A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10 th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be ________ mm .

Solution

Physics sir · 3 steps

Same slits, same screen, same fringe number: the position just scales with wavelength.

1. Bright fringe position. The nth bright fringe sits at yₙ = (nλ D)/d from the central maximum. Here n, D and d stay the same, so yₙ ∝ λ.

   y₂/y₁ = λ₂/λ₁

2. Put in numbers. y₂ = 10 mm × 660/600 = 10 × 1.1 = 11 mm.
3. So. The 10th bright fringe moves out to 11 mm from the central maximum.

The common slip: 9.09. Inverting the ratio, 10 × 600/660, gives about 9.09 mm. A longer wavelength spreads the fringes out, so the fringe must move farther away.

The idea to keep. In YDSE, fringe positions and fringe width are proportional to λ D/d. Change only one of these and use a ratio; you never need the actual slit spacing or screen distance.

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