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In Young's double slit experiment, the m...

In Young's double slit experiment, the maximum intensity is `I_(0)`. What is the intensity at a point on the screen where the path difference between the interfering waves is `lamda/4` ?

A

`sin^(-1)((lambda)/(d))`

B

`sin^(-1)((lambda)/(3d))`

C

`sin^(-1)((lambda)/(2d))`

D

`sin^(-1)((lambda)/(4d))`

Text Solution

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The correct Answer is:
B
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Youngs Double slit experiment | variation of intensity on screen

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 a Young's double slit experiment the amplitudes of the two waves incident on the two slits are A and 2A . If I_(0) is the maximum intensity, then the intensity at a spot on the screen where the phase difference between the two interfering waves is phi .

    A
    `I_(0)cos^(2)(phi//2)`
    B
    `(I_(0))/3 sin^(2)(phi//2)`
    C
    `((I_(0))/9(5+4cos(phi))`
    D
    `(I_(0))/9(5+8 cos (phi))`

  • 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
    `3K//4`
    D
    `K`

  • In the young's double slit experiment the intensity of light at a point on the screen where the path difference is lamda is K, ( lamda being the wavelength of light used). The intensity at a point where the path difference is lamda//4 . Will be:

    A
    k
    B
    `k/4`
    C
    `k/2`
    D
    Zero

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