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Step-by-Step Solution
Step 1: Identify the given data
• The prism is equilateral, so its refracting angle $A = 60^\circ$.
• The refractive index of the prism material is $n = \sqrt{3}$.
• The light undergoes minimum deviation, so the angle of incidence $i$ equals the angle of emergence $e$, and the two angles of refraction inside the prism are equal: $r_1 = r_2$.
Step 2: Express the angles of refraction
For an equilateral triangular prism of angle $A$, when the light passes through it in the condition of minimum deviation, we have
$r_1 = r_2 = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ.$
Step 3: Apply Snell’s law
Snell’s law at the first refracting surface is given by
$ n_1 \sin i = n_2 \sin r_1, $
where $n_1 = 1$ (air), $n_2 = \sqrt{3}$ (prism), $i$ is the angle of incidence, and $r_1$ is the angle of refraction. Substituting $n_1 = 1$, $n_2 = \sqrt{3}$, and $r_1 = 30^\circ$ gives:
$\sin i = \sqrt{3} \, \sin 30^\circ.$
Step 4: Solve for the angle of incidence
Since $\sin 30^\circ = \frac{1}{2}$,
$\sin i = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}.$
Therefore,
$i = 60^\circ.$
Final Answer
The angle of incidence is $60^\circ$.
