Question

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 .

Answer 1

22.916136degrees.

Step-by-step explanation:

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

`n_1sin(\alpha )=n_2sin(\beta )`

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

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