JEE Main 2024 Physics question
Monochromatic light of wavelength 500 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 4ᵗʰ bright fringe. The thickness of the glass-plate is ...................μm.
Answer: 4
Solution
The shift corresponds to a path difference (μ − 1)t = nλ with n = 4: (1.5 − 1)t = 4 × 500 × 10⁻⁹ m, so t = 4 × 10⁻⁶ m = 4 μm.