In young's double slit experiment, distance between the slit `S_1 and S_2` is d and the distance between slit and screen is D . Then longest wavelength that will be missing on the screen in front of `S_1` is

Options

A. `d^2/D`
B. `d^2/(2D)`
C. `D/d^2`
D. None of these

Answer

To solve the problem of finding the longest wavelength that will be missing on the screen in front of \( S_1 \) in Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two slits \( S_1 \) and \( S_2 \) separated by a distance \( d \).
- The distance from the slits to the screen is \( D \).
- We need to find the longest wavelength that will be missing (i.e., where destructive interference occurs) at a point directly in front of \( S_1 \).

2. **Path Difference for Destructive Interference**:
- For destructive interference, the path difference \( \Delta \) must satisfy the condition:
\[ \Delta = (2n + 1) \frac{\lambda}{2} \]
- Here, \( n \) is an integer (0, 1, 2, ...).

3. **Calculate the Path Difference**:
- The path difference \( \Delta \) between the light coming from \( S_1 \) and \( S_2 \) to a point \( P \) on the screen directly in front of \( S_1 \) can be expressed as:
\[ \Delta = S_1P - S_2P \]
- Using geometry, we can find that:
\[ S_1P = D \quad \text{and} \quad S_2P = \sqrt{D^2 + d^2} \]
- Therefore, the path difference becomes:
\[ \Delta = D - \sqrt{D^2 + d^2} \]

4. **Approximate the Path Difference**:
- For small angles (which is often the case in double slit experiments), we can use the approximation:
\[ \Delta \approx \frac{d^2}{2D} \]
- This approximation comes from the binomial expansion of \( \sqrt{D^2 + d^2} \).

5. **Set the Path Difference Equal to the Condition for Destructive Interference**:
- We set the path difference equal to the condition for the first minimum (using \( n = 0 \)):
\[ \frac{d^2}{2D} = \frac{\lambda}{2} \]

6. **Solve for the Wavelength**:
- Rearranging gives:
\[ \lambda = \frac{d^2}{D} \]

7. **Conclusion**:
- The longest wavelength that will be missing on the screen in front of \( S_1 \) is:
\[ \lambda = \frac{d^2}{D} \]

### Final Answer:
\[ \text{Longest wavelength missing on the screen in front of } S_1 = \frac{d^2}{D} \]