A ray of light passes through as equilateral glass prism such that the refracted ray inside the prism is parallel to its base- How is the angle of emergence -i-2- related to the angle of incidence -i-1-

Question

Toppr Admin · Accepted Answer

In equilateral triangle all angles are of $60^{0}$
and it is given that refracted ray is parallel to base of triangle
then
$∠ r_{1} + ∠ r_{2} = 60^{0}$
and $∠ r_{1} = ∠ r_{2} = 30^{0}$
by snell's law
At first surface AB
$μ = \frac{s i n i_{1}}{s i n 30^{0}} (1)$

At surface AC
$\frac{1}{μ} = \frac{s i n 30^{0}}{s i n i_{2}} (2)$
by solving 1 and 2 we get
$∠ i_{1} = ∠ i_{2}$

A ray of light passes through as equilateral glass prism such that the refracted ray inside the prism is parallel to its base. How is the angle of emergence ′i′2 related to the angle of incidence ′i′1

In equilateral triangle all angles are of 600and it is given that refracted ray is parallel to base of triangle then ∠r1+∠r2=600and ∠r1=∠r2=300by snell's lawAt first surface ABμ=sini1sin300(1)At surface AC1μ=sin300sini2(2)by solving 1 and 2 we get∠i1=∠i2

A ray of light passes through as equilateral glass prism such that the refracted ray inside the prism is parallel to its base. How is the angle of emergence $^{'} i_{2}^{'}$ related to the angle of incidence $^{'} i_{1}^{'}$