We discuss the options as follows:

$$\bullet$$ For option (A) : For the angle of incidence i₁ = A, we have minimum deviation and the ray inside the prism is parallel to the base of the prism.

Hence, option (A) is correct.

$$\bullet$$ For option (C) : The angle of deviation is given as

$$\delta$$ = i₁ + i₂ $$-$$ A

At minimum deviation, i₁ = i₂ and r₁ = r₂. Therefore,

i₁ = A ....... (1)

Also, we know that

r₁ + r₂ = A

As r₁ = r₂, we get

2r₁ = A $$\Rightarrow$$ r₁ = $${A \over 2}$$

From Eq. (1), we get

$${r_1} = {{{i_1}} \over 2}$$

Hence, option (C) is correct.

According to Snell's law, we have

$${{\sin {i_2}} \over {\sin {r_2}}} = \mu $$

For emergent ray to be tangential to the surface, we have

i₂ = 90$$^\circ$$

$$\Rightarrow$$ sin90$$^\circ$$ = $$\mu$$sinr₂ $$\Rightarrow$$ $$\mu$$sinr₂ = 1

sinr₂ = $${1 \over \mu }$$ $$\Rightarrow$$ r₂ = sin^$$-$$1$$\left( {{1 \over \mu }} \right)$$ ....... (2)

From the relation r₁ + r₂ = A, we get

r₁ = A $$-$$ r₂

$$\Rightarrow$$ sinr₁ = sin(A $$-$$ r₂)

Using sin(a $$-$$ b) = sinacosb $$-$$ sinbcosa, we have

$$\Rightarrow$$ sinr₁ = sinAcosr₂ $$-$$ sinr₂cosA ......... (3)

Using Eq. (2), we have

sinr₂ = $${{1 \over \mu }}$$

Squaring it, we get

sin²r₂ = $${{1 \over {{\mu ^2}}}}$$

Using the relation sin²x + cos²x = 1, we get

sin²x = 1 $$-$$ cos²x

$$\Rightarrow$$ 1 $$-$$ cos² r₂ = $${{1 \over {{\mu ^2}}}}$$

$$\Rightarrow$$ cos² r₂ = 1 $$-$$ $${{1 \over {{\mu ^2}}}}$$

That is,

cosr₂ = $$\sqrt {1 - {1 \over {{\mu ^2}}}} $$

Substituting the values of sinr₂ and cosr₂ in Eq. (3), we get

sinr₁ = sin A$$\sqrt {1 - {1 \over {{\mu ^2}}}} $$ $$-$$ $${{1 \over \mu }}$$cos A

According to Snell's law, we have

$${{\sin {i_1}} \over {\sin {r_1}}} = \mu $$

That is,

$$\sin {i_1} = \mu \sin {r_1} = \mu \left( {\sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - {1 \over \mu }\cos A} \right)$$

$$\sin {i_1} = \mu \left( {\sin A{{\sqrt {{\mu ^2} - 1} } \over \mu } - {{\cos A} \over \mu }} \right)$$

$$\sin {i_1} = \sin A\sqrt {{\mu ^2} - 1} - \cos A$$ ....... (4)

Now, at minimum deviation, we have A = i₁ and r₁ = $${A \over 2}$$. Thus, from Snell's law, we get

$$\mu = {{\sin {i_1}} \over {\sin {r_1}}} = {{\sin A} \over {\sin (A/2)}}$$

Using $$\sin x = 2\sin {x \over 2}\cos {x \over 2}$$, we get

$$\mu = {{2\sin A/2\cos A/2} \over {\sin A/2}} \Rightarrow \mu = 2\cos {A \over 2}$$ ....... (5)

Now, Eq. (4) becomes

$$\sin {i_1} = \sin A\sqrt {{{\left( {2\cos {A \over 2}} \right)}^2} - 1 - \cos A} $$

$$ = \sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A$$

$${i_1} = {\sin ^{ - 1}}\left[ {\sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A} \right]$$

Hence, option (D) is correct.

$$\bullet$$ For option (B) : From Eq. (5), we have

$$\mu = 2\cos {A \over 2}$$

$$\cos {A \over 2} = {\mu \over 2}$$

$$ \Rightarrow {A \over 2} = {\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$$

$$ \Rightarrow A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$$

Hence, option (B) is incorrect.