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A person stands at height 10 m on the bank of a river 50 m wide. On the opposite bank stands a brightly lit tower 40 m tall. Looking through a polarizer at light from the tower that reflects off the river surface, the person finds the intensity is least (fully polarized) for light reflecting from a point a horizontal distance X from his building, and this corresponds to bulbs at height Y on the tower. Given the refractive index of water = 4/3, the values of X and Y are approximately:

1. 13 m, 27 m
2. 22 m, 13 m
3. 25 m, 10 m
4. 17 m, 29 m

Correct answer: 13 m, 27 m

Solution

At Brewster's angle tan(theta_B) = 4/3, so the angle from the vertical (normal to water) is theta_B with tan(theta_B) = 4/3. The observer's eye is 10 m up; the ray from the reflection point to the eye makes angle theta_B with the vertical, giving horizontal distance X = 10*tan(theta_B) = 10*(4/3) = 13.3 ~ 13 m. The reflected ray then travels to the tower at the same Brewster angle over the remaining horizontal distance (50 - 13.3 = 36.7 m): height risen = 36.7/tan(theta_B)... using the incidence geometry gives Y ~ 27 m. Thus X ~ 13 m, Y ~ 27 m.

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