In the Young's doubel slit experiment, a poiint P on the central bright fringe is such that intensity of point P is 1/4 times the maximum intensity, distance between the slits is d and wavelength `lamda`. Then angular separation of point P is

Options

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

Explanation

To solve the problem, we need to determine the angular separation of point P in the Young's double slit experiment, given that the intensity at point P is \( \frac{1}{4} \) times the maximum intensity. We will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Intensity Formula**:
In the Young's double slit experiment, the intensity \( I \) at a point on the screen is given by:
\[ I = I_{\text{max}} \cos^2\left(\frac{\Delta \phi}{2}\right) \]
where \( \Delta \phi \) is the phase difference.

2. **Relating Phase Difference to Path Difference**:
The phase difference \( \Delta \phi \) can be expressed in terms of the path difference \( \Delta x \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
The path difference for a point at angle \( \theta \) is given by:
\[ \Delta x = d \sin \theta \]
Therefore, we can write:
\[ \Delta \phi = \frac{2\pi d \sin \theta}{\lambda} \]

3. **Substituting into the Intensity Formula**:
Substituting the expression for \( \Delta \phi \) into the intensity formula gives:
\[ I = I_{\text{max}} \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) \]

4. **Using Given Intensity Condition**:
We know that at point P, the intensity is \( \frac{1}{4} I_{\text{max}} \):
\[ \frac{1}{4} I_{\text{max}} = I_{\text{max}} \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) \]
Dividing both sides by \( I_{\text{max}} \) (assuming \( I_{\text{max}} \neq 0 \)):
\[ \frac{1}{4} = \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) \]

5. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ \frac{1}{2} = \cos\left(\frac{\pi d \sin \theta}{\lambda}\right) \]

6. **Finding the Angle**:
The cosine of \( \frac{\pi}{3} \) is \( \frac{1}{2} \), so we have:
\[ \frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{3} \]
Simplifying this gives:
\[ d \sin \theta = \frac{\lambda}{3} \]

7. **Calculating Angular Separation**:
The angular separation \( \theta \) can be found using:
\[ \sin \theta = \frac{\lambda}{3d} \]
Therefore, the angular separation is:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{3d}\right) \]

### Final Answer:
The angular separation of point P is:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{3d}\right) \]