Intensity in Young's Double Slit Experiment Questions in English

(B) Let the intensities of the two waves be $I_1$ and $I_2$.
Given that the ratio of intensities is $I_1 : I_2 = 9 : 1$. Let $I_1 = 9x$ and $I_2 = x$.
The formula for 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 the values,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{9x} + \sqrt{x}}{\sqrt{9x} - \sqrt{x}} \right)^2$.
This simplifies to $\frac{I_{\max}}{I_{\min}} = \left( \frac{3\sqrt{x} + \sqrt{x}}{3\sqrt{x} - \sqrt{x}} \right)^2 = \left( \frac{4\sqrt{x}}{2\sqrt{x}} \right)^2$.
$\frac{I_{\max}}{I_{\min}} = (2)^2 = 4$.
Thus,the ratio of maximum to minimum intensity is $4 : 1$.

(C) Given the ratio of intensities of two waves is $\frac{I_1}{I_2} = \frac{9}{1}$.
Let the amplitudes of the two waves be $A_1$ and $A_2$. Since intensity $I \propto A^2$,the ratio of amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{9}{1}} = \frac{3}{1}$.
The ratio of maximum 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:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{3 + 1}{3 - 1} \right)^2 = \left( \frac{4}{2} \right)^2 = (2)^2 = \frac{4}{1}$.
Thus,the ratio is $4:1$.

(D) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = \frac{100}{1}$.
Let the amplitudes be $A_1$ and $A_2$,such that $\frac{I_1}{I_2} = \frac{A_1^2}{A_2^2} = 100$,which implies $\frac{A_1}{A_2} = 10$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 = \left( \frac{\frac{A_1}{A_2} + 1}{\frac{A_1}{A_2} - 1} \right)^2$.
Substituting the value $\frac{A_1}{A_2} = 10$:
$\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{10 + 1}{10 - 1} \right)^2 = \left( \frac{11}{9} \right)^2 = \frac{121}{81}$.

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

(A) The intensity $I$ of a wave is proportional to the square of its amplitude $a$,i.e.,$I \propto a^2$.
Given the ratio of amplitudes $\frac{a_1}{a_2} = \frac{1}{9}$.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = \left( \frac{\frac{a_1}{a_2} + 1}{\frac{a_1}{a_2} - 1} \right)^2$.
Substituting the given ratio $\frac{a_1}{a_2} = \frac{1}{9}$:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\frac{1}{9} + 1}{\frac{1}{9} - 1} \right)^2 = \left( \frac{\frac{10}{9}}{-\frac{8}{9}} \right)^2 = \left( -\frac{10}{8} \right)^2 = \left( -\frac{5}{4} \right)^2 = \frac{25}{16}$.

(A) Given that the intensity of the first slit $I_1 = 2I_2$,where $I_2$ is the intensity of the second slit.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$
Substituting $I_1 = 2I_2$ into the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{2I_2} + \sqrt{I_2}}{\sqrt{2I_2} - \sqrt{I_2}} \right)^2 = \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)^2$
Rationalizing the denominator:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} \right)^2 = (\sqrt{2} + 1)^4 = (2 + 1 + 2\sqrt{2})^2 = (3 + 2\sqrt{2})^2$
Calculating the value:
$\frac{I_{\max}}{I_{\min}} = 9 + 8 + 12\sqrt{2} = 17 + 12(1.414) \approx 17 + 16.968 \approx 33.968 \approx 34$.

(A) The visibility $V$ of interference fringes is defined as $V = \frac{I_{max} - I_{min}}{I_{max} + I_{min}}$.
Given the intensities of two coherent sources are $I_1$ and $I_2$,we have $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Substituting these into the formula for $V$:
$V = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2} = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$.
Dividing the numerator and denominator by $I_2$:
$V = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1}$.
Given the ratio $p = I_1/I_2$,we substitute $p$ into the equation:
$V = \frac{2\sqrt{p}}{1 + p}$.

(A) Given the ratio of intensities: $\frac{I_1}{I_2} = \frac{9}{1}$.
Since intensity $I \propto A^2$,the ratio of amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{9}{1}} = \frac{3}{1}$.
Let $A_1 = 3x$ and $A_2 = x$.
The maximum amplitude is $A_{max} = A_1 + A_2 = 3x + x = 4x$.
The minimum amplitude is $A_{min} = |A_1 - A_2| = |3x - x| = 2x$.
The ratio of maximum amplitude to minimum amplitude is $\frac{A_{max}}{A_{min}} = \frac{4x}{2x} = \frac{2}{1}$.

