In a Young's double slit experiment, interference fringes were produced on a screen placed 1.8 m from the two slits 0.3 mm apart and illuminated by light of wavelength 600 nm. The fringe width is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.6 mm
Given:
Distance between slits, $d = 0.3\ \text{mm} = 3 \times 10^{-4}\ \text{m}$
Distance of screen, $D = 1.8\ \text{m}$
Wavelength, $\lambda = 600\ \text{nm} = 6 \times 10^{-7}\ \text{m}$
Fringe width formula: $\beta = \frac{\lambda D}{d}$
$\beta = \frac{6 \times 10^{-7} \times 1.8}{3 \times 10^{-4}}$
$\beta = \frac{1.08 \times 10^{-6}}{3 \times 10^{-4}}$
$\beta = 3.6 \times 10^{-3}\ \text{m} = 3.6\ \text{mm}$
∴ Fringe width = 3.6 mm