In Young's double-slit experiment, white light is used. The separation between the slits is d. The screen is at a distance D (D » d) from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are

Options

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

Answer

To solve the problem regarding the missing wavelengths in Young's double-slit experiment using white light, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
In Young's double-slit experiment, we have two slits separated by a distance \( d \), and the screen is placed at a distance \( D \) from the slits, where \( D \gg d \).

2. **Path Difference Calculation**:
The path difference \( \Delta x \) at a point on the screen located at a distance \( y \) from the central maximum is given by:
\[ \Delta x = \frac{d \cdot y}{D} \]
For a point directly in front of one of the slits (let's say the upper slit), the distance \( y \) can be approximated as \( \frac{D}{2} \) (since it is halfway to the screen).

3. **Substituting for \( y \)**:
Thus, the path difference becomes:
\[ \Delta x = \frac{d \cdot \frac{D}{2}}{D} = \frac{d}{2} \]

4. **Condition for Minima**:
For destructive interference (minima) to occur, the path difference must satisfy:
\[ \Delta x = \left(m + \frac{1}{2}\right) \lambda \]
where \( m \) is an integer (0, 1, 2, ...).

5. **Setting Up the Equation**:
Setting the path difference equal to the condition for minima:
\[ \frac{d}{2} = \left(m + \frac{1}{2}\right) \lambda \]

6. **Solving for \( \lambda \)**:
Rearranging the equation gives:
\[ \lambda = \frac{d}{2\left(m + \frac{1}{2}\right)} \]
This indicates that for different integer values of \( m \), we will get different wavelengths that are missing.

7. **Finding Specific Wavelengths**:
- For \( m = 0 \):
\[ \lambda = \frac{d}{2(0 + \frac{1}{2})} = d \]
- For \( m = 1 \):
\[ \lambda = \frac{d}{2(1 + \frac{1}{2})} = \frac{d}{3} \]
- Continuing this way, we can find other wavelengths.

8. **Conclusion**:
The wavelengths that are missing exactly in front of one slit are given by:
\[ \lambda = \frac{d}{2(m + \frac{1}{2})} \]
for integer values of \( m \).