Question

A prism is made with a material of refractive index \( \mu = 2 \). The angle of prism is 60°. The angle of minimum deviation produced by the prism is:

• A. \( 30^\circ \)
• B. \( 45^\circ \)
• C. \( 60^\circ \)
• D. \( 90^\circ \)

Hint:

When calculating the angle of minimum deviation in a prism, ensure to use the refractive index and angle of the prism correctly in the formula.

Answer 1

The Correct Option is A

Use the formula for the angle of minimum deviation.
For a prism, the relation between the angle of minimum deviation \( D \), the angle of the prism \( A \), and the refractive index \( \mu \) is given by: \[ \mu = \frac{\sin\left( \frac{A + D}{2} \right)}{\sin\left( \frac{A}{2} \right)} \] Substituting the values \( A = 60^\circ \) and \( \mu = 2 \), we get: \[ 2 = \frac{\sin\left( \frac{60^\circ + D}{2} \right)}{\sin\left( 30^\circ \right)} \] Using \( \sin(30^\circ) = \frac{1}{2} \), \[ 2 = \frac{\sin\left( \frac{60^\circ + D}{2} \right)}{\frac{1}{2}} \quad \Rightarrow \quad \sin\left( \frac{60^\circ + D}{2} \right) = 1 \] Thus, \[ \frac{60^\circ + D}{2} = 90^\circ \quad \Rightarrow \quad 60^\circ + D = 180^\circ \quad \Rightarrow \quad D = 120^\circ \]