The angle of minimum deviation produced by a thin prism in air is $\delta_1$. If it is immersed in water the…

The angle of minimum deviation produced by a thin prism in air is $\delta_1$. If it is immersed in water the angle of minimum deviation is $\left[\mathrm{a}_{\mathrm{g}}=\frac{3}{2}, \mathrm{a}_{\mathrm{w}}=\frac{4}{3}\right]$
  1. $2 \delta_1$
  2. $\frac{\delta_1}{2}$
  3. $\frac{\delta_1}{3}$
  4. $\frac{\delta_1}{4}$

Solution

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

Asked in: MHT CET 2024 (15 May Shift 2)

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