Angle of deviation `(delta)` by a prism (refractive index = `mu` and supposing the angle of prism A to be small) can be given by

Minimum deviation of prism having refractive index mu and small angle of prism A is shown by .....

Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then the angle of prism is ...... (sin 48^@36. = 0.75)

Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then, the angle is prism is…. (sin 48^(@)36'=0.75)

Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then, the angle is prism is…. (sin 48^(@)36'=0.75)

If delta_min is the minimum deviation through a prism and mu is refractive index and A is angle of prism