Answer:
a
The path of light ray through the glass is shown on the first uploaded image
First surface:
Angle of incidence is [tex]i_1 = 39.0^o[/tex]
Angle of refraction is [tex]r_1= 25.23^o[/tex]
Second surface:
Angle of incidence is [tex]i_2 = 34.77^o[/tex]
Angle of refraction is [tex]r_2= 58.80^o[/tex]
b
Since the angle of incidence is equal to the angle of reflection
Then at the first surface the angle of reflection is [tex]R = 39.8^o[/tex]
And at the first surface the angle of reflection is [tex]R_2 = 34.77^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]i_1 = 39.0^o[/tex]
The refractive index of the prism is [tex]n_{p} = 1.5[/tex]
The angle of the prism is [tex]A = 60^o[/tex]
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
[tex]n_{air} sin( i_1) = n_{p} sin( r_1)[/tex]
The refractive index of air [tex]n_{air}[/tex] has a constant value of 1
Now making the angle of refraction at the first surface of the prism [tex]r_1[/tex] the subject
[tex]r_1 = sin^{-1}[\frac{sin (i_1) }{n_{p}} ][/tex]
[tex]= sin^{-1}[\frac{sin(39.8)}{1.5} ][/tex]
[tex]r_1= 25.23^o[/tex]
For the second surface of the prism
looking at the diagram on the first uploaded image the angle of incidence is mathematically evaluated as
[tex]i_2 = A - r_1[/tex]
Substituting values
[tex]i_2 = 60 -25.23[/tex]
[tex]=34.77^o[/tex]
According to Snell's law
[tex]n_{p} sin( i_2) = n_{air} sin( r_2)[/tex]
Now making the angle of refraction at the second surface of the prism [tex]r_2[/tex] the subject
[tex]r_2 = sin^{-1} [\frac{n_p sin(i_2)}{n_air} ][/tex]
Substituting values into the equation
[tex]r_2 = sin^{-1} [\frac{1.5 * sin(34.77)}{1}][/tex]
[tex]r_2 =58.8^o[/tex]
