1.

The wavelength of the light used in Young's double slit experiment is lambda.The intensity at a point on the screen is I, where the path difference is (lambda)/(6). If I_(0) denotes the maximum intensity, then the ratio of I and I_(0) is :

Answer»

`0.866`
`0.5`
`0.707`
`0.75`

Solution :Phase difference `PHI = (2pi)/(lambda) xx` path difference
` =(2pi)/(lambda) xx (lambda)/(6) = (pi)/(3) = 60^(@)`
Intensity, `I = I_(0)COS^(2)((phi)/(2))`
`(I)/(I_(0)) = cos^(2)(30^(@)) = ((sqrt3)/(2))^(2) = 0.75`.


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