As we know that Deviation , δ = i_{1} + i_{2} – A

From (i ~ δ) graph we conclude that deviation δ is minimum when, ray of light passes symmetrically through the prism , i_{1} = i_{2} = i (say)

⇒ δ_{m} = 2i − A

⇒ δ_{m} + A = 2 i

$ \displaystyle i = \frac{\delta_m +A}{2}$ …(i)

Since i_{1} = i_{2}

⇒ r_{1} = r_{2} = r (say)

r_{1} + r_{2} = A

2r = A

⇒ r = A/2 …(ii)

Putting i & r in terms of A and δ_{m} in Snell’s Law

$ \displaystyle n = \frac{sini}{sin r} $

we obtain an expression for minimum deviation δ_{m} as

$ \displaystyle n = \frac{sin(\delta_m + A)/2}{sin A/2} $

Note : If the angle of prism is very small, sin A ≈ A

⇒ δ_{m} = (n −1)A

This equation is valid only when the refracting angle of the prism is very small.

Since prisms generally are made of optically denser medium w.r.t. air, can the phenomenon of total internal reflection take place over here ?

The phenomenon of total internal internal reflection can also be observed in case of prisms. This can be understood by the following cases.

When the angle of incidence for a prism is 90 degree then the ray just grazes along the surface of the prism . This is known as grazing incidence and in this case the angle of refraction at the first refracting surface is equal to the critical angle of the material of the prism.

Similarly, when the angle of emergence is equal to 90 degree (i.e. ray is incident at the critical angle at the second refracting face) then the case is of grazing emergence.

Both, in case of grazing emergence or grazing incidence, the angle of deviation for the prism is maximum.

Recall: The wavelength is inverse the proportional to the refractive index.