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In Young's double slit experiment intens...

In Young's double slit experiment intensity at a point is `((1)/(4))` of the maximum intensity. Angular position of this point is

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
If `delta` is phase difference between waves at this point, then resultant intensity can be written as follows:
`I = 4I_0 cos^2 (delta)/2`
Maximum intensity is `4 I_0` and hence at this point intensity is `1//4` or we can say `I_0`
`implies I_0 = 4I_0 cos^2 (delta)/2`
`implies cos (delta)/2 = 1/2`
`implies delta = 2pi//3`.
Let there be a point at a distance y from the centre, then phase difference at this point can be written as follows:
`delta = (2pi)/(lambda) xx (yd)/D`
`implies (2pi)/3 = (2pi)/lambda xx (yd)/(D)`
`implies y/D = lambda/(3d) " "....(1)`
Let angular separation of this point is `theta`, then
`sin theta = tan theta = y/D`
Using equation (1) we can write the following:
`sin theta = lambda/(3d)`
`implies theta = sin^(-1) (lambda/(3d))`
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Knowledge Check

  • In Young's double slit experiment intensity at a point is ((1)/(4)) of the maximum intersity. Angular position of this point is

    A
    `sin^(-1)((lambda)/(d))`
    B
    `sin^(-1)((lambda)/(2d))`
    C
    `sin^(-1)((lamda)/(3d))`
    D
    `sin^(-1)((lamda)/(4d))`

  • In Young's double - slit experiment intensity at a point is (3//4)^("th") of the maximum intensity. The possible angular position of this point is

    A
    `sin^(-1)((lambda)/(3d))`
    B
    `sin^(-1)((lambda)/(2a))`
    C
    `sin^(-1)((lambda)/(6d))`
    D
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  • In interference experiment, intensity at a point is (1/4)^(th) of the maximum intensity. The angular position of this point is at (sin 30^(@)=cos60^(@)=0.5, lambda= wavelength of light, d=slitwidth)

    A
    `sin^(-1)(lambda//3d)`
    B
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    C
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