Question Detail

First, find the angle of refraction in glass at the vertical face: n_air sin 39° = n_glass sin r. n_air = 1, so sin r = sin 39 / 1.489 = 0.6293/1.489 = 0.4226. r = 25°. The ray is then incident on the glass-liquid interface at X. Since total internal reflection occurs, the critical angle for glass-liquid is reached. The angle of incidence at X is the critical angle. From the geometry, the angle of incidence at X is 90° - r? Let's trace: The ray enters the vertical face and bends. At X, the ray is incident on the horizontal glass-liquid interface. The angle of incidence at X is the angle between the ray and the normal. If the ray inside the glass is at angle r to the vertical, then it makes an angle of (90° - r) with the horizontal normal. So i_c = 90° - 25° = 65°. The critical angle condition: n_liquid = n_glass sin i_c = 1.489 × sin 65° = 1.489 × 0.9063 = 1.349. This is close to 1.35. But option D is 1.35. Let's check the options: A:1.25, B:1.40, C:1.45, D:1.35. So D is 1.35.