The angle of deviation produced by a thin prism when placed in air is ' $\delta_1$ ' and that when immersed…

The angle of deviation produced by a thin prism when placed in air is ' $\delta_1$ ' and that when immersed in water is ' $\delta_2$ '. The refractive index of glass and water are $\frac{3}{2}$ and $\frac{4}{3}$ respectively. The ratio $\delta_1: \delta_2$ is
  1. $1: 2$
  2. $2: 1$
  3. $1: 4$
  4. $4: 1$

Solution

For thin prism, $\delta=(\mu-\mathrm{i}) \mathrm{A}$ Given, $\mu_1=\frac{\mu_{\text {glass }}}{\mu_{\text {air }}}=\frac{3}{2}$ and $\frac{\mu_{\text {water }}}{\mu_{\text {air }}}=\frac{4}{3}$ $\begin{array}{ll} \therefore & \mu_2=\frac{\mu_{\text {glass }}}{\mu_{\text {water }}}=\frac{\frac{3}{2}}{\frac{4}{3}}=\frac{9}{8} \\ \therefore & \frac{\delta_1}{\delta_2}=\frac{\mu_1-1}{\mu_2-1}=\frac{\frac{3}{2}-1}{\frac{9}{8}-1} \\ \therefore & \frac{\delta_1}{\delta_2}=4 \end{array}$

Asked in: MHT CET 2023 (14 May Shift 1)

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