A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. In the water what angle does the beam make with the surface

θ₂ = 40.5º

Explanation:

For this exercise we must use the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for the incident ray and index 2 is for the refracted ray

in this case the incident ray has an angle of θ₁ = 60º and the refractive index of the water is

          n₂ = 1,333

        sin θ₂ =

let's calculate

        sin  θ₂ = 1 / 1.3333 sin 60

        sin θ₂ = 0.64968

        θ₂ = sin⁻¹ (0.64968)

        θ₂ = 40.5º

i already answered it by 34/s

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