Intensity in Young's Double Slit Experiment Questions in English

(N/A) In the absence of a polariser, the intensity of the principal maxima is $I_0 = 4I$, where $I$ is the intensity of light from each slit.
When a polariser is placed in one of the paths (say path $2$), the light passing through it becomes linearly polarised. Let the amplitude of the light from each slit be $a$. The intensity $I = a^2$.
For the slit without the polariser, the light remains unpolarised. For the slit with the polariser, the light becomes polarised.
When these two beams superpose, the unpolarised light can be considered as two incoherent components of intensity $I/2$ each, vibrating in mutually perpendicular directions.
The polarised light from the second slit has intensity $I' = I/2$ (after passing through the polariser).
At the principal maxima, the two beams are in phase. The intensity of the unpolarised beam is $I$ and the polarised beam is $I/2$. The resultant intensity is $I_{max} = I + I/2 + 2\sqrt{I \cdot I/2} \cdot \cos(0) = I + I/2 + \sqrt{2}I = I(1.5 + 1.414) \approx 2.914I$. Since $I_0 = 4I$, $I = I_0/4$. Thus, $I_{max} = (2.914/4)I_0 = 0.7285I_0$.
At the first minima, the phase difference is $\pi$. The intensity is $I_{min} = I + I/2 - 2\sqrt{I \cdot I/2} = I(1.5 - 1.414) = 0.086I = 0.086(I_0/4) = 0.0215I_0$.

(B) $(i)$ At $R_1$,the path difference between waves from $A$ and $B$ is $\lambda/2$,leading to destructive interference $(y_A + y_B = 0)$. Similarly,waves from $C$ and $D$ interfere destructively at $R_1$. Thus,the net signal at $R_1$ is zero. At $R_2$,the path differences are different,leading to constructive interference. Therefore,$R_2$ picks up the larger signal.
$(ii)$ If $B$ is turned off,the destructive interference at $R_1$ is removed,resulting in a non-zero signal. $R_2$ also experiences a change,but $R_1$ now receives a stronger resultant signal compared to its previous zero state. Thus,$R_1$ picks up the larger signal.
$(iii)$ If $D$ is turned off,the destructive interference at $R_1$ is removed,resulting in a non-zero signal. $R_1$ picks up the larger signal.
$(iv)$ Since the path differences for $B$ and $D$ are different relative to $R_1$ and $R_2$,the change in signal intensity at the receivers will be unique for each source being turned off. Thus,both receivers can distinguish which source is turned off.

(D) Given the ratio of intensities $I_1 : I_2 = 9 : 4$.
Let $I_1 = 9k$ and $I_2 = 4k$.
The amplitudes are proportional to the square root of intensities,so $a_1 = \sqrt{I_1} = 3\sqrt{k}$ and $a_2 = \sqrt{I_2} = 2\sqrt{k}$.
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:
$\frac{I_{max}}{I_{min}} = \left( \frac{3\sqrt{k} + 2\sqrt{k}}{3\sqrt{k} - 2\sqrt{k}} \right)^2 = \left( \frac{5\sqrt{k}}{1\sqrt{k}} \right)^2 = 5^2 = 25$.
Thus,the ratio is $25 : 1$.

(D) Let the intensity of light on the screen due to each slit be $I_0$.
The maximum intensity at the centre of the screen is $I_{max} = 4I_0$.
The intensity $I$ at a point where the phase difference is $\phi$ is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 4I_0 \cos^2(\phi/2)$.
We are given that the intensity becomes half of the maximum intensity,so $I = \frac{I_{max}}{2} = 2I_0$.
Thus,$2I_0 = 4I_0 \cos^2(\phi/2) \implies \cos^2(\phi/2) = 1/2 \implies \cos(\phi/2) = 1/\sqrt{2}$.
This gives $\phi/2 = \pi/4$,so the phase difference $\phi = \pi/2$.
The phase difference is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
So,$\frac{2\pi}{\lambda} \Delta x = \frac{\pi}{2} \implies \Delta x = \frac{\lambda}{4}$.
For small angles,the path difference $\Delta x = d \sin \theta \approx d \tan \theta = d(y/D)$.
Equating the two,we get $d(y/D) = \lambda/4 \implies y = \frac{\lambda D}{4d}$.
Given $\lambda = 5000 \ \mathring{A} = 5 \times 10^{-7} \ m$,$d = 1.0 \ mm = 10^{-3} \ m$,and $D = 1.0 \ m$.
$y = \frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}} = 1.25 \times 10^{-4} \ m = 125 \times 10^{-6} \ m$.
Therefore,the distance is $125$.

