$, where $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = (2\pi/\lambda) \Delta x$.
For the first point, path difference $\Delta x_1 = 0$, so $\phi_1 = 0$. Intensity $I_1 = 4I_0 \cos^2(0) = 4I_0$.
For the second point, path difference $\Delta x_2 = \lambda/4$, so $\phi_2 = (2\pi/\lambda) \times (\lambda/4) = \pi/2$. Intensity $I_2 = 4I_0 \cos^2(\pi/4) = 4I_0 \times (1/\sqrt{2})^2 = 4I_0 \times (1/2) = 2I_0$.
The ratio of intensities is $I_1/I_2 = (4I_0)/(2I_0) = 2/1$ or $2:1$.

(C) The fringe width is given by $\beta = 0.4 \text{ mm}$.
For the $n^{th}$ bright band, the position is $y_n = n\beta$.
For the $m^{th}$ dark band, the position is $y'_m = (m - 0.5)\beta$.
Position of the $6^{th}$ bright band: $y_6 = 6\beta = 6 \times 0.4 = 2.4 \text{ mm}$.
Position of the $4^{th}$ dark band: $y'_4 = (4 - 0.5)\beta = 3.5 \times 0.4 = 1.4 \text{ mm}$.
The distance between them is $|y_6 - y'_4| = |2.4 - 1.4| = 1.0 \text{ mm}$.

(B) In Young's double slit experiment, the path difference $\Delta x$ between the light waves reaching point $P$ from slits $A_1$ and $A_2$ is given by $\Delta x = A_2P - A_1P$.
Given that $A_1P$ is perpendicular to $A_1A_2$, we have a right-angled triangle $\Delta A_1A_2P$ where $A_1A_2 = a$ and $A_1P = D$.
Using the Pythagorean theorem, $A_2P = \sqrt{A_1A_2^2 + A_1P^2} = \sqrt{a^2 + D^2}$.
Thus, the path difference is $\Delta x = \sqrt{a^2 + D^2} - D$.
For the first minimum (destructive interference), the path difference must be equal to $\frac{\lambda}{2}$.
So, $\sqrt{a^2 + D^2} - D = \frac{\lambda}{2}$.
$\sqrt{a^2 + D^2} = D + \frac{\lambda}{2}$.
Squaring both sides, $a^2 + D^2 = (D + \frac{\lambda}{2})^2 = D^2 + \lambda D + \frac{\lambda^2}{4}$.
$a^2 = \lambda D + \frac{\lambda^2}{4}$.
Since $\lambda$ is very small compared to $D$, $\frac{\lambda^2}{4}$ can be neglected.
Therefore, $a^2 \approx \lambda D$, which gives $a = \sqrt{\lambda D}$.

(A) Let the width of the first slit be $w_1 = w$ and the second slit be $w_2 = 2w$. Since intensity $I \propto \text{width}$, we have $I_1 = I_0$ and $I_2 = 2I_0$.
Since $I \propto A^2$, the amplitudes are $A_1 = a$ and $A_2 = a\sqrt{2}$.
The resultant intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi = I_0 + 2I_0 + 2\sqrt{2I_0^2} \cos \phi = 3I_0 + 2\sqrt{2} I_0 \cos \phi$.
The maximum intensity $I_m$ occurs when $\cos \phi = 1$, so $I_m = 3I_0 + 2\sqrt{2} I_0 = I_0(3 + 2\sqrt{2})$.
Using $\cos \phi = 2 \cos^2 \frac{\phi}{2} - 1$, we have $I = 3I_0 + 2\sqrt{2} I_0 (2 \cos^2 \frac{\phi}{2} - 1) = I_0(3 - 2\sqrt{2} + 4\sqrt{2} \cos^2 \frac{\phi}{2})$.
Since $I_0 = \frac{I_m}{3 + 2\sqrt{2}}$, we rationalize: $I_0 = I_m(3 - 2\sqrt{2})$.
Substituting this, $I = I_m(3 - 2\sqrt{2})(3 - 2\sqrt{2} + 4\sqrt{2} \cos^2 \frac{\phi}{2}) = I_m(1 + 8 \cos^2 \frac{\phi}{2}) / 9$.

