The angle of deviation through a prism is minimum when

(A) Incident ray and emergent ray are symmetric to the prism
(B) The refracted ray inside the prism becomes parallel to its base
(C) Angle of incidence is equal to that of the angle of emergence
(D) When angle of emergence is double the angle of incidence
Choose the correct answer from the options given below:

We need to determine the correct conditions for a prism under which the angle of deviation $$\delta$$ is at its minimum value.

The angle of deviation $$\delta$$ through a triangular prism depends on the angle of incidence $$i$$ and follows a characteristic curve that reaches a single minimum point, known as the minimum deviation position $$\delta_m$$.

At the position of minimum deviation, the light ray passes through the prism symmetrically. This symmetry leads to several key physical conditions:

1. The angle of incidence $$i$$ becomes exactly equal to the angle of emergence $$e$$: $$i = e$$. This satisfies statement (C).

2. Because the angles are equal at both faces ($$i = e$$ and consequently the internal refraction angles $$r_1 = r_2$$), the incident ray and the emergent ray are completely symmetric with respect to the prism. This satisfies statement (A).

3. For a standard principal section of an isosceles or equilateral prism, this internal geometric symmetry forces the refracted ray inside the prism to travel parallel to the base of the prism. This satisfies statement (B).

4. There is no physical law or condition stating that the angle of emergence must be double the angle of incidence ($$e = 2i$$) at minimum deviation; in fact, they must be equal ($$e = i$$). Thus, statement (D) is incorrect.

Since statements (A), (B), and (C) are true, the correct combination matching the criteria is represented by Option A.

Therefore, the correct answer is Option A: Statements (A), (B) and (C) are true.