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. `3536 xx 10^(-8) cm`
B. `7071 xx 10^(-8) cm`
C. `2887 xx 10^(-8) cm`
D. `3660 xx 10^(-8) cm`

Explanation

To solve the problem, we need to use the conditions for diffraction minima in a single-slit diffraction pattern. The positions of the minima are determined by the formula:

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

where:
- \( A \) is the slit width,
- \( \theta \) is the angle of the minimum,
- \( n \) is the order of the minimum (1 for the first minimum, 2 for the second minimum),
- \( \lambda \) is the wavelength of the light.

### Step-by-Step Solution:

1. **Identify the given data:**
- Wavelength \( \lambda_1 = 5000 \times 10^{-8} \) cm
- Angle for the first minimum of \( \lambda_1 \) is \( \theta_1 = 30^\circ \)
- Angle for the second minimum of \( \lambda_2 \) is \( \theta_2 = 45^\circ \)

2. **Write the equations for the minima:**
- For the first minimum of \( \lambda_1 \):
\[ A \sin(30^\circ) = 1 \cdot \lambda_1 \]
- For the second minimum of \( \lambda_2 \):
\[ A \sin(45^\circ) = 2 \cdot \lambda_2 \]

3. **Calculate the sine values:**
- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)

4. **Substitute the sine values into the equations:**
- From the first equation:
\[ A \cdot \frac{1}{2} = \lambda_1 \quad \Rightarrow \quad A = 2\lambda_1 \]
- From the second equation:
\[ A \cdot \frac{1}{\sqrt{2}} = 2 \lambda_2 \]

5. **Substitute \( A \) from the first equation into the second equation:**
\[ 2\lambda_1 \cdot \frac{1}{\sqrt{2}} = 2\lambda_2 \]

6. **Simplify the equation:**
\[ \lambda_1 \cdot \frac{1}{\sqrt{2}} = \lambda_2 \]

7. **Substitute the value of \( \lambda_1 \):**
\[ \lambda_2 = \frac{5000 \times 10^{-8}}{\sqrt{2}} \]

8. **Calculate \( \lambda_2 \):**
\[ \lambda_2 \approx \frac{5000 \times 10^{-8}}{1.414} \approx 3536 \times 10^{-8} \text{ cm} \]

### Final Answer:
The value of \( \lambda_2 \) is approximately \( 3536 \times 10^{-8} \) cm.