(C) The intensity $I$ of a wave is proportional to the square of its amplitude $A$, i.e.,$I \propto A^2$.
Let the amplitudes of the two coherent sources be $A_1$ and $A_2$.
The maximum intensity is $I_{max} \propto (A_1 + A_2)^2$ and the minimum intensity is $I_{min} \propto (A_1 - A_2)^2$.
Given the ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{36}{1}$.
Therefore, $\frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{36}{1}$.
Taking the square root on both sides, we get $\frac{A_1 + A_2}{A_1 - A_2} = \frac{6}{1}$.
By applying componendo and dividendo, $\frac{(A_1 + A_2) + (A_1 - A_2)}{(A_1 + A_2) - (A_1 - A_2)} = \frac{6 + 1}{6 - 1}$.
This simplifies to $\frac{2A_1}{2A_2} = \frac{7}{5}$.
Thus, the ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{7}{5}$.

(B) The width of a certain region on the screen is $L$. The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance between the slits and the screen, and $d$ is the distance between the slits.
The number of fringes $n$ in a region of width $L$ is given by $n = \frac{L}{\beta} = \frac{Ld}{\lambda D}$.
Since $L, d,$ and $D$ are constant, we have $n \propto \frac{1}{\lambda}$, which implies $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 8$, $\lambda_1 = 6600 \text{ Å}$, and $\lambda_2 = 4400 \text{ Å}$.
Substituting the values: $8 \times 6600 = n_2 \times 4400$.
$n_2 = \frac{8 \times 6600}{4400} = \frac{8 \times 66}{44} = \frac{8 \times 3}{2} = 12$.
Therefore, the number of fringes observed is $12$.

(C) In Young's double slit experiment, the condition for destructive interference (dark fringe) is that the path difference $\Delta x$ must be an odd multiple of half the wavelength, i.e.,$\Delta x = (2n - 1) \frac{\lambda}{2}$ for $n = 1, 2, 3, \dots$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta x$.
Substituting the value of $\Delta x$ for the $n^{th}$ dark fringe:
$\Delta \phi = \frac{2\pi}{\lambda} \cdot (2n - 1) \frac{\lambda}{2}$.
Simplifying the expression:
$\Delta \phi = (2n - 1)\pi$ radians.
Thus, the correct option is $C$.

(A) The position of the $m^{th}$ bright fringe for wavelength $\lambda_1$ is given by $y_m = m \frac{D \lambda_1}{d}$.
The position of the $n^{th}$ dark fringe for wavelength $\lambda_2$ is given by $y_n = (n - \frac{1}{2}) \frac{D \lambda_2}{d} = \frac{(2n - 1)}{2} \frac{D \lambda_2}{d}$.
Since the fringes coincide, $y_m = y_n$:
$m \frac{D \lambda_1}{d} = \frac{(2n - 1)}{2} \frac{D \lambda_2}{d}$.
Simplifying the equation, we get $m \lambda_1 = \frac{(2n - 1)}{2} \lambda_2$.
Therefore, the ratio $\frac{\lambda_2}{\lambda_1} = \frac{2m}{2n - 1}$.

(A) The condition for constructive interference (maxima) is $d \sin \theta = n\lambda$, where $n$ is an integer.
Since $|\sin \theta| \le 1$, we have $|n| \le \frac{d}{\lambda}$.
Given $d = 2.5\lambda$, the condition becomes $|n| \le 2.5$.
The possible integer values for $n$ are $0, \pm 1, \pm 2$.
These correspond to $n = -2, -1, 0, 1, 2$.
Counting these values, there are $5$ maxima in total.