Angie of deviation produced by prism: Let light ray PQ is incident on one of the refracting faces of the prism. The angles of incidence and refraction at the first face AB are i1 and r1. The path of the light inside the prism is QR. The angle of incidence and refraction at the second face AC is r2 and i2 respectively. RS is the ray emerging from p the second face. Angle i2 is also called angle of emergence. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
The two normals drawn at the point of incidence Q and emergence R are QN and RN. They meet at point N. The incident ray and the emergent ray meet at a point M.
The deviation d at the surface AB is,
In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
∠A + ∠QNR = 180° …(5)
From the triangle ∆QNR,
r1 + r2 ∠QNR = 180° …(6)
Comparing these two equations (5) and (6) we get
r1 + r2 = A …(7)
Substituting this in equation (4) for angle of deviation,
d= i1 + i2 -A …(8)
Thus, the angle of deviation depends on the angle of incidence angle of emergence and the angle for the prism.
Refractive index of the material of the prism:
At minimum deviation,
i1 = i2 = i and r1 = r2 = r
Now, the equation (8) becomes,
D = i1 + i2 -A (or) i =
The equation (7) becomes,
r1 + r2 = A ⇒ 2r = A (or) r =
Substituting i and r in Snell’s law,
The above equation is used to find the refractive index of the material of the prism.
