Q.156. Refraction of Light

Question 156.
Ryan Baloney stands beside his swimming pool completely full of water. (the pool, not Ryan!) (Refer to Fig.)
His eyes are located 1.6 m directly above the edge as shown, and he looks directly across the 7.00 m wide pool. Light from the bottom corner opposite him arrives at his eye in such a direction that the apparent depth of the water, h, is 0.30 m. If the index of refraction for water is 4/3 calculate H, the actual depth of the pool. Answer in m.

Answer 156.
Refr to the following figure for the additional construction of refracted ray from the bottom corner of the pool.

Using the property of similar triangles,
h / x = 1.6 / (7 - x)
=> (0.30) (7 - x) = 1.6x
=> 1.9 x = 2.1
=> x = (2.1) / (1.9) m

sin i = x / √(x^2 + h^2)

sinr = x / √(x^2 + H^2)

sini / sinr = 4/3 = √(x^2 + H^2) / √(x^2 + h^2)

=> 16 (x^2 + h^2) = 9 (x^2 + H^2)

=> 9H^2 = 7x^2 + 16h^2

=> H

= (1/3) √(7x^2 + 16h^2)

= (1/3) √[7 * (2.1/1.9)^2 + 16 * (0.3)^2]

= (1/3) √(8.5512 + 1.44)

= 1.054 m.

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