Question

Rectangular glass slab ABCD, of refractive index n₁, is immersed in water of refractive index n₂ (n₁> n₂). A ray of light in incident at the surface AB of the slab as shown. The maximum value of the angle of incidence a_max, such that the ray comes out only from the other surface CD is given by

• A. sin^–1$\left[ \frac { \mathrm { n } _ { 1 } } { \mathrm { n } _ { 2 } } \cos \left( \sin ^ { - 1 } \frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } } \right) \right]$
• B. sin^–1
• C. sin^–1$\left( \frac { \mathrm { n } _ { 1 } } { \mathrm { n } _ { 2 } } \right)$
• D. sin^–1 $\left( \frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } } \right)$

Answer 1

Answer: (A)

sin^–1$\left[ \frac { \mathrm { n } _ { 1 } } { \mathrm { n } _ { 2 } } \cos \left( \sin ^ { - 1 } \frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } } \right) \right]$

Answer 2

Ray comes out from CD, means rays after refraction from AB get, total internal reflected at AD

n₂ sina_max = n₁sinr₁

sina_max = $\frac { \mathrm { n } _ { 1 } } { \mathrm { n } _ { 2 } }$ sinr₁ ...... (1)

max. value of a = r₂ = c

r₁ + c = 90

r₁ = 90 – c

sinr₁ = sin(90 – c) = cos c

sinr₁ = cos $\left\{ \sin ^ { - 1 } \frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } } \right\}$ Q sin c = $\frac { \mu _ { \mathrm { r } } } { \mu _ { \mathrm { d } } }$ = $\frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } }$

c = $\sin ^ { - 1 } \left( \frac { n _ { 2 } } { n _ { 1 } } \right)$

Putting in I^st equation

$\alpha _ { \max } = \sin ^ { - 1 } \left[ \frac { \mathrm { n } _ { 1 } } { \mathrm { n } _ { 2 } } \cos \left\{ \sin ^ { - 1 } \frac { \mathrm { n } _ { 2 } } { \mathrm { n } _ { 1 } } \right\} \right]$