(D) The intensity at any point in a Young's double slit experiment is given by $I = I_{max} \cos^2\left(\frac{\Delta \phi}{2}\right)$,where $\Delta \phi$ is the phase difference.
Given that $I = \frac{I_{max}}{4}$,we have $\frac{I_{max}}{4} = I_{max} \cos^2\left(\frac{\Delta \phi}{2}\right)$.
This simplifies to $\cos^2\left(\frac{\Delta \phi}{2}\right) = \frac{1}{4}$,which means $\cos\left(\frac{\Delta \phi}{2}\right) = \frac{1}{2}$.
Thus,$\frac{\Delta \phi}{2} = \frac{\pi}{3}$,which gives the phase difference $\Delta \phi = \frac{2\pi}{3}$.
The phase difference is related to the path difference $\Delta x$ by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$. Since $\Delta x = \frac{yd}{D}$,we have $\Delta \phi = \frac{2\pi}{\lambda} \left(\frac{yd}{D}\right)$.
Equating the two expressions for $\Delta \phi$: $\frac{2\pi}{\lambda} \left(\frac{yd}{D}\right) = \frac{2\pi}{3}$.
Solving for $y$: $y = \frac{\lambda D}{3d}$.
Substituting the given values: $y = \frac{600 \times 10^{-9} \ m \times 1.0 \ m}{3 \times 1.0 \times 10^{-3} \ m} = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} \ m = 200 \times 10^{-6} \ m$.
Since $1 \ \mu m = 10^{-6} \ m$,the distance $y = 200 \ \mu m$.

(B) The intensity at any point in the interference pattern is given by $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given that the intensity is half of the peak intensity,$I = \frac{I_{max}}{2}$.
Substituting this into the equation: $\frac{I_{max}}{2} = I_{max} \cos^2(\frac{\phi}{2})$.
This simplifies to $\cos^2(\frac{\phi}{2}) = \frac{1}{2}$,which means $\cos(\frac{\phi}{2}) = \pm \frac{1}{\sqrt{2}}$.
Thus,$\frac{\phi}{2} = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \dots$,which implies $\phi = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$.
In general,$\phi = (2n+1) \frac{\pi}{2}$.
Since the path difference $\Delta x = \frac{\lambda}{2\pi} \phi$,we substitute $\phi$:
$\Delta x = \frac{\lambda}{2\pi} \times (2n+1) \frac{\pi}{2} = (2n+1) \frac{\lambda}{4}$.

(B) Let the intensities of the two coherent sources be $I_1$ and $I_2$. The ratio of intensities is given as $b = \frac{I_1}{I_2}$.
We know that $I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$.
The required ratio is $R = \frac{I_{\text{max}} + I_{\text{min}}}{I_{\text{max}} - I_{\text{min}}}$.
Substituting the expressions for $I_{\text{max}}$ and $I_{\text{min}}$:
$R = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}$.
Expanding the terms:
$R = \frac{(I_1 + I_2 + 2\sqrt{I_1I_2}) + (I_1 + I_2 - 2\sqrt{I_1I_2})}{(I_1 + I_2 + 2\sqrt{I_1I_2}) - (I_1 + I_2 - 2\sqrt{I_1I_2})} = \frac{2(I_1 + I_2)}{4\sqrt{I_1I_2}} = \frac{I_1 + I_2}{2\sqrt{I_1I_2}}$.
Dividing numerator and denominator by $I_2$:
$R = \frac{\frac{I_1}{I_2} + 1}{2\sqrt{\frac{I_1}{I_2}}} = \frac{b + 1}{2\sqrt{b}}$.

