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

Rays come out only form CD , means rays after refraction from AB get total internally reflected at AD .

From the figure :
`r_(1) + r_(2) = 90^(@)`
`r_(1) = 90^(@) - r_(2)`
`(r_(1))_("max") = theta_(C)`
[for total internal reflection at AD]
where `"sin"theta_(C) = (n_(2))/(n_(1)) "or" theta_(C) = sin^(-1)((n_(2))/(n_(1)))`
`therefore (r_(1))_("max") = 90^(@) - theta_(C)`
Now applying Snell's law at face AB:
`(n_(1))/(n_(2)) = ("sin"alpha_("max"))/("sin"(r_(1))_("max")) = ("sin"alpha_("max"))/("sin"(90^(@) - theta_(C))) = ("sin"alpha_("max"))/("cos"theta_(C))`
or `"sin" alpha_("max") = (n_(1))/(n_(2))"cos"theta_(C)`
`therefore alpha_("max") = sin^(-1)[(n_(1))/(n_(2)) "cos"theta_(C)] = sin^(-1)[(n_(1))/(n_(2)) "cos"sin^(-1)((n_(2))/(n_(1)))]`