In figure shown,what is the smallest incident angle $\theta$_i for the beam to undergo total internal reflection at surface S of prism having an index of refraction of 1.5?

Question

Quizplus · Accepted Answer

Using Snell's Law at the surface S, we find that sin(F1F1F1S1θ) = (1/1.5)sin(F1F1F10S1U1B0) = sin(40°). To find the smallest incident angle for total internal reflection, we need to find the critical angle at surface S, which is the angle of incidence that results in an angle of refraction of 90°. Using Snell's Law again, we can set sin(F1F1F1S1θ_c) = (1/1.5)sin(90°) = 1/1.5. Solving for θ_c, we get θ_c = sin^(-1)(1/1.5) = 41.8°. Therefore, the smallest incident angle for total internal reflection is slightly less than 41.8°, and the only answer choice that is below this value is B, which corresponds to an incident angle of 33.9°.

In Figure Shown,what Is the Smallest Incident Angle θ\thetaθ i for the Beam to Undergo Total Internal Reflection at for the Beam

In Figure Shown,what Is the Smallest Incident Angle $\theta$ _i for the Beam to Undergo Total Internal Reflection at for the Beam