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

The critical angle is 41.8°.

Step-by-step 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)((n_2)/(n_1) ).`

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

`\theta_1 = sin^(-1)((1.00)/(1.500) ),`

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

Thus, the critical angle is 41.8°.