White light is used to illuminate the two slits in Young's experiment. The separation between the slits is d and the screen is at a distance D(»d) from the slits. At a point directly in front of one of the slits, certain wavelength is missing. The missing wavelength

Options

A. `lambda=d^(2) // D`
B. `lambda=d^(2) // 5D`
C. `lambda=d^(2) // 3D`
D. All of these

Answer

To find the missing wavelength in Young's double-slit experiment when white light is used, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- In Young's experiment, two slits are separated by a distance \(d\), and a screen is placed at a distance \(D\) from the slits. The distance \(D\) is much larger than \(d\).

2. **Identifying the Point of Interest**:
- We are interested in a point directly in front of one of the slits. At this point, we need to find the condition for destructive interference (a dark fringe).

3. **Calculating Path Difference**:
- The path difference (\(\Delta x\)) between the light waves coming from the two slits at this point is given by:
\[ \Delta x = \frac{d \cdot y}{D} \]
- Here, \(y\) is the vertical distance from the midpoint between the slits to the point on the screen. Since we are directly in front of one of the slits, \(y\) is effectively \(0\), but we need to consider the distance from the midpoint to the screen, which is \(D/2\).

4. **Substituting Values**:
- Therefore, substituting \(y = D/2\) into the path difference formula gives:
\[ \Delta x = \frac{d \cdot (D/2)}{D} = \frac{d}{2} \]

5. **Condition for Destructive Interference**:
- For a dark fringe (minimum), the path difference must satisfy the condition:
\[ \Delta x = \left(m + \frac{1}{2}\right) \lambda \]
- Here, \(m\) is an integer (0, 1, 2, ...), and \(\lambda\) is the wavelength of light.

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

7. **Solving for Wavelength**:
- Rearranging this gives:
\[ \lambda = \frac{d}{2\left(m + \frac{1}{2}\right)} \]
- For \(m = 0\):
\[ \lambda = \frac{d}{2 \cdot \frac{1}{2}} = \frac{d}{1} = d \]
- For \(m = 1\):
\[ \lambda = \frac{d}{2 \cdot \frac{3}{2}} = \frac{d}{3} \]
- For \(m = 2\):
\[ \lambda = \frac{d}{2 \cdot \frac{5}{2}} = \frac{d}{5} \]

8. **Conclusion**:
- The missing wavelengths are \(d\), \(\frac{d}{3}\), and \(\frac{d}{5}\). Thus, the answer is that the missing wavelength can be any of these values.