MHT CET · Physics · Ray Optics
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
\(\left[\mathrm{a}_{\mathrm{g}}=\frac{3}{2}, \mathrm{a}_{\mathrm{w}}=\frac{4}{3}\right]\)
- A \(2 \delta_1\)
- B \(\frac{\delta_1}{2}\)
- C \(\frac{\delta_1}{3}\)
- D \(\frac{\delta_1}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\delta_1}{4}\)
Step-by-step Solution
Detailed explanation
For thin prism, \(\delta=(\mu-1) \mathrm{A}\)
Given, \(\mu_1=\frac{\mu_{\text {glass }}}{\mu_{\text {air }}}=\frac{3}{2}\) and \(\mu_2=\frac{\mu_{\text {water }}}{\mu_{\text {air }}}=\frac{4}{3}\)
\(\begin{aligned} & \therefore \quad \mu^{\prime}=\frac{\mu_1}{\mu_2} \frac{\mu_{\text {glass }}}{\mu_{\text {water }}}=\frac{\frac{3}{2}}{\frac{4}{3}}=\frac{9}{8} \\ & \therefore \quad \frac{\delta_1}{\delta_2}=\frac{\mu_1-1}{\mu^{\prime}-1}=\frac{\frac{3}{2}-1}{\frac{9}{8}-1} \quad \therefore \delta_2=\frac{\delta_1}{4}\end{aligned}\)
Given, \(\mu_1=\frac{\mu_{\text {glass }}}{\mu_{\text {air }}}=\frac{3}{2}\) and \(\mu_2=\frac{\mu_{\text {water }}}{\mu_{\text {air }}}=\frac{4}{3}\)
\(\begin{aligned} & \therefore \quad \mu^{\prime}=\frac{\mu_1}{\mu_2} \frac{\mu_{\text {glass }}}{\mu_{\text {water }}}=\frac{\frac{3}{2}}{\frac{4}{3}}=\frac{9}{8} \\ & \therefore \quad \frac{\delta_1}{\delta_2}=\frac{\mu_1-1}{\mu^{\prime}-1}=\frac{\frac{3}{2}-1}{\frac{9}{8}-1} \quad \therefore \delta_2=\frac{\delta_1}{4}\end{aligned}\)
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