A fisherman, 1.8m tall, stands at the edge of a lake, being watched by a suspici

Question

A fisherman, 1.8m tall, stands at the edge of a lake, being watched by a suspicious trout who is some distance from the fisherman in the horizontal direction and a distance 47cm below the surface of the water. Use 1.35 as the index of refraction for water.

What is the angle of incidence of the light from the top of the fisherman's head on the surface of the water if the light hits the water a horizontal distance 4.21m from the shore?

66.9

What is the angle of transmission of the light ray into the water?

42.9

At what angle from the vertical does the fish see the top of the fisherman

66.9

Explanation / Answer

angle of incidence is the angle to the perpendicular of a surface at which light enters into a surface.
here imagine a right angle triangle consisting of
4.21 m horizontal distance and 1.8 m vertical distance
then angle with the horizontal surface=arctan(1.8/4.21)=23.15 degrees

hence angle with vertical=90-23.15=66.85 degrees

b)if angle of transmission is beta,
using snell's law:

1/1.35=sin(beta)/sin(66.85)

beta=42.93 degree

c)now as the transmitted ray entered into water at an angle 42.93 degrees with vertical,

and the fish is 47 cm below,

to see the fisherman, fish should be at a distance 47/tan(42.93)=50.53 cm horizontally from the point where the light touches the surface.

now the angle with horizontal at which the fish will see the fisherman let be alpha.

then tan(alpha)=(1.8+0.47)/(0.5053)

alpha=77.45 degrees

so angle with vertical =90-77.45=12.55 degrees

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