Q89 / 90 · JEE Main 2024 (09 Apr Shift 2) · Physics
Monochromatic light of wavelength $500 \mathrm{~nm}$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $=1.5$ ), the central maximum is shifted to a position previously occupied by the $4^{\text {th }}$ bright fringe. The thickness of the glass-plate is ______ $\mu \mathrm{m}$.
Correct answer: 4
Solution
$\begin{aligned} & (\mu-1) \mathrm{t}=\mathrm{n} \lambda \\ & (1.5-1) \mathrm{t}=4 \times 500 \times 10^{-9} \mathrm{~m} \\ & \mathrm{t}=4000 \times 10^{-9} \mathrm{~m} \\ & \mathrm{t}=4 \mu \mathrm{m}\end{aligned}$
Hence, the correct answer is 4.