In Young's double slit experiment intensity at a point is `((1)/(4))` of the maximum intensity. Angular position of this point is

Options

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

Explanation

To solve the problem of finding the angular position θ in Young's double slit experiment where the intensity at a point is \( \frac{1}{4} \) of the maximum intensity, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that in Young's double slit experiment, the intensity \( I \) at a point on the screen can be expressed in terms of the maximum intensity \( I_{max} \) and the phase difference \( \Delta \phi \) between the two waves arriving at that point.

2. **Intensity Relation**:
The intensity at any point can be given by:
\[ I = I_{max} \cos^2\left(\frac{\Delta \phi}{2}\right) \]
According to the problem, the intensity at the point is \( \frac{1}{4} I_{max} \):
\[ \frac{1}{4} I_{max} = I_{max} \cos^2\left(\frac{\Delta \phi}{2}\right) \]

3. **Cancelling \( I_{max} \)**:
We can cancel \( I_{max} \) from both sides (assuming \( I_{max} \neq 0 \)):
\[ \frac{1}{4} = \cos^2\left(\frac{\Delta \phi}{2}\right) \]

4. **Taking Square Root**:
Taking the square root of both sides gives:
\[ \cos\left(\frac{\Delta \phi}{2}\right) = \frac{1}{2} \]

5. **Finding Phase Difference**:
The cosine value of \( \frac{1}{2} \) corresponds to angles of \( \frac{\pi}{3} \) or \( \frac{5\pi}{3} \). Therefore, we can write:
\[ \frac{\Delta \phi}{2} = \frac{\pi}{3} \quad \text{or} \quad \frac{\Delta \phi}{2} = \frac{5\pi}{3} \]
Thus,
\[ \Delta \phi = \frac{2\pi}{3} \quad \text{or} \quad \Delta \phi = \frac{10\pi}{3} \]

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

7. **Setting the Equations Equal**:
Setting the two expressions for \( \Delta \phi \) equal gives:
\[ \frac{2\pi d \sin \theta}{\lambda} = \frac{2\pi}{3} \]
Simplifying this, we find:
\[ d \sin \theta = \frac{\lambda}{3} \]

8. **Finding \( \sin \theta \)**:
Rearranging gives:
\[ \sin \theta = \frac{\lambda}{3d} \]

9. **Finding \( \theta \)**:
Finally, taking the inverse sine gives:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{3d}\right) \]

### Final Answer:
Thus, the angular position \( \theta \) is:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{3d}\right) \]