Question

In Young’s double-slit experiment with monochromatic light of wavelength 6000 Å, the fringe width is 3 mm. If the distance between the screen and slits is increased by 50\% and the distance between the slits is decreased by 10\%, then the fringe width is

• A. 12 mm
• B. 5 mm
• C. 6 mm
• D. 10 mm

Hint:

Fringe width increases if \( D \) increases and decreases if \( d \) increases. Always apply percentage changes carefully.

Answer 1

The Correct Option is B

Fringe width formula: \[ \beta = \frac{\lambda D}{d} \] Given: \[ D' = 1.5D, \quad d' = 0.9d \] New fringe width: \[ \beta' = \frac{\lambda (1.5D)}{0.9d} = \frac{1.5}{0.9} \beta \] \[ \beta' = \frac{5}{3} \times 3 = 5 \text{ mm} \] Thus, the correct answer is 5 mm.