Question

In a Young’s double slit experiment, light of wavelength 620 nm is used with slit separation 0.3 mm and width of fringe 1.3 mm. The distance of the screen from source will be

• A. 62.9 m
• B. 6.29 m
• C. 0.629 m
• D. 0.0629 m

Hint:

To solve Young's double slit experiment problems, remember to use the fringe width formula: \( \beta = \frac{\lambda D}{d} \), and rearrange to find the unknown variable.

Answer 1

The Correct Option is B

Step 1: Use the formula for fringe width in Young's double slit experiment.
The fringe width (\( \beta \)) in Young's double slit experiment is given by: 
\[ \beta = \frac{\lambda D}{d} \] where: \( \lambda \) is the wavelength of light, \( D \) is the distance of the screen from the slits, \( d \) is the slit separation. 
Step 2: Plug in the given values.

We are given:
\( \lambda = 620 \, \text{nm} = 620 \times 10^{-9} \, \text{m} \), \( d = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} \), \( \beta = 1.3 \, \text{mm} = 1.3 \times 10^{-3} \, \text{m} \). Rearrange the formula to find \( D \): \[ D = \frac{\beta d}{\lambda} \] Substitute the values: \[ D = \frac{(1.3 \times 10^{-3}) \times (0.3 \times 10^{-3})}{620 \times 10^{-9}} \, \text{m} \] \[ D = 6.29 \, \text{m} \] 
Step 3: Conclusion.
Thus, the distance of the screen from the source is 6.29 m. 
Conclusion:
The correct answer is (B) 6.29 m.