Q141 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics
A small angled prism is made of a material of refractive index $\frac{3}{2}$. The ratio of the angles of minimum deviations when the prism is placed in air and in water of refractive index $\frac{4}{3}$ is
- Correct answerA$4: 1$
- B$3: 4$
- C$2: 3$
- D$1: 3$
Solution
Deviation produced by a thin prism
$\delta=\mathrm{A}(\mu-1)$
In air, $\left.8_1=\mathrm{A}_{\mathrm{a}} \mu_{\mathrm{g}}{ }^{-1}\right)=\mathrm{A}\left(\frac{3}{2}-1\right)=\frac{\mathrm{A}}{2}$ In water, ${ }_w \mu_{\mathrm{g}}=\frac{{ }_{\mathrm{a}} \mu_{\mathrm{g}}}{{ }_{\mathrm{a}} \mu_\omega}=\frac{3 / 2}{4 / 3}=\frac{9}{8}$ $\begin{aligned} & \delta_2=A\left({ }_w \mu_{\mathrm{g}}{ }^{-1}\right)=\mathrm{A}\left(\frac{9}{8}-1\right)=\frac{\mathrm{A}}{8} \\ & \frac{\delta_1}{\delta_2}=\frac{\frac{\mathrm{A}}{\frac{\mathrm{~A}}{8}}}{\frac{8}{2}}=\frac{8}{2}=\frac{4}{1} \end{aligned}$
In air, $\left.8_1=\mathrm{A}_{\mathrm{a}} \mu_{\mathrm{g}}{ }^{-1}\right)=\mathrm{A}\left(\frac{3}{2}-1\right)=\frac{\mathrm{A}}{2}$ In water, ${ }_w \mu_{\mathrm{g}}=\frac{{ }_{\mathrm{a}} \mu_{\mathrm{g}}}{{ }_{\mathrm{a}} \mu_\omega}=\frac{3 / 2}{4 / 3}=\frac{9}{8}$ $\begin{aligned} & \delta_2=A\left({ }_w \mu_{\mathrm{g}}{ }^{-1}\right)=\mathrm{A}\left(\frac{9}{8}-1\right)=\frac{\mathrm{A}}{8} \\ & \frac{\delta_1}{\delta_2}=\frac{\frac{\mathrm{A}}{\frac{\mathrm{~A}}{8}}}{\frac{8}{2}}=\frac{8}{2}=\frac{4}{1} \end{aligned}$
Hence, the correct answer is option: 'A'.