Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is `mu`, a ray, incident at an angle `theta`, on the face AB would get transmitted through the face AC of the prism provided:

Options

A. `thetagtcos^(-1)[musin(A+sin^(-1)((1)/(mu))]`
B. `thetaltcos^-1[musin(A+sin^-1((1)/(mu))]`
C. `thetagtsin^-1[musin(A-sin^-1((1)/(mu))]`
D. `thetaltsin^-1[musin(A-sin^-1((1)/(mu))]`

Answer

C

Explanation

When `r_2=C` `angleN_2Rc=90^circ`
Where C=critical angle
As ``sinC=(1)/(v)=sinr_2`
Applying snell's law at `R`
`musinr_2=1sin90^circ`..(i)
Applying snell's law at `Q`
`1xxsintheta=musinr_1`
But `r_1=A-r_2`..(ii)
So, `sintheta=mu sin (A-r_2)`
`sintheta=musinAcosr_2-cosA`...(iii) [using (i)]
From (1)
`cosr_2=sqrt(1-sin^2r_2)=sqrt(1-(1)/(mu^2))`...(iv)
By eq .(iii) and (iv)
`sintheta=musinAsqrt(1-(1)/(mu^2))-cosA`
on further solving we can show for ray not to transmitted through face AC
`theta=sin^-1[usin(A-sin^(-1)((1)/(mu))]`
So, for transmission through face AC
`thetagtsin^(-1)[usin(A-sin^(-1)((1)/(mu))]`