The ray falls on the surface of the water at an angle of 40 ° at what angle should the ray

The ray falls on the surface of the water at an angle of 40 ° at what angle should the ray fall on the surface of the glass for the angle of refraction to be the same?

According to the law of refraction, the refractive index of the medium n is found by the formula n = sinα / sinβ, where α is the angle of incidence of the ray, β is the angle of refraction of this ray. From the condition of the problem it is known that the ray falls on the water surface at an angle α = 40 °, then the refractive index of water n (water) will be: n (water) = sinα / sinβ, we get sinβ = sinα / n (water). To find out at what angle φ the ray should fall on the glass surface so that the angle of refraction β turns out to be the same, we write the same law for the refractive index of glass n (glass) = sinφ / sinβ, then sinβ = sinφ / n (glass), hence, sinα / n (water) = sinφ / n (glass). Then sinφ = sinα ∙ (n (glass) / n (water)). Refractive indices of media are found in reference materials: n (glass) = 1.51; n (water) = 1.33. Substitute the values ​​of physical quantities in the formulas and make calculations: sinφ = sin40 ° ∙ (1.51 / 1.33); sinφ ≈ 0.73, we find φ = arcsin0.73, we obtain φ ≈ 47 °.
Answer: at an angle of ≈ 47 °, the beam must fall on the glass surface for the angle of refraction to be the same.