The minimum angle of deviation, δ in a prism occurs when the incident angle is such that the angle of refraction is at its maximum possible value. This typically happens when the incident angle is equal to the angle of minimum deviation.

As seen in the above image, the angle of deviation is represented by δ.

In the △MPQ, by exterior angle theorem, we get,

where i₂ = e is the emergent angle or the angle of transmission

In △QNP, ∠QNP + r₁ + r₂ = 180^∘

Let ∠QNP be equal to θ. Thus, above equation becomes θ + r₁ + r₂ = 180^∘ … (2)

In quadrilateral APNQ we get,

∠A + 90^∘ + θ + 90^∘ = 360^∘

∠A + θ = 180^∘ … (3)

Combining the equations 2 and 3 we get,

r₁ + r₂ + (180^∘ – A) = 180^∘

or, r₁ + r₂ = A

Placing this value in equation 1 we get,

This is the angle of minimum deviation. Here i₁ = i is the angle of incidence, A is the angle of prism,  and i₂ = e is the angle of emergence.