a swimmer lights a torch under sea water light from the torch is incident on water surface in such a way that incident light makes an angle of 37 degrees with water surface. Find the angle of refraction if absolute refractive indices of water and air are 1.33 and 1.0 respectively.

Dear Student ,
Taking water as "medium 1" and air as "medium 2",we have

  η₁ = 1.33

  η₂ = 1.00

Angle of incidence, θ₁ = 37°

Angle of refraction, θ₂ = ?

Now, (sin θ₁/sin θ₂) = (η₂/η₁)

∴ (sin 37°/sin θ₂) = (1.00/1.33)

∴ sin θ₂ = [ (0.6015 x 1.33)/1.00]

∴ sin θ₂ = 0.8000

∴ θ₂ = 53° 

Hence, angle of refraction = 53°
Regards