In a Young's double slit experiment, D equals the distance of screen and d is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is same as that due to a single slit is equla to

Options

A. `(D lambda)/(d)`
B. `(D lambda)/(2d)`
C. `(D lambda)/(3d)`
D. `(2D lambda)/(d)`

Explanation

To solve the problem, we need to find the distance from the central maximum in a Young's double slit experiment where the intensity is the same as that due to a single slit.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- In a Young's double slit experiment, we have two slits separated by a distance \( d \) and a screen at a distance \( D \) from the slits. The central maximum is the point directly opposite the midpoint of the two slits.

2. **Intensity in Single Slit**:
- The intensity \( I \) at the central maximum for a single slit is given by \( I = I_0 \), where \( I_0 \) is the maximum intensity.

3. **Intensity in Double Slit**:
- The intensity at a point on the screen in a double slit experiment is given by:
\[ I = I_0 \cos^2\left(\frac{\pi d y}{\lambda D}\right) \]
- Here, \( y \) is the distance from the central maximum on the screen, \( \lambda \) is the wavelength of the light used, and \( D \) is the distance from the slits to the screen.

4. **Condition for Equal Intensity**:
- We want to find the point where the intensity in the double slit experiment equals the intensity in the single slit experiment, which is \( I_0 \). Therefore, we set:
\[ I_0 \cos^2\left(\frac{\pi d y}{\lambda D}\right) = I_0 \]
- Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \cos^2\left(\frac{\pi d y}{\lambda D}\right) = 1 \]

5. **Solving for \( y \)**:
- The equation \( \cos^2\left(\frac{\pi d y}{\lambda D}\right) = 1 \) implies:
\[ \frac{\pi d y}{\lambda D} = n\pi \quad (n \text{ is an integer}) \]
- For the nearest point, we take \( n = 0 \):
\[ \frac{\pi d y}{\lambda D} = 0 \implies y = 0 \]
- The next nearest point occurs when \( n = 1 \):
\[ \frac{\pi d y}{\lambda D} = \pi \implies y = \frac{\lambda D}{d} \]

6. **Final Result**:
- Therefore, the distance of the nearest point to the central maximum where the intensity is the same as that due to a single slit is:
\[ y = \frac{\lambda D}{d} \]