A light ray propagates in Material 1 with index of refraction n 1 = 1.23 n1=1.23 , strikes an interface, then passes into Material 2 with index of refraction n 2 = 1.31 n2=1.31 . The angle of incidence at the interface is θ 1 = 24.5 θ1=24.5 °. Find the angle of refraction θ 2 θ2 .

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15-04-2020
Physics

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A light ray propagates in Material 1 with index of refraction n 1 = 1.23 n1=1.23 , strikes an interface, then passes into Material 2 with index of refraction n 2 = 1.31 n2=1.31 . The angle of incidence at the interface is θ 1 = 24.5 θ1=24.5 °. Find the angle of refraction θ 2 θ2 .

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amnamurad28748 amnamurad28748 · Answer 1 · 2020-04-19T04:22:46+02:00

Answer:

22.916136degrees.

Explanation:

Snell's Law relates the refractive indices of two materials and their incidence and reflected angles in following way.

[tex]n_1sin(\alpha )=n_2sin(\beta )[/tex]

Where left side is for incidence ray and the material that it is in and the right side is for refracted ray and its material .

given that

[tex]n_1 = 1.23[/tex] [tex]\beta = 24.501 degrees.[/tex] and [tex]n_2 = 1.31[/tex] substituting these in Snell's law formula and solving for unknown gives, [tex]\beta = 22.916136[/tex] degrees, which is our refracted angle.