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Light of wavelength lambda(0) in air ent...

Light of wavelength `lambda_(0)` in air enters a medium of refractive index n. If two points A and B in this medium lie along the path of this light at a distance x, then phase difference `phi_(0)` between these two point is

A

`phi_(0) = (1)/(n) ((2 pi)/(lambda_(0))) x`

B

`phi_(0) = n ((2 pi)/(lambda_(0))) x`

C

`phi_(0) = (n - 1) ((2 pi)/(lambda_(0))) x`

D

`phi_(0) = (1)/((n - 1)) ((2 pi)/(lambda_(0))) x`

Text Solution

Verified by Experts

The correct Answer is:
b
`q phi_(0) = (2 pi)/(lambda_(0)) ("optical path")`
`phi_(0) = (2 pi)/(lambda_(0)) (nx)`
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Knowledge Check

  • Light of wavelength lambda in air enters a medium of refractive index mu . Two points in this medium, lying along the path of this light, are at a distance of x apartThe phase difference between these point is

    A
    `mu((2pi)/lambda)x`
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    A
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