In Young's double slit experiment, 62 fringes are seen in visible region for sodium light of wavelength `5893Å`. If violet light of wavelength `4358Å`, is used in place of sodium light, then number of fringes seen will be

Options

A. `54`
B. `64`
C. `74`
D. `84`

Explanation

To solve the problem, we need to determine how many fringes will be seen when violet light of wavelength 4358 Å is used instead of sodium light of wavelength 5893 Å.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
In Young's double slit experiment, the number of fringes (n) observed is related to the wavelength (λ) of the light used. The formula for the number of fringes is given by:
\[ n = \frac{D}{\beta} \]
where D is the length of the visible region and β is the fringe width.

2. **Fringe Width Calculation**:
The fringe width (β) is given by:
\[ \beta = \frac{d \cdot \lambda}{D} \]
where d is the distance between the slits and D is the distance from the slits to the screen. Since both d and D remain constant when changing the light source, we can relate the number of fringes for two different wavelengths.

3. **Setting Up the Equation**:
For two different wavelengths (λ1 for sodium light and λ2 for violet light), we can write:
\[ n_1 \cdot \lambda_1 = n_2 \cdot \lambda_2 \]
where \( n_1 \) is the number of fringes for sodium light, \( n_2 \) is the number of fringes for violet light, \( \lambda_1 \) is the wavelength of sodium light, and \( \lambda_2 \) is the wavelength of violet light.

4. **Substituting Known Values**:
Given:
- \( n_1 = 62 \)
- \( \lambda_1 = 5893 \, \text{Å} = 5893 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 4358 \, \text{Å} = 4358 \times 10^{-10} \, \text{m} \)

We can rearrange the equation to solve for \( n_2 \):
\[ n_2 = n_1 \cdot \frac{\lambda_1}{\lambda_2} \]

5. **Calculating \( n_2 \)**:
Substituting the values:
\[ n_2 = 62 \cdot \frac{5893 \times 10^{-10}}{4358 \times 10^{-10}} \]
Simplifying:
\[ n_2 = 62 \cdot \frac{5893}{4358} \]

6. **Performing the Calculation**:
First, calculate \( \frac{5893}{4358} \):
\[ \frac{5893}{4358} \approx 1.352 \]
Now, multiply by 62:
\[ n_2 \approx 62 \cdot 1.352 \approx 84.0 \]

7. **Final Result**:
Thus, the number of fringes seen with violet light will be approximately 84.

### Final Answer:
The number of fringes seen with violet light of wavelength 4358 Å is **84**.