(A) In an interference pattern,the intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For maximum and minimum intensity,$\cos \phi$ is $1$ and $-1$ respectively.
$I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Given $\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9}{1}$,we have $\frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = 9$.
Taking the square root on both sides: $\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = 3$.
Using componendo and dividendo: $\frac{\sqrt{I_1}}{\sqrt{I_2}} = \frac{3+1}{3-1} = \frac{4}{2} = 2$.
Therefore,the ratio of intensities of the sources is $\frac{I_1}{I_2} = (2)^2 = 4$.

(C) The intensity of light $I$ is directly proportional to the width of the slit $W$,so $\frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{4}{9}$.
Let $I_1 = 4k$ and $I_2 = 9k$. The amplitudes are proportional to the square root of intensity,so $A_1 = \sqrt{4k} = 2\sqrt{k}$ and $A_2 = \sqrt{9k} = 3\sqrt{k}$.
The ratio of maximum intensity to minimum intensity is given by the formula $\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2}$.
Substituting the values: $\frac{I_{max}}{I_{min}} = \frac{(2\sqrt{k} + 3\sqrt{k})^2}{(2\sqrt{k} - 3\sqrt{k})^2} = \frac{(5\sqrt{k})^2}{(-\sqrt{k})^2} = \frac{25k}{k} = \frac{25}{1}$.
Thus,the ratio is $25:1$.

(D) The intensity in a 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 $\phi = (2\pi/\lambda) \Delta x$.
Case $1$: Path difference $\Delta x = \lambda$. Then $\phi = (2\pi/\lambda) \times \lambda = 2\pi$.
Intensity $I = 4I_0 \cos^2(2\pi/2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0 = K$.
Case $2$: Path difference $\Delta x = \lambda/4$. Then $\phi = (2\pi/\lambda) \times (\lambda/4) = \pi/2$.
Intensity $I' = 4I_0 \cos^2((\pi/2)/2) = 4I_0 \cos^2(\pi/4) = 4I_0 (1/\sqrt{2})^2 = 4I_0(1/2) = 2I_0$.
Since $K = 4I_0$,then $2I_0 = K/2$.
Therefore,the intensity at the point is $K/2$.

(A) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.
For the first point,the path difference is $\Delta x_1 = \lambda$. Thus,the phase difference is $\phi_1 = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
The intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Since the sources are identical,$I_1 = I_2 = I_0$,so $I = 2I_0 + 2I_0 \cos \phi = 4I_0 \cos^2(\phi/2)$.
For the first point: $K = 4I_0 \cos^2(2\pi/2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Therefore,$I_0 = K/4$.
For the second point,the path difference is $\Delta x_2 = \lambda/4$. The phase difference is $\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity at this point is $I' = 4I_0 \cos^2(\phi_2/2) = 4I_0 \cos^2(\pi/4) = 4I_0 (1/\sqrt{2})^2 = 4I_0 (1/2) = 2I_0$.
Substituting $I_0 = K/4$,we get $I' = 2(K/4) = K/2$ units.

(B) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$.
For a path difference $\Delta x_1 = \lambda$,the phase difference is $\phi_1 = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
The intensity is given by $I = I_{max} \cos^2(\phi/2)$.
For $\phi_1 = 2\pi$,$I_1 = I_{max} \cos^2(\pi) = I_{max} = k$.
For a path difference $\Delta x_2 = \lambda/4$,the phase difference is $\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity at this point is $I_2 = I_{max} \cos^2(\phi_2/2) = k \cos^2(\pi/4)$.
Since $\cos(\pi/4) = 1/\sqrt{2}$,we have $I_2 = k \times (1/\sqrt{2})^2 = k/2$.

(B) The intensity at any point on the screen is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = (2\pi / \lambda) \times \Delta x$.
Case $1$: When path difference $\Delta x = \lambda$,the phase difference is $\phi = (2\pi / \lambda) \times \lambda = 2\pi$.
Substituting this into the intensity formula: $I = 4I_0 \cos^2(2\pi / 2) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Given this intensity is $K$,we have $K = 4I_0$.
Case $2$: When path difference $\Delta x = \lambda / 4$,the phase difference is $\phi = (2\pi / \lambda) \times (\lambda / 4) = \pi / 2$.
Substituting this into the intensity formula: $I' = 4I_0 \cos^2((\pi / 2) / 2) = 4I_0 \cos^2(\pi / 4) = 4I_0 (1 / \sqrt{2})^2 = 4I_0 (1 / 2) = 2I_0$.
Since $K = 4I_0$,then $2I_0 = K / 2$. Therefore,the intensity is $K / 2$.

