Question

Calculate the angle of minimum deviation of an equilateral prism. The refractive index of the prism is \(\sqrt{3}\). Calculate the angle of incidence for this case of minimum deviation also.

Answer 1

Calculation of Angle of Minimum Deviation and Angle of Incidence for a Prism

Given:

• Refractive index of the prism, \( \mu = \sqrt{3} \)
• Angle of the prism, \( A = 60^\circ \) (since the prism is equilateral)

Solution:

The refractive index for minimum deviation \( \mu \) is given by the following formula:

\[ \mu = \frac{\sin \left( \frac{A + D_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \]

Substitute the given values (\( A = 60^\circ \) and \( \mu = \sqrt{3} \)):

\[ \sqrt{3} = \frac{\sin \left( \frac{60^\circ + D_m}{2} \right)}{\sin 30^\circ} \]

Since \( \sin 30^\circ = 0.5 \), the equation becomes:

\[ \sqrt{3} = \frac{\sin \left( \frac{60^\circ + D_m}{2} \right)}{0.5} \]

Multiplying both sides by \( 0.5 \):

\[ \sin \left( \frac{60^\circ + D_m}{2} \right) = \frac{\sqrt{3}}{2} \]

From trigonometry, we know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). So, we have:

\[ \frac{60^\circ + D_m}{2} = 60^\circ \]

Therefore:

\[ 60^\circ + D_m = 120^\circ \quad \Rightarrow \quad D_m = 60^\circ \]

Step 2: Calculate the Angle of Incidence at Minimum Deviation

The angle of incidence at minimum deviation \( i \) is given by the formula:

\[ i = \frac{A + D_m}{2} \]

Substituting \( A = 60^\circ \) and \( D_m = 60^\circ \), we get:

\[ i = \frac{60^\circ + 60^\circ}{2} = 60^\circ \]

Final Answer:

• The angle of minimum deviation, \( D_m = 60^\circ \)
• The angle of incidence for minimum deviation, \( i = 60^\circ \)