A light ray in medium 1 of refractive index 1 travels toward a horizontal interface along the vector AO = 2î-3ĵ. It refracts into medium 2 of refractive index 1.5 and emerges along OB = Cî-4ĵ, as shown in the figure. What is C?

Question

G1 Physics · Accepted Answer

The interface is horizontal, so the normal is vertical. For the incident ray, AO = 2î-3ĵ. Its magnitude is |AO| = √(2²+3²) = √13. The sine of the incidence angle is the ratio of the component parallel to the interface to the vector magnitude: sinα = 2 ÷ √13. For the refracted ray, OB = Cî-4ĵ. Its magnitude is |OB| = √(C²+16). Hence, sinβ = C ÷ √(C²+16). Apply Snell's law: n₁ sinα = n₂ sinβ. Therefore, 1×[2 ÷ √13] = 1.5×[C ÷ √(C²+16)]. Square both sides: 4 ÷ 13 = 2.25C² ÷ (C²+16). Write 2.25 as…