1.

The intensity at the maximum in a Young's double slit experiment is I_(0). Distance between two slits is d=5 lambda where lambda is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D=10d ?

Answer»

`(I_(0))/(4)`
`(3)/(4)I_(0)`
`(I_(0))/(2)`
`I_(0)`

SOLUTION :Path difference `=(dx)/(D)`
`=(dxx(d)/(2))/(10d)`
`=(d)/(20)=(5 lambda)/(20)`
`"sin" theta=(lambda)/(4)`
`:. sin theta=(2pi)/(4) [ :. lambda= 2pi]`
`:. sin theta=(pi)/(2)`
`:. theta=90^(@)`
From Malus, law,
`I=I_(0)"cos"^(2)(phi)/(2)=I_(0)cos^(2)45^(@)`
`:.I=I_(0)XX((1)/(sqrt(2)))^(2) :.I=(I_(0))/(2)`


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