The refractive index of a material is 1.5. Calculate the critical angle at the glass-air interface

Analyze the question: We need the critical angle for light travelling from glass (n = 1.5) into air (n = 1).

Concept: The critical angle, c, is the angle of incidence in the denser medium at which the refracted ray travels along the boundary (angle of refraction = 90°). It is given by:

sin c = 1/n

(Here n is the refractive index of the denser medium relative to air.)

Solution:

sin c = 1/1.5 = 0.6667

c = sin⁻¹(0.6667)

Since sin 41.8° ≈ 0.6665, we get:

c ≈ 41.8° ≈ 42°

Check against the options:

• A (19°) and B (21°) are too small.
• C (39°) would give sin 39° ≈ 0.629, so n ≈ 1.59. That is not 1.5.
• D (42°) gives sin 42° ≈ 0.669, so n ≈ 1.49, which matches 1.5.

Tip: Remember that c = sin⁻¹(1/n). For common materials:

• Glass (n = 1.5) has c ≈ 42°
• Water (n = 1.33) has c ≈ 49°
• Diamond (n = 2.42) has c ≈ 24°

A higher refractive index means a smaller critical angle. If you can recall that glass is about 42°, you can answer this type of question quickly.