In Young's double - slit experiment intensity at a point is `(1//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, we will follow these steps:

### Step 1: Understand the relationship between intensity and phase difference
In Young's double-slit experiment, the intensity \( I \) at a point is given by the formula:
\[ I = I_{\text{max}} \cos^2\left(\frac{\delta}{2}\right) \]
where \( I_{\text{max}} \) is the maximum intensity and \( \delta \) is the phase difference.

### Step 2: Set up the equation for the given intensity
We are given that the intensity at a point is \( \frac{1}{4} \) of the maximum intensity:
\[ I = \frac{I_{\text{max}}}{4} \]
Substituting this into the intensity formula gives:
\[ \frac{I_{\text{max}}}{4} = I_{\text{max}} \cos^2\left(\frac{\delta}{2}\right) \]

### Step 3: Simplify the equation
Dividing both sides by \( I_{\text{max}} \) (assuming \( I_{\text{max}} \neq 0 \)):
\[ \frac{1}{4} = \cos^2\left(\frac{\delta}{2}\right) \]

### Step 4: Solve for the cosine
Taking the square root of both sides, we find:
\[ \cos\left(\frac{\delta}{2}\right) = \pm \frac{1}{2} \]

### Step 5: Determine the phase difference \( \delta \)
The values of \( \frac{\delta}{2} \) that satisfy this equation are:
\[ \frac{\delta}{2} = \frac{\pi}{3} \quad \text{or} \quad \frac{\delta}{2} = \frac{5\pi}{3} \]
Thus, multiplying by 2 gives:
\[ \delta = \frac{2\pi}{3} \quad \text{or} \quad \delta = \frac{10\pi}{3} \]

### Step 6: Find the path difference
The path difference \( \Delta x \) is related to the phase difference by:
\[ \Delta x = \frac{\lambda}{2\pi} \delta \]
Substituting \( \delta = \frac{2\pi}{3} \):
\[ \Delta x = \frac{\lambda}{2\pi} \cdot \frac{2\pi}{3} = \frac{\lambda}{3} \]

### Step 7: Relate path difference to angular position
The path difference is also given by:
\[ \Delta x = d \sin \theta \]
where \( d \) is the distance between the slits and \( \theta \) is the angular position. Setting these equal gives:
\[ d \sin \theta = \frac{\lambda}{3} \]

### Step 8: Solve for \( \theta \)
Rearranging gives:
\[ \sin \theta = \frac{\lambda}{3d} \]
Thus, the angular position \( \theta \) is:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{3d}\right) \]

### Conclusion
The possible angular position of the point where the intensity is \( \frac{1}{4} \) of the maximum intensity is:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{3d}\right) \]