(B) The intensity $I$ of light is directly proportional to the width $W$ of the slit,and also to the square of the amplitude $A$ of the wave.
$\therefore \frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{A_1^2}{A_2^2}$
Given the ratio of widths $\frac{W_1}{W_2} = \frac{1}{25}$,we have:
$\frac{A_1^2}{A_2^2} = \frac{1}{25} \implies \frac{A_1}{A_2} = \sqrt{\frac{1}{25}} = \frac{1}{5}$
The ratio of maximum intensity to minimum intensity is given by:
$\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \left( \frac{\frac{A_1}{A_2} + 1}{\frac{A_1}{A_2} - 1} \right)^2$
Substituting $\frac{A_1}{A_2} = \frac{1}{5}$:
$\frac{I_{max}}{I_{min}} = \left( \frac{\frac{1}{5} + 1}{\frac{1}{5} - 1} \right)^2 = \left( \frac{\frac{6}{5}}{-\frac{4}{5}} \right)^2 = \left( -\frac{6}{4} \right)^2 = \left( -\frac{3}{2} \right)^2 = \frac{9}{4}$

(A) Let the intensities of the two sources be $I_1$ and $I_2$ such that $\frac{I_1}{I_2} = \alpha$. Since $I \propto A^2$,we have $\frac{A_1}{A_2} = \sqrt{\alpha}$.
The maximum and minimum intensities in an interference pattern are given by $I_{max} = (A_1 + A_2)^2$ and $I_{min} = (A_1 - A_2)^2$.
We need to calculate the value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$.
Substituting the expressions for $I_{max}$ and $I_{min}$:
$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{(A_1 + A_2)^2 - (A_1 - A_2)^2}{(A_1 + A_2)^2 + (A_1 - A_2)^2}$
Expanding the squares:
$= \frac{(A_1^2 + A_2^2 + 2A_1A_2) - (A_1^2 + A_2^2 - 2A_1A_2)}{(A_1^2 + A_2^2 + 2A_1A_2) + (A_1^2 + A_2^2 - 2A_1A_2)}$
$= \frac{4A_1A_2}{2(A_1^2 + A_2^2)} = \frac{2A_1A_2}{A_1^2 + A_2^2}$
Dividing the numerator and denominator by $A_2^2$:
$= \frac{2(A_1/A_2)}{(A_1/A_2)^2 + 1}$
Since $\frac{A_1}{A_2} = \sqrt{\alpha}$,we substitute this into the expression:
$= \frac{2\sqrt{\alpha}}{(\sqrt{\alpha})^2 + 1} = \frac{2\sqrt{\alpha}}{\alpha + 1}$.

(B) The fringe visibility $V$ is defined as $V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$.
Given the intensities of the two sources are $I_1$ and $I_2$,such that $\frac{I_1}{I_2} = p$.
The maximum intensity is $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and the minimum intensity is $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Substituting these into the visibility formula:
$V = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2} = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$.
Dividing the numerator and denominator by $I_2$:
$V = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1} = \frac{2\sqrt{p}}{p + 1}$.
Thus,the correct option is $B$.

(B) The intensity at any point in an interference pattern is given by $I = I_{max} \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase difference.
Given the path difference $\Delta = \frac{\lambda}{4}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \Delta = \frac{2\pi}{\lambda} \left( \frac{\lambda}{4} \right) = \frac{\pi}{2}$.
The intensity at this point is $I = I_{max} \cos^2 \left( \frac{\pi/2}{2} \right) = I_{max} \cos^2 \left( \frac{\pi}{4} \right) = I_{max} \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{I_{max}}{2}$.
At the central fringe,the path difference is $0$,so the intensity is $I_{max}$.
Thus,the ratio of the intensity at this point to the intensity at the central fringe is $\frac{I_{max}/2}{I_{max}} = \frac{1}{2}$ or $1:2$.

(A) The ratio of maximum to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \frac{(a+b)^2}{(a-b)^2} = \frac{49}{9}$.
Taking the square root on both sides,we get $\frac{a+b}{a-b} = \frac{7}{3}$.
By cross-multiplying,$3(a+b) = 7(a-b)$,which simplifies to $3a + 3b = 7a - 7b$.
Rearranging the terms,$4a = 10b$,so the ratio of amplitudes is $\frac{a}{b} = \frac{10}{4} = \frac{5}{2}$.
Since intensity $I \propto a^2$,the ratio of intensities of the component waves is $\frac{I_1}{I_2} = \frac{a^2}{b^2} = \left(\frac{a}{b}\right)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25$.

(A) The intensity in an interference pattern is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
At the central bright fringe,the phase difference $\phi = 0$,so $I_1 = 4I_0 \cos^2(0) = 4I_0$.
For a point at a distance $x = \frac{\beta}{4}$ from the centre,where $\beta$ is the fringe width,the phase difference $\phi$ is given by $\phi = \frac{2\pi}{\beta} \cdot x$.
Substituting $x = \frac{\beta}{4}$,we get $\phi = \frac{2\pi}{\beta} \cdot \frac{\beta}{4} = \frac{\pi}{2}$.
The intensity at this point is $I_2 = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4}) = 4I_0 (\frac{1}{\sqrt{2}})^2 = 4I_0 \cdot \frac{1}{2} = 2I_0$.
The ratio of the intensities is $\frac{I_1}{I_2} = \frac{4I_0}{2I_0} = 2$.

