A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the secondary maximum in the pattern for red light of wavelength 6500 Å ?

Options

A. 4400 Å
B. 4100 Å
C. 4642.8 Å
D. 9100 Å

Explanation

To solve the problem of finding the wavelength of light for which the third secondary maximum in the diffraction pattern coincides with the secondary maximum in the pattern for red light (wavelength = 6500 Å), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Condition for Secondary Maximum**:
The condition for the position of the secondary maximum in a single slit diffraction pattern is given by:
\[ A \sin \theta = \left( n + \frac{1}{2} \right) \lambda \]
where \( n \) is the order of the maximum, \( A \) is the slit width, and \( \lambda \) is the wavelength of light.

2. **Identifying the Orders**:
For the third secondary maximum, we set \( n = 3 \):
\[ A \sin \theta = \left( 3 + \frac{1}{2} \right) \lambda = \frac{7}{2} \lambda \]
For red light (wavelength = 6500 Å), the secondary maximum corresponds to \( n = 2 \):
\[ A \sin \theta = \left( 2 + \frac{1}{2} \right) \lambda_{\text{red}} = \frac{5}{2} \lambda_{\text{red}} = \frac{5}{2} \times 6500 \text{ Å} \]

3. **Setting the Positions Equal**:
Since the positions of the maxima must coincide, we can set the two equations equal to each other:
\[ \frac{7}{2} \lambda = \frac{5}{2} \times 6500 \text{ Å} \]

4. **Solving for the Wavelength**:
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = \frac{5 \times 6500 \text{ Å}}{7} \]
Now, calculating the value:
\[ \lambda = \frac{32500 \text{ Å}}{7} \approx 4642.86 \text{ Å} \]

5. **Final Answer**:
The wavelength of light for which the third secondary maximum coincides with the secondary maximum for red light is approximately:
\[ \lambda \approx 4642.86 \text{ Å} \]