Determine the angle of minimum deviation for an equilateral prism...

For an equilateral prism, the angle of the prism is A = 60^{}^{{}^{\circ}}.

The refractive index of the material is given as $\mu = 1.732 = \sqrt{3}$.

The relation between the refractive index, angle of prism, and angle of minimum deviation $\delta_m$ is:

$\mu = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$

Substituting the given values:

\sqrt{3} = \dfrac{\sin\left(\dfrac{60^{}^{{}^{\circ}} + \delta_m}{2}\right)}{\sin\left(\dfrac{60^{}^{{}^{\circ}}}{2}\right)}

\sqrt{3} = \dfrac{\sin\left(30^{}^{{}^{\circ}} + \dfrac{\delta_m}{2}\right)}{\sin(30^{}^{{}^{\circ}})}

\sqrt{3} = \dfrac{\sin\left(30^{}^{{}^{\circ}} + \dfrac{\delta_m}{2}\right)}{\dfrac{1}{2}}

\sin\left(30^{}^{{}^{\circ}} + \dfrac{\delta_m}{2}\right) = \dfrac{\sqrt{3}}{2}

Since \sin(60^{}^{{}^{\circ}}) = \dfrac{\sqrt{3}}{2}, we get:

30^{}^{{}^{\circ}} + \dfrac{\delta_m}{2} = 60^{}^{{}^{\circ}}

\dfrac{\delta_m}{2} = 30^{}^{{}^{\circ}} \Rightarrow \delta_m = 60^{}^{{}^{\circ}}

The corresponding angle of incidence $i$ at minimum deviation is given by:

$i = \dfrac{A + \delta_m}{2}$

i = \dfrac{60^{}^{{}^{\circ}} + 60^{}^{{}^{\circ}}}{2} = 60^{}^{{}^{\circ}}

Thus, the angle of minimum deviation is 60^{}^{{}^{\circ}} and the angle of incidence is 60^{}^{{}^{\circ}}.