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A prism of a certain angle deviates the red and blue rays by $8^{\circ}$ and $12^{\circ}$ respectively. Another prism of the same angle deviates the red and blue rays by $10^{\circ}$ and $14^{\circ}$ respectively. The prisms are small angled and made of different materials. The dispersive powers of the materials of the prisms are in the ratio
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Verified Answer
The correct answer is:
$6: 5$
Dispersive power, $\omega=\frac{\delta_{B}-\delta_{R}}{\delta}$
For 1st prism,
$$
\begin{aligned}
\omega_{1} &=\frac{\delta_{B}-\delta_{R}}{\delta} \\
&=\frac{12^{\circ}-8^{\circ}}{10^{\circ}}=\frac{2}{5}
\end{aligned}
$$
where $\delta=\frac{\delta_{B}+\delta_{R}}{2}$
Similarly, for 2 nd prism,
$$
\begin{aligned}
\omega_{2} &=\frac{14^{\circ}-10^{\circ}}{12}=\frac{1}{3} \\
\therefore \quad \frac{\omega_{1}}{\omega_{2}} &=\frac{2}{5} \times \frac{3}{1}=\frac{6}{5}
\end{aligned}
$$
For 1st prism,
$$
\begin{aligned}
\omega_{1} &=\frac{\delta_{B}-\delta_{R}}{\delta} \\
&=\frac{12^{\circ}-8^{\circ}}{10^{\circ}}=\frac{2}{5}
\end{aligned}
$$
where $\delta=\frac{\delta_{B}+\delta_{R}}{2}$
Similarly, for 2 nd prism,
$$
\begin{aligned}
\omega_{2} &=\frac{14^{\circ}-10^{\circ}}{12}=\frac{1}{3} \\
\therefore \quad \frac{\omega_{1}}{\omega_{2}} &=\frac{2}{5} \times \frac{3}{1}=\frac{6}{5}
\end{aligned}
$$
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