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You shine a flashlight on a pond. The light hits the pond at an angle of 70° to the normal of the pond. What is angle between the refracted ray entering the pond and the surface normal of the pond? The index of refraction of water is 4/3 Select One of the following: (a) 86.99 (b) 23.4° (c) 35 (d) 44.8 (e) Toby

Answer :

Answer:

(d)  44.8°

Explanation:

For refraction,  

Using Snell's law as:

[tex]\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}[/tex]

Where,  

Θ₁ is the angle of incidence  

Θ₂ is the angle of refraction

n₁ is the refractive index of air which is 1

n₂ is the refractive index of water which is 4/3

Given that angle of incidence = 70°

So,  

[tex]\frac {sin\theta_2}{sin70}=\frac {1}{\frac {4}{3}}[/tex]

[tex]{sin\theta_2}=0.7048[/tex]

Angle of refraction = sin⁻¹ 0.7048 = 44.8°.

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