Question

In Young's double slit experiment, the wavelength of monochromatic light is increased by 20% and the distance between the two slits is decreased by 25%. If the initial fringe width is 0.3 mm, then the final fringe width is

• A. 0.72 mm
• B. 0.60 mm
• C. 0.16 mm
• D. 0.48 mm

Hint:

Fringe width in YDSE: \( \beta = \frac{\lambda D}{d} \). If \( \lambda \) changes to \( \lambda' = k_1 \lambda \) and \( d \) changes to \( d' = k_2 d \) (D constant), then the new fringe width \( \beta' = \frac{(k_1 \lambda) D}{(k_2 d)} = \frac{k_1}{k_2} \left(\frac{\lambda D}{d}\right) = \frac{k_1}{k_2} \beta \). Increase by P% means new value = Old value \( \times (1 + P/100) \). Decrease by P% means new value = Old value \( \times (1 - P/100) \).

Answer 1

The Correct Option is D

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 between the slits and the screen, and \( d \) is the distance between the two slits.
Initial state: \( \lambda_1, d_1, \beta_1 = 0.
3 \) mm.
So, \( \beta_1 = \frac{\lambda_1 D}{d_1} = 0.
3 \) mm.
(Assume D is constant).
Changes: Wavelength is increased by 20%: \( \lambda_2 = \lambda_1 + 0.
20 \lambda_1 = 1.
20 \lambda_1 \).
Distance between slits is decreased by 25%: \( d_2 = d_1 - 0.
25 d_1 = 0.
75 d_1 \).
Final fringe width \( \beta_2 \): \[ \beta_2 = \frac{\lambda_2 D}{d_2} = \frac{(1.
20 \lambda_1) D}{(0.
75 d_1)} \] \[ \beta_2 = \left(\frac{1.
20}{0.
75}\right) \left(\frac{\lambda_1 D}{d_1}\right) \] We know \( \frac{\lambda_1 D}{d_1} = \beta_1 = 0.
3 \) mm.
\[ \beta_2 = \left(\frac{1.
20}{0.
75}\right) \beta_1 \] Calculate the ratio \( \frac{1.
20}{0.
75} = \frac{120}{75} \).
Divide by 15: \( \frac{120 \div 15}{75 \div 15} = \frac{8}{5} \).
So, \( \frac{1.
20}{0.
75} = \frac{8}{5} = 1.
6 \).
\[ \beta_2 = 1.
6 \times \beta_1 = 1.
6 \times 0.
3 \, \text{mm} \] \[ \beta_2 = 0.
48 \, \text{mm} \] This matches option (4).