(C) The path difference is given by $\Delta p = \frac{xd}{D}$.
Given $x = 4 \times 10^{-5} \, m$,$d = 0.25 \, cm = 2.5 \times 10^{-3} \, m$,and $D = 100 \, cm = 1 \, m$.
$\Delta p = \frac{(4 \times 10^{-5} \, m) \times (2.5 \times 10^{-3} \, m)}{1 \, m} = 10^{-7} \, m$.
The phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta p$.
Given $\lambda = 6000 \mathring{A} = 6 \times 10^{-7} \, m$.
$\phi = \frac{2\pi}{6 \times 10^{-7}} \times 10^{-7} = \frac{\pi}{3} = 60^{\circ}$.
The resultant intensity $I$ is given by $I = I_0 \cos^2(\phi / 2)$.
$I = I_0 \cos^2(60^{\circ} / 2) = I_0 \cos^2(30^{\circ})$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $I = I_0 \times (\frac{\sqrt{3}}{2})^2 = I_0 \times \frac{3}{4} = \frac{3I_0}{4}$.

(A) The intensity $I$ at a point on the screen in a Young's double slit experiment is given by the formula $I = I_0 \cos^2 \left( \frac{\phi}{2} \right)$,where $I_0$ is the maximum intensity and $\phi$ is the phase difference.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the path difference $\Delta x = \frac{\lambda}{6}$,we calculate the phase difference:
$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
Now,substitute $\phi$ into the intensity formula:
$\frac{I}{I_0} = \cos^2 \left( \frac{\pi/3}{2} \right) = \cos^2 \left( \frac{\pi}{6} \right)$.
Since $\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$,we have:
$\frac{I}{I_0} = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}$.

(A) Let the amplitudes be $a_1 = a$ and $a_2 = 2a$. The intensities are $I_1 = a^2$ and $I_2 = (2a)^2 = 4a^2 = 4I_1$.
The resultant intensity $I$ is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Substituting the values: $I = I_1 + 4I_1 + 2\sqrt{I_1(4I_1)} \cos \phi = 5I_1 + 4I_1 \cos \phi$.
The maximum intensity $I_m$ occurs when $\cos \phi = 1$,so $I_m = (a_1 + a_2)^2 = (a + 2a)^2 = 9a^2 = 9I_1$.
Thus,$I_1 = \frac{I_m}{9}$.
Substituting $I_1$ into the expression for $I$:
$I = \frac{5I_m}{9} + \frac{4I_m}{9} \cos \phi = \frac{I_m}{9}(5 + 4 \cos \phi)$.
Using the identity $\cos \phi = 2\cos^2 \frac{\phi}{2} - 1$:
$I = \frac{I_m}{9}(5 + 4(2\cos^2 \frac{\phi}{2} - 1)) = \frac{I_m}{9}(5 + 8\cos^2 \frac{\phi}{2} - 4) = \frac{I_m}{9}(1 + 8\cos^2 \frac{\phi}{2})$.

(B) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $\beta = \frac{I_1}{I_2}$.
Maximum intensity $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Minimum intensity $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
We need to calculate $X = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$.
Substituting the expressions:
$I_{\max} - I_{\min} = (\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2 = 4\sqrt{I_1 I_2}$.
$I_{\max} + I_{\min} = (\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2 = 2(I_1 + I_2)$.
Thus,$X = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$.
Dividing numerator and denominator by $I_2$:
$X = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1} = \frac{2\sqrt{\beta}}{\beta + 1}$.

(D) Let the intensities of the two beams be $I_1 = I$ and $I_2 = I + \Delta I$.
Given that the difference in intensity is $1\%$,we have $\frac{\Delta I}{I} = 1\% = 10^{-2}$.
The intensity of the minima in a $YDSE$ is given by $I_{\text{min}} = (\sqrt{I_2} - \sqrt{I_1})^2$.
Substituting the values: $I_{\text{min}} = (\sqrt{I + \Delta I} - \sqrt{I})^2 = I \left( \sqrt{1 + \frac{\Delta I}{I}} - 1 \right)^2$.
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$:
$I_{\text{min}} \approx I \left( 1 + \frac{1}{2} \frac{\Delta I}{I} - 1 \right)^2 = I \left( \frac{1}{2} \frac{\Delta I}{I} \right)^2$.
$I_{\text{min}} = \frac{I}{4} \left( \frac{\Delta I}{I} \right)^2 = \frac{I}{4} (10^{-2})^2 = \frac{I}{4} (10^{-4})$.

