In YDSE, find the missing wavelength at `y = d`, where symbols have their usual meaning `(take Dgt gt d)`.

Options

A. `(d^(2))/(D)`
B. `(2d^(2))/(7 D)`
C. `(3d^(2))/(D)`
D. `(d^(2))/(3D)`

Explanation

To solve the problem of finding the missing wavelength at \( y = d \) in Young's Double Slit Experiment (YDSE), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( d \) be the distance between the two slits.
- Let \( D \) be the distance from the slits to the screen.
- We are given that \( D \gg d \).

2. **Path Difference Calculation**:
- The path difference \( \Delta x \) between the two waves arriving at a point on the screen can be expressed as:
\[ \Delta x = d \sin \theta \]
- For small angles (which is valid since \( D \gg d \)), we can approximate \( \sin \theta \) using the small angle approximation:
\[ \sin \theta \approx \frac{y}{D} \]
- Therefore, the path difference becomes:
\[ \Delta x = d \cdot \frac{y}{D} \]

3. **Substituting \( y = d \)**:
- We substitute \( y = d \) into the equation for path difference:
\[ \Delta x = d \cdot \frac{d}{D} = \frac{d^2}{D} \]

4. **Condition for Destructive Interference**:
- The condition for destructive interference is given by:
\[ \Delta x = \frac{(2n + 1)\lambda}{2} \]
where \( n \) is an integer (0, 1, 2, ...). For the first missing wavelength, we can take \( n = 0 \), which gives:
\[ \Delta x = \frac{\lambda}{2} \]

5. **Equating Path Difference to Wavelength**:
- Setting the two expressions for path difference equal gives:
\[ \frac{d^2}{D} = \frac{\lambda}{2} \]

6. **Solving for Wavelength**:
- Rearranging the equation to solve for \( \lambda \):
\[ \lambda = \frac{2d^2}{D} \]

7. **Considering Higher Orders of Destructive Interference**:
- For higher orders of destructive interference, we can express the wavelength as:
\[ \lambda = \frac{2d^2}{(2n + 1)D} \]
- This gives us the wavelengths for \( n = 0, 1, 2, \ldots \) as:
- For \( n = 0 \): \( \lambda_1 = \frac{2d^2}{D} \)
- For \( n = 1 \): \( \lambda_2 = \frac{2d^2}{3D} \)
- For \( n = 2 \): \( \lambda_3 = \frac{2d^2}{5D} \)

### Final Answer:
The missing wavelengths at \( y = d \) are:
- \( \lambda = \frac{2d^2}{D} \)
- \( \lambda = \frac{2d^2}{3D} \)
- \( \lambda = \frac{2d^2}{5D} \)