Relation between Refractive index (m) Angle of Prism (A) and angle of deviation (d)
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Draw LO and MO at L and M respectively. Extend KL and MN to meet at P.



i.e. d = i1 r1 + i2 r2
d = i1 + i2 (r1 + r2)
(exterior angle is sum of interior opposite angles)


In the quadrilateral ALOM





(From Snell's law)
As the angle of incidence is increased, angle of deviation 'd' decreases and reaches minimum value. If the angle of incidence is further increased, the angle of deviation is increased. Let dm be the angle of minimum deviation. The refracted ray in the prism in that case will be parallel to the base.

For minimum deviation position the incident ray and emergent rays are symmetrical with respect to the refracting surface and LM is parallel to BC.
i1 = i2 = i
and r1 = r2 = r
2r1 = A , r1= A/2
dm = 2i1 - 2r1
dm = 2i1 - A
or 2i1 = dm + A

From Snell's Law


This is the Prism formula when the prism is in the minimum deviation position.
For thin prism A is very small and if the light is incident at a small angle then i1, r1, r2, i2 are so small.
That d= (m-1) A