In a Young's double slit experiment, two slits are separated by 3 mm and the screen is placed 1.5 m away. The distance between two consecutive minima for a light of wavelength 420 nm is

0.21 mm

The correct answer is Option (4) → 0.21 mm

Given:

Distance between slits, $ d = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m} $

Distance between slits and screen, $ D = 1.5\ \text{m} $

Wavelength, $ \lambda = 420\ \text{nm} = 420 \times 10^{-9}\ \text{m} $

For interference pattern:

Distance between two consecutive bright or dark fringes (fringe width)

$ \beta = \frac{\lambda D}{d} $

Substituting values,

$ \beta = \frac{420 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} $

$ \beta = \frac{630 \times 10^{-9}}{3 \times 10^{-3}} $

$ \beta = 2.1 \times 10^{-4}\ \text{m} $

$ \beta = 0.21\ \text{mm} $

Therefore, the distance between two consecutive minima is $ 0.21\ \text{mm} $.