derive an expression for the angle of deavition for aray of light passing through an equatral prism of reflax index angleA
Answers
Given, A = 60
0
μ=1.732
Since, angle of minimum deviation is given by,
μ=
sin
2
A
sin
2
A+δm
$$1.732 \times \frac{1}{2} = sin ( 30 + \frac{\delta m}{2})$$
sin
−1
(0.8666)=30+
2
δm
60
0
=30
2
δm
δm=60
0
Now,
δm=i+i
′
−A
60
0
=i+i
′
−60
0
( δ=60
0
minimum deviation )
= > i = 60
0
.
so, the angle of incidence must be 60
0
.
Answer:
Monochromatic light enters an equilateral prism made of glass in such a way that after refraction at the first surface light travels parallel to the base of the prism and is refracted on the second surface. This implies that light is refracted twice.
The angle of deviation as indicted in the image diagram attached is the angle through which the emergent ray deviates from the direction of the incident ray.
In the image you will find an explanation of the movement of these rays and the derivation of the formulae for getting the refractive index.
