AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Consider the figure of equilateral prism.According to Snell's law for incident ray is.

sin‌i

sin‌r1

=√2
Substitute the value in above expression we get.

sin‌0

sin‌r1

=√2
sin‌r1=0
r1=0
According to Snell's law for emitted ray is.

sin‌r2

sin‌e

=

1

µ

sin‌e=√2‌sin‌r2
The angles of an equilateral triangle is 60°.
sin‌e=√2‌sin‌60°
sin‌e=√2×

√3

2

>1
Thus, the light is incidenting is more than critical angle and total internally reflected.
The angle of deviation is given as,
δd=(i+e)−(r1+r2)
Substitute the valuc in above expression.
δd=0−60°=−60°
|δd|=60°