In Young's double-slit experiment, the intensity at a point P on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is `lambda` and d is the distance between the slits, the angular separation between point P and the center of the screen is

Options

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

Answer

To solve the problem, we need to find the angular separation between point P and the center of the screen in Young's double-slit experiment, given that the intensity at point P is half the maximum intensity.

### Step-by-Step Solution:

1. **Understanding the Intensity Relation**:
In Young's double-slit experiment, the intensity \( I \) at any point on the screen can be expressed as:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_0 \) is the maximum intensity and \( \phi \) is the phase difference.

2. **Setting Up the Equation**:
Given that the intensity at point P is half the maximum intensity:
\[ I = \frac{I_0}{2} \]
Substituting this into the intensity equation gives:
\[ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right) \]

3. **Simplifying the Equation**:
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} = \cos^2\left(\frac{\phi}{2}\right) \]
Taking the square root of both sides:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]

4. **Finding the Phase Difference**:
The angle whose cosine is \( \frac{1}{\sqrt{2}} \) is:
\[ \frac{\phi}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \phi = \frac{\pi}{2} \]

5. **Relating Phase Difference to Path Difference**:
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Setting \( \phi = \frac{\pi}{2} \):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]
Solving for the path difference \( \Delta x \):
\[ \Delta x = \frac{\lambda}{4} \]

6. **Finding the Angular Separation**:
The path difference \( \Delta x \) can also be expressed in terms of the angle \( \theta \) and the distance \( d \) between the slits:
\[ \Delta x = d \sin \theta \]
Setting this equal to \( \frac{\lambda}{4} \):
\[ d \sin \theta = \frac{\lambda}{4} \]
Solving for \( \sin \theta \):
\[ \sin \theta = \frac{\lambda}{4d} \]

7. **Final Expression for Angular Separation**:
To find \( \theta \):
\[ \theta = \sin^{-1}\left(\frac{\lambda}{4d}\right) \]

### Final Answer:
The angular separation between point P and the center of the screen is:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{4d}\right) \]