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Question: In a young’s double slit experiment, (slit distance d) monochromatic light of wavelength \(\lambda\)...

In a young’s double slit experiment, (slit distance d) monochromatic light of wavelength λ\lambdais used and the fringe pattern observed at a distance D from the slits. The angular position of the bright fringes are:

A

sini(Nλd)\sin^{- i}\left( \frac{N\lambda}{d} \right)

B

sin1((N+12)λd)\sin^{- 1}\left( \frac{(N + \frac{1}{2})\lambda}{d} \right)

C

sin1NλD\sin^{- 1}\frac{N\lambda}{D}

D

sin1((N+12)λD)\sin^{- 1}\left( \frac{(N + \frac{1}{2})\lambda}{D} \right)

Answer

sini(Nλd)\sin^{- i}\left( \frac{N\lambda}{d} \right)

Explanation

Solution

The condition for bright fringes is, path difference,

δ=dsinθbright=Nλ\delta = d\sin\theta_{bright} = N\lambda

Where N=0,±1,±2N = 0, \pm 1, \pm 2

The angular position of the bright fringes is

θbright=sin1(Nλd)\theta_{bright} = \sin^{- 1}\left( \frac{N\lambda}{d} \right)