In Young's double - slit experiment intensity at a point is `(3//4)^("th")` of the maximum intensity. The possible angular position of this point is

Options

A. `sin^(-1)((lambda)/(3d))`
B. `sin^(-1)((lambda)/(2a))`
C. `sin^(-1)((lambda)/(6d))`
D. `sin^(-1)((lambda)/(4d))`

Answer

To solve the problem step by step, we will analyze the situation in Young's double-slit experiment where the intensity at a point is given as \( \frac{3}{4} \) of the maximum intensity. We need to find the possible angular position of this point.

### Step 1: Understand the relationship between intensity and phase difference
In Young's double-slit experiment, the intensity \( I \) at a point on the screen can be expressed in terms of the maximum intensity \( I_{\text{max}} \) and the phase difference \( \Delta \Phi \) as follows:
\[ I = I_{\text{max}} \cos^2\left(\frac{\Delta \Phi}{2}\right) \]

### Step 2: Set up the equation for the given intensity
We are given that the intensity \( I \) is \( \frac{3}{4} \) of the maximum intensity \( I_{\text{max}} \):
\[ I = \frac{3}{4} I_{\text{max}} \]
Substituting this into the intensity equation gives:
\[ \frac{3}{4} I_{\text{max}} = I_{\text{max}} \cos^2\left(\frac{\Delta \Phi}{2}\right) \]

### Step 3: Simplify the equation
Dividing both sides by \( I_{\text{max}} \) (assuming \( I_{\text{max}} \neq 0 \)):
\[ \frac{3}{4} = \cos^2\left(\frac{\Delta \Phi}{2}\right) \]

### Step 4: Solve for the phase difference
Taking the square root of both sides:
\[ \cos\left(\frac{\Delta \Phi}{2}\right) = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \]
This gives us two possible angles:
\[ \frac{\Delta \Phi}{2} = \frac{\pi}{6} + n\pi \quad \text{or} \quad \frac{\Delta \Phi}{2} = \frac{5\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]
Thus, the phase difference \( \Delta \Phi \) can be:
\[ \Delta \Phi = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \Delta \Phi = \frac{5\pi}{3} + 2n\pi \]

### Step 5: Relate phase difference to path difference
The phase difference \( \Delta \Phi \) is also related to the path difference \( \Delta x \) by:
\[ \Delta \Phi = \frac{2\pi}{\lambda} \Delta x \]
The path difference \( \Delta x \) can be expressed as:
\[ \Delta x = d \sin \theta \]
where \( d \) is the distance between the slits and \( \theta \) is the angle from the central maximum.

### Step 6: Substitute and solve for \( \theta \)
Substituting for \( \Delta x \):
\[ \Delta \Phi = \frac{2\pi}{\lambda} (d \sin \theta) \]
Setting this equal to the two possible values of \( \Delta \Phi \):
1. For \( \Delta \Phi = \frac{\pi}{3} \):
\[ \frac{\pi}{3} = \frac{2\pi}{\lambda} (d \sin \theta) \implies \sin \theta = \frac{\lambda}{6d} \]
2. For \( \Delta \Phi = \frac{5\pi}{3} \):
\[ \frac{5\pi}{3} = \frac{2\pi}{\lambda} (d \sin \theta) \implies \sin \theta = \frac{5\lambda}{6d} \]

### Step 7: Conclusion
The possible angular positions \( \theta \) can be found using:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{6d}\right) \quad \text{or} \quad \theta = \sin^{-1}\left(\frac{5\lambda}{6d}\right) \]