In Young's double slit experiment, slits are illuminated by light of wavelength 650 nm. The slits are 0.13 cm apart and the screen is placed 1.2 m away. What will be the separation of the adjacent maxima?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.6 mm
Given:
$\lambda = 650 \, nm = 650 \times 10^{-9} \, m$
$d = 0.13 \, cm = 0.13 \times 10^{-2} = 1.3 \times 10^{-3} \, m$
$D = 1.2 \, m$
Fringe separation: $\beta = \frac{\lambda D}{d}$
$\beta = \frac{650 \times 10^{-9} \times 1.2}{1.3 \times 10^{-3}}$
$\beta = \frac{7.8 \times 10^{-7}}{1.3 \times 10^{-3}}$
$\beta = 6.0 \times 10^{-4} \, m$
$\beta = 0.6 \, mm$
Answer: 0.6 mm