(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 $4: 1$.

(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 $\frac{x}{2}$.

(B) The intensity $I$ at any point on the screen 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 $\phi = \frac{2\pi}{\lambda} \times (\text{path difference} \Delta x)$.
For point $P$,path difference $\Delta x_P = 0$,so $\phi_P = 0$. Intensity $I_P = I_{max} \cos^2(0) = I_{max}$.
For point $Q$,path difference $\Delta x_Q = \frac{\lambda}{4}$,so $\phi_Q = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} = 90^{\circ}$.
Intensity $I_Q = I_{max} \cos^2(\frac{90^{\circ}}{2}) = I_{max} \cos^2(45^{\circ}) = I_{max} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_{max}}{2}$.
The ratio of intensities is $\frac{I_P}{I_Q} = \frac{I_{max}}{I_{max}/2} = 2: 1$.

(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 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 $I = \beta (1/2)^2 = \beta/4$.

(B) The intensity in Young's double slit experiment is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,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 $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
Given intensity $K = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
Now,for 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 $K'$ is $K' = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $K' = 4I_0 \times (\frac{1}{\sqrt{2}})^2 = 4I_0 \times \frac{1}{2} = 2I_0$.
Since $K = 4I_0$,then $2I_0 = \frac{K}{2}$.
Therefore,the intensity is $\frac{K}{2}$.

(D) 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 wave and $\phi$ 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 $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
Given intensity $K = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
For path difference $\Delta x = \frac{\lambda}{4}$,the phase difference is $\phi' = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I'$ is $I' = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $I' = 4I_0 \cdot (\frac{1}{\sqrt{2}})^2 = 4I_0 \cdot \frac{1}{2} = 2I_0$.
Since $K = 4I_0$,then $2I_0 = \frac{K}{2}$.
Therefore,the intensity is $\frac{K}{2}$.

(C) The resultant intensity $I_R$ in Young's double slit experiment is given by $I_R = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual slit and $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Case $1$: When $\Delta x = \lambda$,$\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$. The intensity $I = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(-1)^2 = 4I_0$.
Case $2$: When $\Delta x = \frac{\lambda}{4}$,$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I' = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4}) = 4I_0 (\frac{1}{\sqrt{2}})^2 = 4I_0 \times \frac{1}{2} = 2I_0$.
Since $I = 4I_0$,we have $I_0 = \frac{I}{4}$.
Substituting $I_0$ into the expression for $I'$,we get $I' = 2(\frac{I}{4}) = \frac{I}{2}$.

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

(D) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Given the path difference $\Delta x = \frac{\lambda}{6}$,we calculate the phase difference: $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} = 60^{\circ}$.
The intensity $I$ at any point on the screen is given by $I = I_{0} \cos^{2}\left(\frac{\phi}{2}\right)$,where $I_{0}$ is the maximum intensity.
Substituting the value of $\phi$: $\frac{I}{I_{0}} = \cos^{2}\left(\frac{60^{\circ}}{2}\right) = \cos^{2}(30^{\circ})$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $\frac{I}{I_{0}} = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4} = 0.75$.

(C) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = 2x$.
We know that $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
The expression to evaluate is $V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$.
Substituting the expressions for $I_{\max}$ and $I_{\min}$:
$V = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2}$.
Using the algebraic identities $(a+b)^2 - (a-b)^2 = 4ab$ and $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$:
$V = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}$.
Dividing the numerator and denominator by $I_2$:
$V = \frac{2\sqrt{I_1/I_2}}{I_1/I_2 + 1}$.
Substituting $\frac{I_1}{I_2} = 2x$:
$V = \frac{2\sqrt{2x}}{2x + 1}$.

