In Young's double slit experiment using sodium light (`lamda=5898Å`) 92 fringes are seen if given colour `(lamda=5461Å)` is used how many fringes will be seen

Options

A. 62
B. 67
C. 85
D. 99

Explanation

To solve the problem of how many fringes will be seen in Young's double slit experiment when using a different wavelength of light, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between the number of fringes and wavelength**: In Young's double slit experiment, the number of fringes (n) observed on the screen is directly proportional to the wavelength (λ) of the light used. The relationship can be expressed as:
\[ n \propto \lambda \]
This means that if the wavelength increases, the number of fringes seen will also increase, and vice versa.

2. **Set up the equation for the two scenarios**: Let \( n_1 \) be the number of fringes seen with the first wavelength \( \lambda_1 \) (sodium light) and \( n_2 \) be the number of fringes seen with the second wavelength \( \lambda_2 \) (given color). We can express the relationship as:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

3. **Substitute the known values**: From the problem, we know:
- \( n_1 = 92 \) (number of fringes with sodium light)
- \( \lambda_1 = 5898 \, \text{Å} \) (wavelength of sodium light)
- \( \lambda_2 = 5461 \, \text{Å} \) (wavelength of the given color)

Plugging these values into the equation gives:
\[ 92 \times 5898 = n_2 \times 5461 \]

4. **Solve for \( n_2 \)**: Rearranging the equation to find \( n_2 \):
\[ n_2 = \frac{92 \times 5898}{5461} \]

5. **Calculate \( n_2 \)**: Performing the calculation:
\[ n_2 = \frac{92 \times 5898}{5461} \approx \frac{542736}{5461} \approx 99 \]

6. **Conclusion**: Therefore, when the wavelength of light is 5461 Å, the number of fringes observed will be approximately 99.

### Final Answer:
The number of fringes seen with the wavelength of 5461 Å is **99**.

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