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In Young's double-slit experiment, the i...

In Young's double-slit experiment, the intensity of light in front of one of the slits on a screen is `I_(0)//2` where `I_(0)` is the maximum intensity. The distance between the slits is `5lambda` where `lambda` is the wavelength of monochromatic light. How far away is the screen from the slit?

A

`20lambda`

B

`25lambda`

C

`40lambda`

D

`50lambda`

Text Solution

Verified by Experts

The correct Answer is:
D

`(dx)/(D) = Deltax`
`2I = I +I +2I cos theta`
`cos theta = 0`
`phi = pi//2`
`(2pi)/(lambda) Delta x = (pi)/(2)`
`Delta x = (lambda)/(4)`
`(d xx x)/(D) = (lambda)/(4)`
`(5lambda)/(D) xx (5lambda)/(2) = (lambda)/(4)`
`D = 50 lambda`
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In a Young's double slit experiment, the intensity of light at a point on the screen where the path difference is lambda is K units. Find the intensities at a point, where path diff. is (i) lambda//4 (ii) lambda//2 .

Knowledge Check

  • In Young's double slit experiment the intensity of light at a point on the screen where the path difference lambda is K. The intensity of light at a point where the path difference is (lambda)/(6) [ lambda is the wavelength of light used] is

    A
    `K//4`
    B
    `K//3`
    C
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    B
    K / 3
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  • In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is lambda=l . The intensity of light at a point where the path difference becomes lambda//3 is

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