Question

Problem 3:  Consider the situation in the figure - a student diver in a pool and an instructor on the edge, outside the water. Since the indices of refraction of air and water are different, the light rays coming from the diver and the instructor are refracted at the surface, changing their apparent position with respect to each other. The diver sees the instructor at an apparent angle of θ_a = 26°, measured from the normal to the interface.

Randomized Variables:  θ_a = 26°

Part (a) Find the height of the instructor’s head above the water in meters, noting that you will first have to calculate the angle of incidence (you can take the indices of refraction to be n_a = 1 for air and n_w = 1.33 for water).
Numeric   : A numeric value is expected and not an expression.
h = __________________________________________


Part (b) Find the apparent depth of the diver’s head below water as seen by the instructor in meters, assuming the diver and his image both have the same horizontal distance along the surface of the water.
Numeric   : A numeric value is expected and not an expression.
x = __________________________________________

-2.0 m d=2.0 ml

Answer 1

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air (a) from the snells relation sin() nwater 1.33 sin() 1.0 o = angles of incidence of light ray from instructor head in air medium 0 = angle of refraction of light rayinto water 1.33 =>sin(0)=; sin(@) 1.0 we plug the know values in the relation gives =35.66° if the actual height of theinstructor head =h h tan( 90 - )= - we plug the know values in the relation gives 2.0 m 1 h=2.787 m (b) if the actual depth of diver head = d and apparent depth of diver head = X d 1.33 na 1.0 2.Om -= 1.504 m 1.33 IL = = =>X=