A beam of parallel rays of width b propagates in glass at an angle`theta`to its plane face .The beam width after it goes over to air through this faceis…….if the refractive index of glass is `mu.v`

Answer

`(b(1-mu^(2)os^(2)theta)^(1//2))/(sin theta)`

Explanation

From snell's law
`mu sin(90^(@)-theta)=1 sin(90^(@)-alpha)`
`mu cos theta=1-cos alpha`
`cos alpha=mu cos theta …….(1)`
From right angle triangle `ABN`
`(b)/(AB)=cos(90- theta)`
`AB=(b)/(sin theta).....(2)`
From right angle triangle `ABN'`
`(b')/(AB)=sin alpha`
`b'=AB sinalpha....(3)`
from (2)and(3)
`b'=(b)/(sin theta)sqrt(1-cos^(2)alpha)...........(4)`
from(1)&(4)
`b'(b)/(sintheta)sqrt(1-mu^(2)costheta^(2))`