JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 21)
In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ____________ nm.
Svar
450
Forklaring
$$y = {d \over 2}$$,
$$\therefore$$ $$\Delta x = y{d \over D}$$
$$ \Rightarrow {{{d^2}} \over {2D}} = {\lambda \over 2}$$
$$ \Rightarrow \lambda = {{{{(0.6 \times {{10}^{ - 3}})}^2}} \over {0.8}}$$
= 450 nm