(B) The intensity at any point in an interference pattern is given by $I = I_0 \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase difference.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the 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:
$I = I_0 \cos^2 \left( \frac{\pi/3}{2} \right) = I_0 \cos^2 \left( \frac{\pi}{6} \right)$.
Since $\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$,we have:
$I = I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = I_0 \times \frac{3}{4}$.
Therefore,the ratio $\frac{I_0}{I} = \frac{4}{3}$.

(C) The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.
Given amplitudes are $A$ and $2A$,the intensities are $I_1 = I$ and $I_2 = 4I$.
The maximum intensity $I_0$ is given by $I_0 = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.
Thus,$I = \frac{I_0}{9}$.
The resultant intensity $I'$ at a point with phase difference $\phi$ is given by $I' = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Substituting the values: $I' = I + 4I + 2\sqrt{I \cdot 4I} \cos \phi = 5I + 4I \cos \phi$.
Substituting $I = \frac{I_0}{9}$,we get $I' = \frac{I_0}{9} (5 + 4 \cos \phi)$.

(D) Let the intensities of the two coherent beams be $I_1 = I$ and $I_2 = I + \Delta I$.
Given that the intensities differ by $1\%$,we have $\frac{\Delta I}{I} = 1\% = 10^{-2}$.
The intensity of the minima in a $YDSE$ setup is given by $I_{\text{min}} = (\sqrt{I_2} - \sqrt{I_1})^2$.
Substituting the values,$I_{\text{min}} = (\sqrt{I + \Delta I} - \sqrt{I})^2 = I \left( \sqrt{1 + \frac{\Delta I}{I}} - 1 \right)^2$.
Using the binomial approximation $(1+x)^n \approx 1 + nx$ for small $x$,where $x = \frac{\Delta I}{I} = 10^{-2}$:
$I_{\text{min}} \approx I \left( (1 + \frac{1}{2} \cdot \frac{\Delta I}{I}) - 1 \right)^2 = I \left( \frac{1}{2} \cdot \frac{\Delta I}{I} \right)^2$.
$I_{\text{min}} = I \cdot \frac{1}{4} \cdot \left( \frac{\Delta I}{I} \right)^2 = \frac{I}{4} (10^{-2})^2 = \frac{I}{4} (10^{-4})$.

(B) The resultant intensity in Young's double slit experiment is given by $I_R = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$,where $I = 4I_0$ is the maximum intensity.
Given the intensity at a point is $I' = \frac{I}{4} = \frac{4I_0}{4} = I_0$.
Substituting this into the formula: $I_0 = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$.
$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \implies \cos\left(\frac{\phi}{2}\right) = \frac{1}{2}$.
This implies $\frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3}$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Since $\Delta x = d \sin \theta$,we have $\phi = \frac{2\pi}{\lambda} d \sin \theta$.
Equating the two expressions for $\phi$: $\frac{2\pi}{\lambda} d \sin \theta = \frac{2\pi}{3}$.
Solving for $\sin \theta$: $\sin \theta = \frac{\lambda}{3d}$.
Therefore,$\theta = \sin^{-1}\left(\frac{\lambda}{3d}\right)$.

(C) Given the ratio of amplitudes $A_1 : A_2 = 2 : 3$.
Since intensity $I \propto A^2$,the ratio of intensities is $\frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.
Let $I_1 = 4k$ and $I_2 = 9k$.
The intensity of the minimum is $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{4k} - \sqrt{9k})^2 = (2\sqrt{k} - 3\sqrt{k})^2 = (-\sqrt{k})^2 = k$.
The intensity of the maximum is $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{4k} + \sqrt{9k})^2 = (2\sqrt{k} + 3\sqrt{k})^2 = (5\sqrt{k})^2 = 25k$.
Therefore,the ratio $\frac{I_{\min}}{I_{\max}} = \frac{k}{25k} = \frac{1}{25}$.

(B) The intensity at any point in $YDSE$ is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $I_{max}$ is the intensity of the central bright fringe.
Given $I_{max} = 8 \, mW/m^2$ and path difference $\Delta x = \frac{\lambda}{6}$.
The phase difference $\phi$ is calculated as $\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
Substituting the values into the intensity formula:
$I = 8 \cos^2(\frac{\pi/3}{2}) = 8 \cos^2(\frac{\pi}{6})$.
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have $\cos^2(\frac{\pi}{6}) = \frac{3}{4}$.
Therefore,$I = 8 \times \frac{3}{4} = 6 \, mW/m^2$.

