Question

The refracting angle of a prism is A, and refractive index of the material of the prism is cot $\left(\frac{A}{2}\right)$. The angle of minimum deviation is :

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

${\nu }_1=\omega \sqrt{A^2-x_1^2}\therefore {\nu }_1^2={\omega }^2(A^2-x_1^2)$

${\nu }_2=\omega \sqrt{A^2-x_2^2}\therefore {\nu }_2^2={\omega }^2(A^2-x_2^2)$
$\therefore {\nu }_1^2-{\nu }_1^2={\omega }^2(x_1^2-x_2^2)\therefore {\omega }^2=\frac{{\nu }_1^2-{\nu }_2^2}{x_1^2-x_2^2}$
$\therefore \omega =\sqrt{\frac{{\nu }_1^2-{\nu }_2^2}{x_2^2-x_1^2}}\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{x_2^2-x_1^2}{{\nu }_1^2-{\nu }_2^2}}$