Question

in a couple of lines of math, derive an expression for the critical minimum angle at which total internal reflection that can occur when light is incident from higher refraction index to lower refraction index material. answer should be in terms of indicies of refractions of these materials

Answer 1

Consider a surface XY seperating the rarer medium a (having low refractive index) from the denser medium b (having high refractive index). We know, when the light travels from the denser to rarer medium, it moves away from the normal. As the angle of incidence of light in the denser medium increases, the angle of refraction goes on increasing.

There comes a stage when corresponding to a certain angle of incidence, angle of refraction becomes 90o. This particular angle of incidence for which angle of refraction becomes 900 is what is known as Critical Angle. If the angle of incidence is further increased beyond this critical angle,the ray of light undergoes reflection rather than refraction and this phenomenon is known as Total internal Reflection.

According to Snell's law, refractive index bua of the rarer medium a w.r.t denser medium b is given by

bua = Sin i/Sin r .......eqn 1

where i denotes angle of incidence while r denotes angle of refraction.

As mentioned above, at critical angle of incidence C; angle of refraction (r) becomes 900 ;

Hence, eqn 1 can be written as

bua = Sin C/Sin 90o

bua = Sin C ......eqn 2 (Sin 90o = 1)

It must be noted that since the light is travelling from denser to rarer medium, so we need the refractive index of denser medium (b) w.r.t rarer medium (a )

According to principle of reversibility of the light;

aub* bua = 1

Hence, refractive index (aub) of denser medium b with respect to rarer medium a is given by;

aub = 1/bua ......eqn 3

Using eqn 2 in eqn 3

aub = 1/Sin C

or Sin C = 1/aub

C = Sin-1 (1/aub) .......eqn 4

This gives the required expression for the Critical Angle C in terms of  refractive index bua required to observe total internal reflection.