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In Young's double slit experiment the in...

In Young's double slit experiment the intensity of light on the screen where the where the path difference is `lambda` is k ( `lambda` being the wavelength of light used). The intensity at a point where the path difference is `(lambda)/(4)` will be

A

k

B

`(k)/(4)`

C

`(k)/(2)`

D

zero

Text Solution

Verified by Experts

We know I = `I_(0) "cos"^(2) (pi delta)/(lambda)`
When path difference is `lambda` then I=k= `I_(0) "cos"^(2) pi =I_(0)`
When path difference is `(lambda)/(4)` then I = `I_(0) "cos" ^(2) (pi)/(4)= I_(0)/(2)=(k)/(2)`
The option C is correct.
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Knowledge Check

  • If young' s double slit experiment is performed using white light, then

    A
    the central fringe will be white
    B
    no fringe will be completely dark
    C
    the fringe adjacent to the central one will be red
    D
    the fringe adjacent to the central one will be violet

  • In Young's double slit experiment the distance between the two slits is d and wavelength of the light used is lambda . The angular width of the fringe

    A
    `(d)/(lambda)`
    B
    `(lambda)/(d)`
    C
    `(2lambda)/(d)`
    D
    `(lambda)/(2d)`

  • In Young's double slit experiment the distance of the screen from the plane of the slits is 1.0 m. The wavelength of light used and width of the fringe are 6000 Å and 2 mm respectively. Distance between the slits is

    A
    0.5 mm
    B
    0.4 mm
    C
    0.2 mm
    D
    0.3 mm

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