Question

Light rays in a material with index of refraction 1.31 can undergo total internal reflection when they strike the interface with another material at a critical angle of incidence. Find this material\'s index of refraction when the required critical angle is 78.3°.

Answer 1

Answer:

The refractive index of the material is 1.28.

Explanation:

It is given that,

Refractive index of medium 1, n₁ = 1.31

Critical angle, $\theta_c=78.3$

At critical angle rays will reflects at 90 degrees. Using Snell's law as:

$n_1sin\theta_1=n_2sin\theta_2$

At critical angle, $n_1sin\ \theta_c=n_2\ sin90$

$n_1sin\ \theta_c=n_2$

$n_2=1.31\times sin(78.3)$

$n_2=1.28$

So, the refractive index of the material is 1.28. Hence, this is the required solution.