# A ray in glass arrives at the glass-water interface at an angle of 48° with the normal. The refracted ray, in water, makes a 72° angle with the normal. The index of refraction of water is 1.33. Then the ray in glass is redirected so its new angle of incidence is 37°. What is the new angle of refraction in the water? Show all work.

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The new angle of refraction in the water is [tex]50.35^{\circ}[/tex].Given data:The angle made by ray at glass-water interface is, [tex]\theta _{1} = 48^{\circ}[/tex].The angle made by the refracted ray with the normal is, [tex]\theta_{2} = 72^{\circ}[/tex].The index of refraction of water is, [tex]n=1.33[/tex].The angle of incidence for the redirected glass is, [tex]\theta_{3} = 37^{\circ}[/tex].The entire problem is based on the concepts of Snell's law, which says that the ratio of sine of angle of incidence to sine of angle of refraction is equal to the ratio of refractive index and incident index.So on applying the Snell's law as,[tex]n' \times sin \theta_{1} = n \times sin \theta_{2}[/tex]Here, n' is the index of refraction of glass.Solving as,[tex]n' \times sin 48 = 1.33 \times sin 72\\\\n' = \dfrac{ 1.33 \times sin 72}{sin 48} \\\\n' =1.702[/tex]For redirected condition, again apply the Snell' law as,[tex]n' \times sin \theta_{3} = n \times sin \theta_{4}[/tex]Here, [tex]\theta_{4}[/tex] is the new angle of refraction in the water.Solving as, [tex]1.702 \times sin 37 = 1.33 \times sin \theta_{4}\\\\sin \theta_{4} = \dfrac{1.702 \times sin 37}{1.33} \\\\\theta_{4} =sin^{-1}(0.770)\\\\\theta_{4} =50.35^{\circ}[/tex]Thus, we can conclude that the new angle of refraction in the water is [tex]50.35^{\circ}[/tex].Learn more about the Snell's law here:https://brainly.com/question/24181353

Answer:50.4°Explanation:Snell's law states:n₁ sin θ₁ = n₂ sin θ₂where n is the index of refraction and θ is the angle of incidence (relative to the normal).When θ₁ = 48°:n sin 48° = 1.33 sin 72°n = 1.702When θ₁ = 37°:1.702 sin 37° = 1.33 sin θθ = 50.4°