1.

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference is lambda is K. (K being the wavelength of light used). The intensity at 2 a point where the path difference is (lambda)/(4) will be.....

Answer»

K
`(K)/(4)`
`(K)/(2)`
zero

Solution :In `I=I_(0) cos^(2)[(K(r_(1)-r_(2)))/(2)],r_(1)-r_(2)=1 rArr I=K`
`:. K=I_(0) cos^(2)[(2pi)/(lambda)xx(lambda)/(2)]`
`:. K=I_(0)cos^(2)pi`
`:. K=I_(0)=....(I) "" [ :. Cos^(2)pi=(-1)^(2)=1]`
Now again `I.=I_(0)cos^(2)[(K(r_(1)-r_(2)))/(2)],r_(1)-r_(2)=(lambda)/(4)`
and `K=(2pi)/(lambda)`
`:.I.= I_(2)cos^(2)[(2pi)/(lambda)xx(lambda)/(8)]`
`=I_(0)cos^(2)(pi)/(4)`
`=(I_(0))/(2) [ :."cos"^(2)(pi)/(4)=((1)/(sqrt(2)))^(2)=(1)/(2)]`
`:. I=(K)/(2) [ :. I_(0)=` Kfrom EQUATION (1)]


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