Light exits a Sylgard 184 silicone elastomer, index of refraction = 1.4, into air, index of refraction = 1, with an angle of incidence of 30°. What is the angle of refraction in the air on the other side of the Sylgard/air interface?

a) 30°

b) 45°

c) 60°

d) No refraction, the light is totally reflected internally.

Explanation

This problem is solved using Snell’s law: η₁sinθ₁ = η₂sinθ₂. On the MCAT, if you are asked to perform a Snell’s law problem, the angles will be angles with which you are familiar: think 30/60/90 triangles and 45/45/90 triangles. Sin30º=1/2, and 1.4 is roughly √2, therefore the right side of the equation is √2/2. The left side of the equation is sinθ₁. The angle for which the sin is √2/2 is 45°.

Total internal reflection occurs when the angle of refraction exceeds 90°, and sin90° = 1. For this problem, this would occur at an incident angle of 1.4sinθ_i = 1sin90°. Sinθ_i = 1/1.4 = 0.71. θ_i for total internal reflection is 45°.

a) 30°, incorrect, At an interface between two different indices of refraction the angle of refraction cannot equal the angle of incidence.

b) 45°, correct.

c) 60°, incorrect.

d) No refraction, the light is totally reflected internally. Incorrect, see explanation above.

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