Question

A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5) at an angle of incidence of 39.8°. (a) Trace the path of the light ray through the glass, and find the angles of incidence and refraction at each surface.

Answer 1

a

The path of light ray through the glass is shown on the first uploaded image

First surface:

Angle of incidence is
`i_1 = 39.0^o`

Angle of refraction is
`r_1= 25.23^o`

Second surface:

Angle of incidence is
`i_2 = 34.77^o`

Angle of refraction is
`r_2= 58.80^o`

b

Since the angle of incidence is equal to the angle of reflection

Then at the first surface the angle of reflection is
`R = 39.8^o`

And at the first surface the angle of reflection is
`R_2 = 34.77^o`

Step-by-step explanation:

From the question we are told that

The angle of incidence is
`i_1 = 39.0^o`

The refractive index of the prism is
`n_(p) = 1.5`

The angle of the prism is
`A = 60^o`

The path of light ray through the glass is shown on the first uploaded image

For the first surface of the prism

According to Snell's law

`n_(air) sin( i_1) = n_(p) sin( r_1)`

The refractive index of air
`n_(air)` has a constant value of 1

Now making the angle of refraction at the first surface of the prism
`r_1` the subject

`r_1 = sin^(-1)[(sin (i_1) )/(n_(p)) ]`

`= sin^(-1)[(sin(39.8))/(1.5) ]`

`r_1= 25.23^o`

For the second surface of the prism

looking at the diagram on the first uploaded image the angle of incidence is mathematically evaluated as

`i_2 = A - r_1`

Substituting values

`i_2 = 60 -25.23`

`=34.77^o`

According to Snell's law

`n_(p) sin( i_2) = n_(air) sin( r_2)`

Now making the angle of refraction at the second surface of the prism
`r_2` the subject

`r_2 = sin^(-1) [(n_p sin(i_2))/(n_air) ]`

Substituting values into the equation

`r_2 = sin^(-1) [(1.5 * sin(34.77))/(1)]`

`r_2 =58.8^o`