9a) An opaque cylindrical tank with an open top has a diameter of 3.00 m and is
ID: 1401640 • Letter: 9
Question
9a) An opaque cylindrical tank with an open top has a diameter of 3.00 m and is completely filled with water. When the afternoon Sun reaches an angle of 28.0 degree above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank? (Hint: make a drawing) Ans. 3.39 m 9b(Variation). The same tank of the previous problem (diam = 3.00 m; h = 3.39 m) has now no water. Find the angle above the horizon below which the afternoon Sun cannot reach the bottom of the tank at all.Explanation / Answer
9a)from the given data,
angle of incidence, theta_i = 90 - 28 = 62 degrees
let theta_r is the angle of regraction.
n1 = 1(for air)
n2 = 1.33(for water)
Apply snell's law
n1*sin(theta_i) = n2*sin(theta_r)
1*sin(62) = 1.33*sin(theta_r)
sin(theta_r) = 0.664
theta_r = sin^-1(0.664)
= 41.6 degrees
let h is the depth.
tan(41.6) = d/h
==> h = d/tan(41.6)
= 3/tan(41.6)
= 3.38 m <<<<<<<-----Answer
9b)
angle above the horizonatal, theta = tan^-1(h/d)
= tan^-1(3.39/3)
= 48.5 degrees <<<<<<<-----Answer
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