A rectangular slab ABCD, of refractive index `n_1` , is immersed in water of refractive index `n_2(n_1 lt n_2)`. A ray of light is incident at the surface AB of the slab as shown in Fig. Find the maximum value of angle of incidence `alpha_(max)`, such that the ray comes out only from the other surface CD.

Options

A. `sin^(-1)[(n_(1))/(n_(2))cos(sin^(-1).(n_(2))/(n_(1)))]`
B. `sin^(-1)[n_(1)cos(sin^(-1).(1)/(n_(2)))]`
C. `sin^(-1)(n_(1)/(n_(2)))`
D. `sin^(-1)(n_(2)/(n_(1)))`

Answer

A

Explanation

Ray comes out from `CD`,means rays after refraction from `AB` get, total internally reflected at `AD`
`(n_(1))/(n_(2))=(sina_(max))/(sinr_(1))impliesa_(max)=sin^(-1)[(n_(1))/(n_(2))sinr_(1)]` ...(i)
Also, `r_(1)+r_(2)=90^(@)impliesr_(1)=90-r_(2)=90-C`
`r_(1)=90-sin^(-1)((1)/(2mu_(2)))impliesr_(1)=90-sin^(-1)((n_(2))/(n_(1)))` ....(ii)
Hence from equation (i) and (ii)
`a_(max)=sin^(-1)[(n_(1))/(n_(2))sin{90-sin^(-1).(n_(2))/(n_(1))}]`
`=sin^(-1)[(n_(1))/(n_(2))cos(sin^(-1).(n_(2))/(n_(1)))]`