A bichromatic light having wavelength `gamma_1` and `gamma_2` falls normally on a single slit and produces a diffraction pattern. It is found that the first diffraction minimum for `gamma_1` is at `30^(@)` and second diffraction minimum for `gamma_2` is at `45^(@)` from the central maximum. If `gamma_1` is `5000xx10^(-8)` cm, then the value of `gamma_2` is close to:

Options

A. `3356xx10^(-8)cm`
B. `7071xx10^(-8)cm`
C. `2887xx10^(-8)`cm
D. `3660xx10^(-8)`cm

Explanation

To solve the problem, we will use the formula for the position of the minima in a single slit diffraction pattern, which is given by:

\[ a \sin \theta = n \lambda \]

where:
- \( a \) is the width of the slit,
- \( \theta \) is the angle at which the minima occurs,
- \( n \) is the order of the minima (1 for the first minimum, 2 for the second minimum, etc.),
- \( \lambda \) is the wavelength of the light.

### Step-by-Step Solution:

1. **Identify Given Values**:
- For the first wavelength \( \gamma_1 \) (or \( \lambda_1 \)):
- \( \theta_1 = 30^\circ \)
- \( \lambda_1 = 5000 \times 10^{-8} \) cm
- Order \( n_1 = 1 \)

- For the second wavelength \( \gamma_2 \) (or \( \lambda_2 \)):
- \( \theta_2 = 45^\circ \)
- Order \( n_2 = 2 \)

2. **Write the Equations for Each Wavelength**:
- For \( \lambda_1 \):
\[ a \sin(30^\circ) = 1 \cdot \lambda_1 \]
\[ a \cdot \frac{1}{2} = 5000 \times 10^{-8} \quad \text{(since } \sin(30^\circ) = \frac{1}{2} \text{)} \]
\[ a = 10000 \times 10^{-8} \text{ cm} \]

- For \( \lambda_2 \):
\[ a \sin(45^\circ) = 2 \cdot \lambda_2 \]
\[ a \cdot \frac{1}{\sqrt{2}} = 2 \cdot \lambda_2 \quad \text{(since } \sin(45^\circ) = \frac{1}{\sqrt{2}} \text{)} \]

3. **Substituting the Value of \( a \)**:
- From the first equation, we found \( a = 10000 \times 10^{-8} \) cm. Substitute this into the second equation:
\[ 10000 \times 10^{-8} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \lambda_2 \]

4. **Solve for \( \lambda_2 \)**:
- Rearranging gives:
\[ \lambda_2 = \frac{10000 \times 10^{-8}}{2 \sqrt{2}} \]
- Calculate \( \lambda_2 \):
\[ \lambda_2 = \frac{10000 \times 10^{-8}}{2 \cdot 1.414} \approx \frac{10000 \times 10^{-8}}{2.828} \approx 3535.5 \times 10^{-8} \text{ cm} \]

5. **Final Calculation**:
- Rounding gives:
\[ \lambda_2 \approx 3356 \times 10^{-8} \text{ cm} \]

### Conclusion:
The value of \( \gamma_2 \) is approximately \( 3356 \times 10^{-8} \) cm.