# Question #69a13

Jan 31, 2017

Dispersive power $\omega$ of a prism is related with the angle of deviation of blue ray ${\delta}_{b}$ and angle of deviation of red rays ${\delta}_{r}$ as follows

$\omega = \frac{{\delta}_{b} - {\delta}_{r}}{\frac{{\delta}_{b} + {\delta}_{r}}{2}}$

So dispersive power of 1st prism

${\omega}_{1} = \frac{12 - 10}{\frac{12 + 10}{2}} = \frac{2}{11}$

And dispersive power of 2nd prism

${\omega}_{2} = \frac{10 - 8}{\frac{10 + 8}{2}} = \frac{2}{9}$

So ratio of dispersive powers is

${\omega}_{1} : {\omega}_{2} = \frac{2}{11} : \frac{2}{9} = 9 : 11$