Question

(8%) 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 e, - 33", measured from the normal to the interface Randomized Variables = 33 2.0 m d20m 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, - 1 for air and - 1.33 for water). Grade Summary Daductions 096 Potential 10096 ET sin cotan atan cosh Submissions Attempts remaining: 10 (3 per attempt diledi d co10 tan asino acos acotano sinho tanh cotanho) Degrees O Radians 4 S6 1 2 3 0 VO Submit Hints: 09 deduction per hint. Hints remaining 3 Feedback 02 deduction per fondback 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.

Answer 1

Na=1, nw=133 Oa=33 h=? now sina = na sine sin r = Nw sin 33 = 1.33 sin 33 na 9 = sin' 07243) = 46.41° il oa 2m. 2m m . In EX 2-90°-46.4 = 43,599 now tanah tom 43.59° = h 2 2 h=2x0.9519 = 1.gom [h=igom a) b) tano a = x 2 x X= 2too 33 x = 1.298m as given image & diver both have the same hostizontal distance along the surface of the water. So, image is also am away from normal pa n toma=h' Xtand h' 1 = | 298 tom 43. S9 = 23S

h=1.235m