(A) In Young's double slit experiment, the intensity at any point on the screen is given by $I = I_{max} \cos^2 \left( \frac{\phi}{2} \right)$, where $\phi$ is the phase difference.
Given that the intensity of the central fringe is $I_0$, we have $I_{max} = I_0$.
The phase difference $\phi$ is related to the path difference $\Delta p$ as $\phi = \frac{2\pi}{\lambda} \Delta p$.
For a point at distance $x$ from the central fringe, the path difference is $\Delta p = d \sin \theta \approx d \tan \theta = d \left( \frac{x}{D} \right)$.
Thus, $\phi = \frac{2\pi}{\lambda} \left( \frac{dx}{D} \right)$.
We know that the fringe width is $\beta = \frac{\lambda D}{d}$, which implies $\frac{d}{\lambda D} = \frac{1}{\beta}$.
Substituting this into the phase difference expression, we get $\phi = 2\pi \left( \frac{x}{\beta} \right)$.
Now, substituting $\phi$ into the intensity formula: $I = I_0 \cos^2 \left( \frac{2\pi x / \beta}{2} \right) = I_0 \cos^2 \left( \frac{\pi x}{\beta} \right)$.

(D) The intensity in a double slit experiment is given by $I = I_{max} \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference.
We are given that $I = 0.5 I_{max}$, so $0.5 I_{max} = I_{max} \cos^2(\frac{\phi}{2})$.
This implies $\cos^2(\frac{\phi}{2}) = 0.5$, or $\cos(\frac{\phi}{2}) = \frac{1}{\sqrt{2}}$.
Thus, $\frac{\phi}{2} = \frac{\pi}{4}$, which gives $\phi = \frac{\pi}{2}$.
The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting $\phi = \frac{\pi}{2}$, we get $\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x$, which means $\Delta x = \frac{\lambda}{4}$.
The path difference for a point at distance $y$ from the central maximum is $\Delta x = \frac{yd}{D}$.
Equating the two, $\frac{yd}{D} = \frac{\lambda}{4}$, so $y = \frac{\lambda D}{4d}$.
Given $\lambda = 6600 \, Å = 6.6 \times 10^{-7} \, m$, $D = 4 \, m$, and $d = 0.6 \, mm = 6 \times 10^{-4} \, m$.
$y = \frac{(6.6 \times 10^{-7} \, m) \times (4 \, m)}{4 \times (6 \times 10^{-4} \, m)} = \frac{6.6 \times 10^{-7}}{6 \times 10^{-4}} \, m = 1.1 \times 10^{-3} \, m = 1.1 \, mm$.

(B) Given: Intensities of two beams of monochromatic light are $I_1 = 64 \ mW$ and $I_2 = 4 \ mW$.
The formula for the resultant intensity of two interfering waves is given by $I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi$.
When the intensity is reduced to $84 \ mW$,we substitute the known values into the equation:
$84 = 64 + 4 + 2 \sqrt{64 \times 4} \cos \phi$
$84 = 68 + 2 \times 8 \times 2 \cos \phi$
$84 - 68 = 32 \cos \phi$
$16 = 32 \cos \phi$
$\cos \phi = \frac{16}{32} = \frac{1}{2}$
Therefore,$\phi = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ$.

(D) The intensity at any point in an interference pattern is given by $I_p = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given that $I_{max} = I$ (intensity of the central bright fringe).
The relationship between phase difference $\phi$ and path difference $\Delta x$ is $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given $\Delta x = \frac{\lambda}{3}$,we have $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$.
Substituting this into the intensity formula:
$I_p = I \cos^2(\frac{2\pi/3}{2}) = I \cos^2(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $I_p = I (\frac{1}{2})^2 = \frac{I}{4}$.

