Question

What is the critical angle θcritθcrittheta_crit for light propagating from a material with index of refraction of 1.50 to a material with index of refraction of 1.00?

Answer 1

Answer:

The critical angle is 41.8°.

Explanation:

The critical angle is $\theta_1$ for which the angle of refraction $\theta_2$ is 90°. From Snell's law we have

$n_1sin (\theta_1 ) = n_2 sin(\theta_2)$

$n_1sin (\theta_1 ) = n_2 sin(90^o)$

$sin (\theta_1 ) = n_2/n_1,$

$\theta_1 = sin^{-1}(\dfrac{n_2}{n_1} ).$

Putting in $n_2 =1.00,$ and $n_1 = 1.50$ we get:

$\theta_1 = sin^{-1}(\dfrac{1.00}{1.500} ),$

$\boxed{\theta_1 = 41.8^o}$

Thus, the critical angle is 41.8°.