Consider the situation in the figure- a student diver in a pool and an instructo

Question

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 28 degrees, measured from the normal to the interface.

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 na=1 for air and nw= 1.33 for water)

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

(13%) 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 ,-28°, measured from the normal to the interface Randomized Variables 2.0 m = 28 = 2.0 m ©theexpertta.com 50% 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 na-1 for air and nw-133 for water) × 50% 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 Grade Summary Deductions Potential x- 2.0l 4% 96% sinO cotan0 asinO acosO atan0 acotan) sinh0 cosh0 tanh0 conho Submissions Attempts remaining: 9 (4% per attempt) detailed view tan() | | ( 78 9HOME 4% 0 END Degrees O Radians O BACKSPACE DEL CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback

Explanation / Answer

theta-a = 28 deg ; na = 1 ; nw = 1.33

a)Let, angle of ref in air be theta-r and h be the height of the instructor.

Using Snell's law

nw sin(theta-a) = na sin(theta-r)

1.33 x sin28 = 1 sin(theta-r)

theta-r = sin -1(1.33 x 0.469) = 38.64 deg

2/h = tan(theta-r)

h = 2/tan(theta-r) = 2/tan38.64 = 2.5 m

Hence, h = 2.5 m

b)Apparent depth will be given by

H(app) = h - H'

H' = 2 x tan(theta-a)/tan(theta-r) = 2 x tan28/tan38.64 = 1.33

H(app) = 2.5 - 1.33 = 1.17 m

Hence, H(app) = 1.17 m

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