(A) The intensity of light in a double slit experiment is given by $I = I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference.
Alternatively,using the formula $I = I_{max} \cos^2(\frac{\pi \Delta x}{\lambda})$,where $\Delta x$ is the path difference.
For the first point,$\Delta x_1 = 2000 \ Å$ and $\lambda = 6000 \ Å$.
Phase difference $\phi_1 = \frac{2\pi}{\lambda} \Delta x_1 = \frac{2\pi}{6000} \times 2000 = \frac{2\pi}{3} = 120^{\circ}$.
Intensity $I_1 = I_{max} \cos^2(\frac{120^{\circ}}{2}) = I_{max} \cos^2(60^{\circ}) = I_{max} (1/2)^2 = I_{max}/4$.
For the second point,$\Delta x_2 = 1000 \ Å$.
Phase difference $\phi_2 = \frac{2\pi}{\lambda} \Delta x_2 = \frac{2\pi}{6000} \times 1000 = \frac{\pi}{3} = 60^{\circ}$.
Intensity $I_2 = I_{max} \cos^2(\frac{60^{\circ}}{2}) = I_{max} \cos^2(30^{\circ}) = I_{max} (\sqrt{3}/2)^2 = 3I_{max}/4$.
Therefore,the ratio $\frac{I_1}{I_2} = \frac{I_{max}/4}{3I_{max}/4} = \frac{1}{3}$.
Thus,$I_1: I_2 = 1: 3$.

(C) The resultant intensity $I_R$ at any point is given by $I_R = I_{\max} \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase difference.
Given that $I_R = 75 \% \text{ of } I_{\max} = 0.75 I_{\max} = \frac{3}{4} I_{\max}$.
Substituting this into the formula:
$\frac{3}{4} I_{\max} = I_{\max} \cos^2 \left( \frac{\phi}{2} \right)$
$\cos^2 \left( \frac{\phi}{2} \right) = \frac{3}{4}$
Taking the square root on both sides:
$\cos \left( \frac{\phi}{2} \right) = \frac{\sqrt{3}}{2}$
Since $\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$,we have:
$\frac{\phi}{2} = \frac{\pi}{6}$
$\phi = \frac{\pi}{3}$

(C) The intensity of light $I$ is directly proportional to the width $w$ of the slit,i.e.,$I \propto w$.
Given that the width of the first slit $w_1 = 4w_2$,the intensities are related as $I_1 = 4I_2$.
Let $I_2 = I$,then $I_1 = 4I$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{4I} + \sqrt{I})^2}{(\sqrt{4I} - \sqrt{I})^2} = \frac{(2\sqrt{I} + \sqrt{I})^2}{(2\sqrt{I} - \sqrt{I})^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(3\sqrt{I})^2}{(\sqrt{I})^2} = \frac{9I}{I} = \frac{9}{1}$
Thus,the ratio is $9: 1$.

(B) Let the intensities of light from the two slits be $I_1$ and $I_2$. Given that $I_1 = 1.5 I_2$,we have the ratio $\frac{I_1}{I_2} = 1.5 = \frac{3}{2}$.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$.
Dividing the numerator and denominator by $\sqrt{I_2}$,we get:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1/I_2} + 1}{\sqrt{I_1/I_2} - 1} \right)^2$.
Substituting $\frac{I_1}{I_2} = \frac{3}{2} = 1.5$:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{1.5} + 1}{\sqrt{1.5} - 1} \right)^2 \approx \left( \frac{1.225 + 1}{1.225 - 1} \right)^2 = \left( \frac{2.225}{0.225} \right)^2 \approx (9.88)^2 \approx 97.7 \approx 98$.

(B) The path difference $\Delta x$ for a point on the screen at a vertical distance $y$ from the center is given by $\Delta x = d \sin \theta \approx dy/D$.
For a point exactly opposite to one of the slits,the vertical distance $y = d/2$.
Substituting this into the path difference formula,we get $\Delta x = d(d/2) / D = d^2 / (2D)$.
Given $D = 10d$,the path difference becomes $\Delta x = d^2 / (20d) = d/20$.
Given $d = 5\lambda$,we substitute to find $\Delta x = 5\lambda / 20 = \lambda/4$.
The phase difference $\phi$ is calculated as $\phi = (2\pi/\lambda) \Delta x = (2\pi/\lambda) \times (\lambda/4) = \pi/2$.
The intensity $I$ at this point is given by $I = I_0 \cos^2(\phi/2)$.
Substituting $\phi = \pi/2$,we get $I = I_0 \cos^2(\pi/4) = I_0 (1/\sqrt{2})^2 = I_0/2$.