In Young's double slit experiment performed using a monochromatic light of wavelength λ, when a glass plate (μ = 1.5) of thickness xλ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be :

Solution

Physics sir · 3 steps

The glass adds (μ - 1)t of path; the old centre stays bright only if that equals nλ.

1. Extra path. Light inside the glass travels an optical path of μ t in place of t. So the plate adds an extra path difference of (μ - 1)t between the two beams at the old central point.
2. Condition. The intensity there stays the same, so the point must still be a bright fringe. That needs the extra path to be a whole number of wavelengths.

   (μ - 1)t = nλ ⇒ (1.5 - 1) xλ = nλ ⇒ x = 2n

   Option	Extra path 0.5 xλ	At the old centre
   (a) x = 3	1.5λ	dark fringe
   (b) x = 2	λ, a whole wavelength	bright, intensity unchanged
   (c) x = 1.5	0.75λ	half the maximum intensity
   (d) x = 0.5	0.25λ	half the maximum intensity

3. So. x must be an even number, and only (b) gives that: x = 2 adds exactly one wavelength, so the central point stays bright.

The idea to keep. A slab of thickness t always adds (μ - 1)t of path, not μ t. Set that equal to nλ for a bright fringe and to an odd multiple of λ/2 for a dark one.

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