The tip of a triangle is held 12.0 cm above the surface of a flat pool of water. A submerged swimmer in the pool sees the tip of the triangle at what distance above the water? Let the indices of refraction nwater = 1.33 and nair = 1.00.

Question

Physics

Katie Henson

5 October, 04:19

The tip of a triangle is held 12.0 cm above the surface of a flat pool of water. A submerged swimmer in the pool sees the tip of the triangle at what distance above the water? Let the indices of refraction nwater = 1.33 and nair = 1.00.

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Answers (1)

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Madeline Coleman · Answer 1

Answer: 9cm

Explanation:

Refractive index can also be defined as the ratio of the real depth to the apparent depth.

Given that the

Real depth = 12 m

Refractive index of water = 1.33

Refractive index of air = 1.00

nair/nwater = real depth/apparent depth

Substitute all the parameters into the formula

1.33/1 = 12 / apparent depth

Cross multiply

1.33 Apparent depth = 12

Apparent depth = 12/1.33

Apparent depth = 9.02 cm

Therefore, A submerged swimmer in the pool sees the tip of the triangle at 9cm approximately distance above the water.