A ray of light is incident on a prism in the minimum deviation position as shown and it suffers a deviation of 40 degree. what will be the deviation suffered if the colored half of prism is removed?

The answer to this question is 20°
Solution:
∂(min)=2i-A(which is the initial prism angle)
r₁+r₂  = A
So    40= 2i-A-------------------------(1)
           i =20+(A/2)
Now after the colored side has been removed light will emerge out parallel to the base of the prism after refraction from 1st face
Therefore r₂=0 -------->    r₁=A/2
Now new prism angle is A/2
therefore
∂(new) = i₁ + i₂  –  (A/2)
              = i + 0 – (A/2)
             =20 +(A/2) – (A/2)
             = 20° Which is the new angle of deviation
Hope it helps..
Regards,
Kenny Toijam