A ray of light incident on one face of equilateral glass prism is refracted in such a way that it emerges from opposite surface at an angle of 90 to the normal. Calculate the angle of incidence​

Respuesta :

Answer:

i ≅ [tex]28^{0}[/tex]

Explanation:

Let the angle of incidence be represented by i, angle of emergence be represented by e, and angle of refraction by r.

Snell's law states that;

n = [tex]\frac{sin i}{sin r}[/tex] ................ 1

where n is the refractive index of the prism.

Given that emergence = [tex]90^{0}[/tex]

But from a ray diagram for the given question, we have;

[tex]60^{0}[/tex] + ([tex]90^{0}[/tex] - r) + ([tex]90^{0}[/tex] - [tex]r^{I}[/tex]) = [tex]180^{0}[/tex] (sum of angles in a triangle) .................. 2

([tex]90^{0}[/tex] - r ) + ([tex]90^{0}[/tex] - [tex]r^{I}[/tex] ) =  [tex]180^{0}[/tex] - [tex]60^{0}[/tex]

180° - (r + [tex]r^{I}[/tex]) =  [tex]180^{0}[/tex] - [tex]60^{0}[/tex]

r + [tex]r^{I}[/tex]  = [tex]60^{0}[/tex]

⇒  r = [tex]60^{0}[/tex] - [tex]r^{I}[/tex]  ........................ 3

The refractive index of the equilateral prism = 1.5.

Applying Snell's law to the refracting surface,

 [tex]\frac{sinr^{I} }{sin e}[/tex] = [tex]\frac{1}{n}[/tex]

[tex]\frac{sinr^{I} }{sin 90^{0} }[/tex] = [tex]\frac{1}{1.5}[/tex]

⇒   [tex]r^{I}[/tex]  = [tex]41.81^{0}[/tex]

From equation 3,

r = [tex]60^{0}[/tex] - [tex]r^{I}[/tex]

r = [tex]60^{0}[/tex] - [tex]41.81^{0}[/tex]

r = [tex]18.19^{0}[/tex]

So that ;

n = [tex]\frac{sin i}{sin r}[/tex]

1.5 = [tex]\frac{sin i}{sin18.19^{0} }[/tex]

sin i = 0.4683

i = [tex]27.92^{0}[/tex] ≅ [tex]28^{0}[/tex]