1.

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 :

Answer»

`(lambda)/(2theta)`
`(2theta)/(theta)`
`(lambda)/(theta)`
`2lambda sin theta`

Solution :
`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~thetabeta=(lambda)/(theta)`

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


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