Question

An opaque cylindrical tank with an open top has a diameter of 4.00 m and is completely filled with water. When the afternoon Sun reaches an angle of 26.5° above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank?

Answer 1

open top tank has a diameter D = 4.00 m

When the afternoon Sun reaches above the horizon the an angle of = 26.5°

Deep of the tank H = ?

Tan = H/D

Tan 26.5 = H/4.0

H = 4.0 xTan 26.5 = 1.99 m

If refractive index of the water is consider then

Illumination angel/ incident angle i = 26.5°

Refraction angle r = ?

Refractive index n = sini/sinr

But for water n = 1.33

Then sinr = sin26.5/1.33 = 0.3355

Now r = Sin-1 (0.3355) = 19.60

Therefore depth of water H = Dx Tanr = 4xTan 19.60 = 1.424 m

Deep of Tank id H = 1.424 m