Question

In searching the bottom of a pool at night, a watchman shines a narrow beam of light, 1.5 m above the water level, onto the surface of the water at a point 2.5 m from his foot at the edge of the pool. 1.5 m 01 2.5 m 2 m a) (10 Points) Calculate the angle of incidence, 01 (Submit a PDF file with a maximum size of 8 MB) Choose File no file selected b) (10 Points) Calculate the angle of refraction, 02 (Submit a PDF file with a maximum size of 8 MB) Choose File no file selected c) (5 Points) Where does the spot of light hit the bottom of the 2-meter deep pool? Measure from the bottom of the wall beneath his foot. (Submit a PDF file with a maximum size of 8 MB) Choose File no file selected

Answer 1

Answers :

Part (a) Answer : θ₁ = 59.04^o

Part (b) Answer : θ₂ = 40.14^o

Part (c) Answer : d = 3.789 m

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Solution :

Here the angle made by the incident ray with the surface will be :

$\theta =tan^{-1}\left [ \frac{1.5\ m}{2.5\ m} \right ]=30.96^o$

.

Therefore, Angle of incidence will be : θ₁ = 90^o - θ = 90^o - 30.96^o = 59.04^o

^.

And, Since refractive index of water is : n = 1.33

So, According to the Snell's Law : n_air sinθ₁ = n_water sinθ₂

∴ (1.00) sin(59.04) = (1.33) sinθ₂

∴ θ₂ = 40.14^o

^.

And, The distance where the spot of light hits the bottom will be :

d = (2.5 m) + (2 m) sinθ₂ = (2.5 m) + (2 m) sin(40.14)= 3.789 m