Question

A beam of light is incident from air on the surface of a liquid. If the angle of incidence is 25.5° and the angle of refraction is 15.6°, find the critical angle for the liquid when surrounded by air.

Answer 1

Фc = 38.66°

Step-by-step explanation:

from Snell's law:

n1×sin(Ф1) = n2×sin(Ф2)

for air, n1 = 1.0

1.0×sin(Ф1) = n2×sin(Ф2)

n2 = 1.0×sin(Ф1)/sin(Ф2)

n2 = 1.0×sin(25.5)/sin(15.6) = 1.601

For the critical angle: Ф2 = 90° and n2 = 1.601

n2×sin(Фc) = n1×sin(Ф2) = n1

sin(Фc) = n1/(n2×sin(Ф2))

sin(Фc) = (1.0)/(1.601×sin(90))

sin(Фc) = 0.625

Фc = 38.66°

Therefore, the critical angle is 38.66°