(A) Let $I_1$ and $I_2$ be the intensities of the two sources. The maximum intensity is $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and the minimum intensity is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
The average intensity is $I_{avg} = I_1 + I_2$.
The variation in intensity is given as $5\%$ of the average intensity,so $I_{max} - I_{min} = 0.05 I_{avg}$.
We know $I_{max} - I_{min} = 4\sqrt{I_1 I_2}$ and $I_{avg} = I_1 + I_2$.
Thus,$4\sqrt{I_1 I_2} = 0.05(I_1 + I_2)$.
Let $x = I_1/I_2$. Then $4\sqrt{x} = 0.05(x + 1)$.
$80\sqrt{x} = x + 1 \Rightarrow x - 80\sqrt{x} + 1 = 0$.
Solving for $\sqrt{x}$ using the quadratic formula: $\sqrt{x} = \frac{80 \pm \sqrt{6400 - 4}}{2} = 40 \pm \sqrt{1599} \approx 40 \pm 39.987$.
Taking the larger root,$\sqrt{x} \approx 79.987$,so $x \approx 6398$. However,using the provided solution logic where $I_{max} = 1.05 I_{avg}$ and $I_{min} = 0.95 I_{avg}$:
$\frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \frac{1.05}{0.95} = \frac{21}{19}$.
Taking the square root: $\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = \sqrt{\frac{21}{19}} \approx 1.0513$.
Solving for $r = \sqrt{I_1/I_2}$,we get $r = \frac{1.0513 + 1}{1.0513 - 1} \approx 41$. Thus $x = r^2 \approx 1681$.

(A) At the centre of a bright fringe $(P_1)$,the intensity is maximum,$I_1 = I_{max} = 4I_0$,where $I_0$ is the intensity of each individual slit.
The path difference $\Delta x$ at a distance $y$ from the centre is given by $\Delta x = \frac{yd}{D}$.
Given $y = \frac{\beta}{4}$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.
Thus,$\Delta x = \frac{(\lambda D / 4d) \cdot d}{D} = \frac{\lambda}{4}$.
The phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity at any point is given by $I = I_{max} \cos^2(\frac{\phi}{2})$.
For point $P_2$,$I_2 = I_1 \cos^2(\frac{\pi/2}{2}) = I_1 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $\cos^2(\frac{\pi}{4}) = \frac{1}{2}$.
Therefore,$I_2 = I_1 \cdot \frac{1}{2}$,which implies $\frac{I_1}{I_2} = 2$.

(B) Let the intensity of each individual slit be $I_0$. The resultant intensity is given by $I_R = 4I_0 \cos^2(\frac{\Delta \phi}{2})$.
For a path difference $\Delta x = \lambda$,the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
Given $I_R = M$ at $\Delta x = \lambda$,we have $M = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$. Thus,$I_0 = \frac{M}{4}$.
For a path difference $\Delta x = \frac{\lambda}{3}$,the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$.
The resultant intensity is $I_R' = 4I_0 \cos^2(\frac{2\pi/3}{2}) = 4I_0 \cos^2(\frac{\pi}{3})$.
Substituting $I_0 = \frac{M}{4}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I_R' = 4(\frac{M}{4}) \cdot (\frac{1}{2})^2 = M \cdot \frac{1}{4} = \frac{M}{4}$.

(C) Let the intensities of the two coherent sources be $I_1$ and $I_2$. Given the ratio $I_1/I_2 = \beta/1$,we have $I_1 = \beta I$ and $I_2 = I$.
The fringe visibility (or contrast) $V$ is defined as the ratio of the difference between maximum and minimum intensity to their sum:
$V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$
We know that $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Substituting these into the formula:
$V = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2} = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$
Substituting $I_1 = \beta I$ and $I_2 = I$:
$V = \frac{2\sqrt{(\beta I)(I)}}{\beta I + I} = \frac{2\sqrt{\beta} I}{I(\beta + 1)} = \frac{2\sqrt{\beta}}{1 + \beta}$

(A) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula: $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Given $\Delta x = \frac{\lambda}{8}$,we have $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4}$.
The intensity $I$ at any point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$. Assuming $I_1 = I_2 = I_0$,we get $I = 2I_0 + 2I_0 \cos(\frac{\pi}{4}) = 2I_0(1 + \frac{1}{\sqrt{2}})$.
The intensity at the central fringe (where $\Delta x = 0$) is $I_{max} = (\sqrt{I_0} + \sqrt{I_0})^2 = 4I_0$.
The ratio of the intensities is $\frac{I}{I_{max}} = \frac{2I_0(1 + 0.707)}{4I_0} = \frac{1.707}{2} = 0.8535 \approx 0.853$.

(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 $A$,the phase difference $\phi_A = \frac{\pi}{2}$.
$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos\left(\frac{\pi}{2}\right) = 5I + 2(2I)(0) = 5I$.
At point $B$,the phase difference $\phi_B = 2\pi$.
$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(2\pi) = 5I + 2(2I)(1) = 5I + 4I = 9I$.
The difference between the resultant intensities is $I_B - I_A = 9I - 5I = 4I$.

