# Wondering about angle of incidence [closed]

Hi I was wondering if someone could help me please find the angle of incidence of light passing through a glass slab with a refractive index of 1.52 and undergoing one TIR at an angle of 42 and emerging out of the slab.Here is a visual of the problem I am trying to find a solution to

Here is a visual of the problem I am trying to solve

Is there a way to find this angle out? I have tried snells law but I keep getting an error. Thank you!

Since there are no arrows, how the light propagates, I assume you want to know the angle marked with a ? on the left.

Since we know that the sum of all 3 angles within a plane triangle is $$180°$$, we can infer the missing angle of the imaginative triangle within the glass slab with identifying the 90° angle on the upper edge:

$$180° - 90°- 42° = 48°$$

Now we can infer the angle to the complementary, perpenticular (dotted) line to the slot as $$90° - 48° = 42°$$. With is consistent with our understanding that the original angle and this outgoing angle are 'similar angles'.

Next, assuming the outer environment is air and hence $$n_{air} \approx 1$$, we can calculate the incident angle through Snell's law:

$$\phi_{inc} = arcsin \left(n_{glass} \, sin(\phi_{out}) \right) = arcsin\left( 1.52 \cdot sin( 42° ) \right) = \ldots$$

which gives you an error, because the inclination angle is too large and the refractive index of the glass with $$1.52$$ is unison also too high for refraction to occur in the first place. Rather what happens, is total internal reflection.