After a long day of meetings a business woman takes a late-night swim in the mot

Question

After a long day of meetings a business woman takes a late-night swim in the motel swimming pool. When she returns to her room, she realizes she has lost the room key in the pool. She borrows a powerful flashlight and walks around the pool, shining the light into the pool. The light shines on the key, which is lying on the bottom of the pool, when the flashlight is held 1.2 meters above the water surface and is directed at the suface 1.5 meters from the edge of the pool.

a) knowing that the index refraction for water is 1.33, and that the water level in the pool is exactly 4.0 meters, how far is the key from the edge of the pool???

b) What is the speed that the light beam is traveling while it is in the water?

Explanation / Answer

Data: Height of the flash light = 1.2 m Distance of focus = 1.5 m Refractive index of water = 1.33 Height of water level = 4 m Let i = angle of incidence,
r = angle of reflection

Solution:
tan i = 1.5/1.2         = 5/4
Hence, sin i = 0.78

Refractive index of water = 1.33
So, sin i/sin r = 1.33
      0.78/sin r = 1.33
               sin r = 0.78/1.33                       r = 35.9 deg
               tan r = 0.724 --------------(1)
Let x = horizontal distance of key from the point at       which the light is incident on water surface
Then, tan r = x/depth of water tan r = x/4 ----------------------------(2)
From (1) and (2)
x/4 = 0.724
x = 4 * 0.724 = 2.9 m
Horizontal distance of key = Horizontal distance of incidence of light + x
= 1.5 + 2.9 = 4.4 m Ans: Horizontal distance = 4.4 m (b) Speed of light in water = Speed of light in air / refractive index                                         = 3 x 10^8 / 1.33                                         = 2.25 x 10^8 m/s Ans: Speed of light in water = 2.25 x 10^8 m/s

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