Q134 / 150 · MHT CET 2024 (15 May Shift 2) · Physics
A glass prism ' $A$ ' deviates the red and blue rays through $10^{\circ}$ and $12^{\circ}$ respectively. A second prism ' B ' deviates them through $8^{\circ}$ and $10^{\circ}$ respectively. The ratio of their dispersive powers is (A to B)
- A9:13
- B$4: 5$
- Correct answerC$9: 11$
- D$8: 9$
Solution
Dispersive power
$\omega=\frac{\delta_v-\delta_R}{\delta_y}, \delta_y=\frac{\delta_v+\delta_R}{2}...(i)$
For prism A: $\delta_y=\frac{12+10}{2}=11$ $\omega_{\mathrm{A}}=\frac{12-10}{11}=\frac{2}{11}$ For prism B: $\delta_y=\frac{8+10}{2}=9$ $\begin{aligned} & \omega_{\mathrm{B}}=\frac{10-8}{9}=\frac{2}{9} \\ \therefore \quad & \frac{\omega_{\mathrm{A}}}{\omega_{\mathrm{B}}}=\frac{9}{11} \end{aligned}$
For prism A: $\delta_y=\frac{12+10}{2}=11$ $\omega_{\mathrm{A}}=\frac{12-10}{11}=\frac{2}{11}$ For prism B: $\delta_y=\frac{8+10}{2}=9$ $\begin{aligned} & \omega_{\mathrm{B}}=\frac{10-8}{9}=\frac{2}{9} \\ \therefore \quad & \frac{\omega_{\mathrm{A}}}{\omega_{\mathrm{B}}}=\frac{9}{11} \end{aligned}$
Hence, the correct answer is option: 'C'.