(D) 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 that the intensity of each wave is $I_0$,we have $I_1 = I_0$ and $I_2 = I_0$.
Substituting these values into the formula:
$I = I_0 + I_0 + 2 \sqrt{I_0 \cdot I_0} \cos \phi$
$I = 2I_0 + 2I_0 \cos \phi$
$I = 2I_0 (1 + \cos \phi)$
Thus,the correct option is $D$.

(B) Let the intensities of the two waves be $I_1$ and $I_2$.
Given the ratio of intensities: $\frac{I_1}{I_2} = \frac{16}{1}$.
The ratio of maximum to minimum intensity in interference is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$.
Dividing the numerator and denominator by $\sqrt{I_2}$:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{\frac{I_1}{I_2}} + 1}{\sqrt{\frac{I_1}{I_2}} - 1} \right)^2$.
Substituting the given value $\frac{I_1}{I_2} = 16$:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{16} + 1}{\sqrt{16} - 1} \right)^2 = \left( \frac{4 + 1}{4 - 1} \right)^2 = \left( \frac{5}{3} \right)^2 = \frac{25}{9}$.

(B) Let the intensities of the two beams be $I_1$ and $I_2$. The ratio is given as $\frac{I_1}{I_2} = \beta$.
Since intensity $I \propto A^2$,we have $\frac{A_1}{A_2} = \sqrt{\beta}$.
The maximum and minimum intensities are given by $I_{\max} = (A_1 + A_2)^2$ and $I_{\min} = (A_1 - A_2)^2$.
We need to calculate the value of $X = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$.
Substituting the expressions for $I_{\max}$ and $I_{\min}$:
$X = \frac{(A_1 + A_2)^2 - (A_1 - A_2)^2}{(A_1 + A_2)^2 + (A_1 - A_2)^2} = \frac{4A_1 A_2}{2(A_1^2 + A_2^2)} = \frac{2A_1 A_2}{A_1^2 + A_2^2}$.
Dividing the numerator and denominator by $A_2^2$:
$X = \frac{2(A_1/A_2)}{(A_1/A_2)^2 + 1} = \frac{2\sqrt{\beta}}{\beta + 1}$.

(A) The intensity at any point in an interference pattern is given by $I' = I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference.
The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = (2\pi / \lambda) \times \Delta x$.
Given the path difference $\Delta x = \lambda / 6$,we calculate the phase difference:
$\phi = (2\pi / \lambda) \times (\lambda / 6) = \pi / 3$.
Now,substitute $\phi$ into the intensity formula:
$I' = I_0 \cos^2(\pi / 6)$.
Since $\cos(\pi / 6) = \sqrt{3} / 2$,we have:
$I' = I_0 (\sqrt{3} / 2)^2 = I_0 (3 / 4)$.
Therefore,the ratio $I'/I_0 = 3/4$.

(A) Let $I_{0}$ be the intensity of each individual slit. The resultant intensity $I$ at any point is given by $I = 4I_{0} \cos^{2}(\phi/2)$,where $\phi$ is the phase difference.
Phase difference $\phi = \frac{2\pi}{\lambda} \times (\text{path difference})$.
For path difference $\Delta x = \lambda$,the phase difference $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
The intensity $I = 4I_{0} \cos^{2}(2\pi/2) = 4I_{0} \cos^{2}(\pi) = 4I_{0}(1)^{2} = 4I_{0}$.
Given that this intensity is $K$,we have $4I_{0} = K$,so $I_{0} = K/4$.
Now,for path difference $\Delta x = \lambda/3$,the phase difference $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
The new intensity $I' = 4I_{0} \cos^{2}(\phi/2) = 4I_{0} \cos^{2}(\frac{2\pi/3}{2}) = 4I_{0} \cos^{2}(\pi/3)$.
Since $\cos(\pi/3) = 1/2$,we have $I' = 4I_{0} \times (1/2)^{2} = 4I_{0} \times (1/4) = I_{0}$.
Substituting $I_{0} = K/4$,we get $I' = K/4$ units.

(N/A) Constructive Interference:
When two waves superimpose at a point,constructive interference occurs if they are in phase. This happens when the path difference between the waves is an integral multiple of the wavelength $\lambda$.
Path difference $\Delta x = n\lambda$,where $n = 0, 1, 2, 3, \dots$
Since a path difference of $\lambda$ corresponds to a phase difference of $2\pi$,the condition for constructive interference in terms of phase difference $\phi$ is:
$\phi = 2n\pi$,where $n = 0, 1, 2, 3, \dots$
At these points,the resultant amplitude is maximum $(2a)$,and the intensity is $I_{max} = (a + a)^2 = 4I_0$.
Destructive Interference:
Destructive interference occurs when the two waves are out of phase by $\pi$ (or an odd multiple of $\pi$). This happens when the path difference is an odd multiple of half the wavelength $\lambda/2$.
Path difference $\Delta x = (2n + 1)\frac{\lambda}{2}$,where $n = 0, 1, 2, 3, \dots$
The condition for destructive interference in terms of phase difference $\phi$ is:
$\phi = (2n + 1)\pi$,where $n = 0, 1, 2, 3, \dots$
At these points,the resultant amplitude is minimum $(a - a = 0)$,and the intensity is $I_{min} = 0$.

