A ray of light mekes normal incidence on the diagonal face of a right angled prism as shown in figure. If `theta=37^@,` then the angle of deviation after second step (from `AB`) is `(sin 37^@=3//5)`

Options

A. (A) `53^@`
B. (B) `74^@`
C. (C)`106^@`
D. (D)`90^@`

Answer

B

Explanation

`i_1=0^@`
`r_2=90^@-theta=53^@`
`theta_C=sin^-1(1/mu)=37^@`
Since, `r_2gttheta_C, T//R` will take place on the face AB
`delta=delta_(AC)+delta_(AB)`
`=0+180^@-2r_2`
`=180^@-2xx53^@=74^@`