In young's experiment when sodium light of wavelength 5893A is used, then 62 fringes are seen in the field of view, instead. If violet light of wavelength 4350A is used, then the number of fringes that will be seen in the field of view will be

Options

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

Answer

To solve the problem, we need to determine how many fringes will be seen in Young's double-slit experiment when violet light of wavelength 4350 Å is used, given that 62 fringes are seen with sodium light of wavelength 5893 Å.

### Step-by-Step Solution:

1. **Understand the relationship between the number of fringes and wavelength**:
In Young's experiment, the number of fringes (n) observed is related to the wavelength (λ) of the light used. The formula for the fringe width (β) is given by:
\[ \beta = \frac{\lambda D}{d} \]
where \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the slits.

2. **Set up the equation for the two scenarios**:
For sodium light (λ1 = 5893 Å) and the number of fringes (n1 = 62):
\[ n_1 \beta_1 = n_2 \beta_2 \]
Since the length of the region is constant, we can equate the two scenarios:
\[ n_1 \left(\frac{\lambda_1 D}{d}\right) = n_2 \left(\frac{\lambda_2 D}{d}\right) \]

3. **Simplify the equation**:
The \(D\) and \(d\) cancel out, leading to:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

4. **Substitute the known values**:
We know:
- \(n_1 = 62\)
- \(\lambda_1 = 5893 \, \text{Å}\)
- \(\lambda_2 = 4350 \, \text{Å}\)

Plugging these values into the equation:
\[ 62 \times 5893 = n_2 \times 4350 \]

5. **Solve for \(n_2\)**:
Rearranging the equation gives:
\[ n_2 = \frac{62 \times 5893}{4350} \]
Now, calculate \(n_2\):
\[ n_2 = \frac{364786}{4350} \approx 84.00 \]

6. **Round to the nearest whole number**:
Since the number of fringes must be a whole number, we round \(n_2\) to 84.

### Final Answer:
The number of fringes that will be seen in the field of view when violet light of wavelength 4350 Å is used is **84**.

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