Two parallel beams of light of wavelength `lambda` inclined to each other at angle `theta (ltlt1)` are incident on a plane at near normal incidence. The fringe width will be :

Options

A. `(lambda)/(2theta)`
B. `(2theta)/(theta)`
C. `(lambda)/(theta)`
D. `2lambda sin theta`

Answer

C

Explanation

`R=beta sinalpha`
`QR` is the difference between the light reaching at `Q` and `P` respectively

`PQ=beta sinalpha`
for given case `alpha=(theta)/(2)`

For `PQ` to be one fringe. the path difference between the interfering light beams will change by `lambda` while moving from `P` to `Q`
`|"path difference at" P-"path difference at" Q|=lambda`
`|(betasin'(theta)/(2)-(-betasin'(theta)/(2)))|=lamda rArr2beta sin'(theta)/(2)=lambda beta=(lambda)/(2sin(theta//2))`
for near normal incidence `sintheta~theta beta=(lambda)/(theta)`

`tan theta//2=(d//2)/(D) (therefore tan theta//2~theta//2)`
`therefore theta=(d)/(D)`
`beta=(lambdaD)/(d)`
`beta=(lambda)/(theta)`