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 __________ μ m.

Solution

Physics sir · 4 steps

The plate adds an extra path of (μ - 1)t. Set that equal to 4 wavelengths and solve for t.

1. Extra path. Inside glass of thickness t, light covers an optical path μ t in place of t in air. So the plate adds an extra path of (μ - 1)t to the light from that slit.
2. Fringe shift. The central maximum is where the total path difference is zero. If it moves onto the old 4ᵗʰ bright fringe, the extra path must cancel a path difference of 4λ.

   (μ - 1) t = nλ

3. Put in numbers. (1.5 - 1) t = 4 × 500 nm, so 0.5 t = 2000 nm and t = 4000 nm.
4. So. t = 4000 nm = 4 μm, so the answer is 4.

The common slip: 1.33. Writing the extra path as μ t in place of (μ - 1)t gives t = 2000/1.5 nm ≈ 1.33 μm, not 4.

The idea to keep. A thin plate in front of one slit shifts the whole pattern by n fringes, where (μ - 1)t = nλ. The fringe width does not change. Only the pattern slides.

Answer from the NTA final answer key. Solution by PrepWiser.