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A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is

A

0

B

`(pi)/(2)`

C

`pi`

D

`2pi`

Text Solution

Verified by Experts

The correct Answer is:
D

The phase difference `(phi)` between the wavelets from the top edge and the bottom edge of the slit is `phi=(2 pi)/(lambda)(d sin theta)`, where d is the slit width.
The first minima of the diffraction pattern occurs at `sin theta=(lambda)/(d)`
So, `phi=(2pi)/(lambda)(d xx (lambda)/(d))=2pi`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-INTERFERENCE AND DIFFRACTION OF LIGHT -Exercise 1 (TOPICAL PROBLEMS)
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