Question

A light beam is traveling inside a glass block that has an index of refraction of 1.46. As this light arrives at the surface of the block, it makes an angle of 53 degrees with the normal. At what angle with the normal in the air will it leave the block?

Answer 1

To solve this problem it is necessary to apply the concepts related to Snell's Law. For this case we know that the internal reflection would be subject to the attempt of the ray of light to pass from a high refractive index to a lower one, therefore, mathematically we have to,

$n_1 sin\theta_1 =n_2sin\theta_2$

$n_{1,2}$ = Refractive Index

$\theta_1$ = Incident angel

$\theta_2$= Refractive angel

Replacing with our values and solving to find \theta_2

$\theta_2 = sin^{-1}(\frac{n_2}{n_1}*sin\theta_1)$

Replacing our values we have that,

$\theta_2 = sin^{-1}(\frac{1.46}{1}*sin(53))$

$\theta_2 = sin^{-1}(1.166)$

Because the angle of refraction is greater than one, there is no possible solution for the ArcSin of the value. This only indicates that the light beam cannot leave the block.