(N/A) Suppose point $G$ as shown in the figure and let the phase difference between the two displacements be $\phi$ at that point.
If the displacement produced by $S_{1}$ at $G$ is $y_{1} = a \cos \omega t$,then the displacement produced by $S_{2}$ at $G$ would be $y_{2} = a \cos (\omega t + \phi)$.
The resultant displacement according to the superposition principle is:
$y = y_{1} + y_{2} = a \cos \omega t + a \cos (\omega t + \phi)$
$y = a [\cos \omega t + \cos (\omega t + \phi)]$
Using the trigonometric identity $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$:
$y = 2a \cos \left(\frac{\phi}{2}\right) \cos \left(\omega t + \frac{\phi}{2}\right)$
The amplitude of the resultant displacement is $A_{res} = 2a \cos \left(\frac{\phi}{2}\right)$.
Since intensity $I \propto A_{res}^2$,we have:
$I \propto 4a^2 \cos^2 \left(\frac{\phi}{2}\right)$
$I = 4I_{0} \cos^2 \left(\frac{\phi}{2}\right)$,where $I_{0} \propto a^2$ is the intensity of each individual source.
For constructive interference (maximum intensity),the phase difference must be $\phi = 0, \pm 2\pi, \pm 4\pi, \dots$ (i.e.,$\phi = 2n\pi$ where $n$ is an integer).
For destructive interference (minimum intensity),the phase difference must be $\phi = \pm \pi, \pm 3\pi, \pm 5\pi, \dots$ (i.e.,$\phi = (2n+1)\pi$ where $n$ is an integer).

(A) The intensity of a light wave is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference between two interfering waves.
The time-averaged value of the intensity over a complete cycle is $\langle I \rangle = \frac{1}{2} I_{max}$.
We are asked to find the phase difference $\phi$ when the intensity $I = \frac{1}{2} I_{max}$.
Substituting this into the intensity formula:
$\frac{1}{2} I_{max} = I_{max} \cos^2(\phi/2)$
$\frac{1}{2} = \cos^2(\phi/2)$
$\cos(\phi/2) = \frac{1}{\sqrt{2}}$
$\phi/2 = \pi/4$
$\phi = \pi/2$.

(N/A) The figure shows the intensity distribution for a double-slit experiment. The overall envelope represents the single-slit diffraction pattern,while the finer fringes within this envelope represent the double-slit interference pattern.
In a double-slit experiment,the resultant intensity pattern on the screen is a superposition of the single-slit diffraction pattern (from each individual slit) and the double-slit interference pattern. The broader diffraction peak contains several narrower interference fringes.
The number of interference fringes that appear within the central diffraction peak is determined by the ratio $\frac{d}{a}$,where $d$ is the distance between the two slits and $a$ is the width of each individual slit.
As the slit width $a$ becomes very small,the diffraction envelope becomes very broad and flat.
It is important to note that while interference is a necessary phenomenon for observing a diffraction pattern,the diffraction phenomenon itself is not required to observe a standard interference pattern.

(D) The situation described is depicted in the figure. According to the problem,the distance of the screen from the slits is $D$,and the distance between the slits is $d$. We are given that $D = d/2$,which implies $d = 2D$.
The path difference between the waves superposing at point $P$ is given by $\Delta x = r_2 - r_1 = S_2P - S_1P$.
From the geometry of the figure:
$S_2P = \sqrt{D^2 + (d/2 + D)^2} = \sqrt{D^2 + (D + D)^2} = \sqrt{D^2 + 4D^2} = \sqrt{5}D$.
$S_1P = \sqrt{D^2 + (d/2 - D)^2} = \sqrt{D^2 + (D - D)^2} = D$.
Therefore,the path difference is $\Delta x = \sqrt{5}D - D = D(\sqrt{5} - 1)$.
For the first order minima (destructive interference),the condition is $\Delta x = \frac{\lambda}{2}$.
Equating the two expressions:
$D(\sqrt{5} - 1) = \frac{\lambda}{2}$.
$D = \frac{\lambda}{2(\sqrt{5} - 1)}$.
Using $\sqrt{5} \approx 2.236$:
$D = \frac{\lambda}{2(2.236 - 1)} = \frac{\lambda}{2(1.236)} = \frac{\lambda}{2.472} \approx 0.404\lambda$.