The unpolarised light is incident on polarisers placed above the other, so what is the angle between these two polarisers so that the than the intensity of the transmitted light is `(1)/(3)` than the intensity of incident light?

Options

A. `54.7^(@)`
B. `35.3^(@)`
C. `0^(@)`
D. `60^(@)`

Answer

B

Explanation

Let the intensity of light is `I_(0)`.
If the intensity of transmitted light from firs polariser `I_(1)`
`I_(1)=I_(0) cos^(2) theta`
`=(I_(0))/(2)` [From illustration 10]
Now the intensity of transmitted light fron second polariser `I_(1)=(I_(0))/(3)` given,
`:.I_(2)=I_(1) cos^(2) theta`
`:.(I_(0))/(3)=(I_(0))/(2) cos^(2) theta`
`(2)/(3)= cos^(2) theta`
`:. cos theta.=sqrt((2)/(3))`
`:. sin (90^(@)-theta)=0.8165`
`:. 90^(@)- theta=54.7^(@)`
`:. theta=90^(@)-54.7^(@)`
` :. theta=35.3^(@)`