Due DatE!8/972018 11:39:00 PM End Date: 8/9/2018 11:59.00 PM (10%) Problem 5: Co

Question

Due DatE!8/972018 11:39:00 PM End Date: 8/9/2018 11:59.00 PM (10%) Problem 5: 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 0 22°, measured from the normal to the interface. Randomized Variables ?.-22? 2.0m -2.0m Otheexpertta.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 a 1 for air and n 1.33 for water). > X 50% Part (b) Fin 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= 2.611 100%

Explanation / Answer

a) using snells law

nw*sin(i) = na*sin(r)

1/33sin22 = 1sin(r)

r = 29.88 degrees

height of the instuctor

2/h = tan(29.88)

h = 3.478 m

b) h' = 2*tan(22)/tan(29.88)

h' = 1.41 m

the apperent depth for the diver head is 1.41 m

h app = 3.478-1.41 = 2.06 m

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