Question

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​

Answer 1

Answer:

i ≅ $28^{0}$

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 = $\frac{sin i}{sin r}$ ................ 1

where n is the refractive index of the prism.

Given that emergence = $90^{0}$

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

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

($90^{0}$ - r ) + ($90^{0}$ - $r^{I}$ ) =  $180^{0}$ - $60^{0}$

180° - (r + $r^{I}$) =  $180^{0}$ - $60^{0}$

r + $r^{I}$  = $60^{0}$

⇒  r = $60^{0}$ - $r^{I}$  ........................ 3

The refractive index of the equilateral prism = 1.5.

Applying Snell's law to the refracting surface,

 $\frac{sinr^{I} }{sin e}$ = $\frac{1}{n}$

$\frac{sinr^{I} }{sin 90^{0} }$ = $\frac{1}{1.5}$

⇒   $r^{I}$  = $41.81^{0}$

From equation 3,

r = $60^{0}$ - $r^{I}$

r = $60^{0}$ - $41.81^{0}$

r = $18.19^{0}$

So that ;

n = $\frac{sin i}{sin r}$

1.5 = $\frac{sin i}{sin18.19^{0} }$

sin i = 0.4683

i = $27.92^{0}$ ≅ $28^{0}$