A fisherman spots a fish underneath the water. It appears that the fish is d_0 =

Question

A fisherman spots a fish underneath the water. It appears that the fish is d_0 = 0.54 m under the water surface at an angle of theta_a = 52 degrees with respect to the normal to the surface of the water. The index of refraction of water is n_w =1.3 and the index of refraction of air is n_a = 1. Randomized Variables d_o = 0.54 m theta_a = 52 degrees The perpendicular distance from the fish to the normal of the water surface is L. Express L in terms of tan theta_a and d_0. L = ________ Solve for the numerical value of L, in meters Express the sine of the angle theta_w, in terms of theta_a, n_w, and n_a. Solve for the numerical value of theta_w in degrees. Express the real depth of the fish, d, in terms of L and tan theta_w. Solve for the numerical value of d, in meters.

Explanation / Answer

a) from the figure,

tan(theta_a) = L/do

==> L = do*tan(theta_a)

b)
L = 0.54*tan(52)

= 0.69 m

c)
Use Snell's law

sin(theta_w)/sin(theta_a) = na/nw

sin(theta_w) = sin(theta_a)*(na/nw)

d)

sin(theta_w) = sin(52)*1/1.3

= 0.606

theta_w = sin^-1(0.606)

= 37 degrees

e) tan(theta_w) = L/d

==> d = L/tan(theta_w)

f) d = 0.69/tan(37)

= 0.916 m

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