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Light of wavelength lambda in air enters...

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`

B

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

C

`(mu-1)((2pi)/lambda)x`

D

`1/(mu-1)((2pi)/lambda)x`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • 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`

  • A light wave enters from air into a medium of refractive index 1.5. The speed of light in the medium will be

    A
    `2 xx 10^(8)m//s`
    B
    `4.5 xx 10^(8) m//s`
    C
    `9 xx 10^(8) m//s`
    D
    `(330//1.5) xx 10^(8) m//s`

  • A light of wavelength 6000 in air enters a medium of refractive index 1.5 . Inside the medium , its frequency is v and it wavelength is lambda .

    A
    `v=5xx10^(14)Hz`
    B
    `v=7.5xx10^(14)Hz`
    C
    `lambda=4000`
    D
    `A=9000`

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