Question

If the angle of prism equals the angle of minimum deviation, and given that the refractive index \( \mu = \sqrt{3} \), then the angle of the prism is:

Answer 1

For a prism with the angle of prism \( A \) and the angle of minimum deviation \( D \), the refractive index \( \mu \) is given by the formula: \[ \mu = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)}. \] Since the angle of prism \( A \) equals the angle of minimum deviation \( D \), we substitute \( A = D \) into the formula: \[ \mu = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)}. \] We are given that \( \mu = \sqrt{3} \), so: \[ \sqrt{3} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)}. \] Solving this equation will give the value of \( A \). After solving, we find \( A = 60^\circ \). Thus, the angle of the prism is \( 60^\circ \).