WAEC Physics 2013 Paper

Concept:The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90^\circ$.Explanation:Apply Snell’s law at the glass-air interface.For critical angle $C$, the refracted ray grazes the surface, so the angle of refraction is $90^\circ$.Thus, $n \sin C = 1 \times \sin 90^\circ = 1$.Given refractive index $n = 1.5$, we get:$$\sin C = \frac{1}{n} = \frac{1}{1.5} = 0.6667$$Now find the inverse sine:$$C = \sin^{-1}(0.6667) \approx 41.8^\circ$$Rounding to the nearest whole degree gives $C \approx 42^\circ$.Answer:The correct option is D. $42^\circ$.