MHT CET · Physics · Ray Optics
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
- A \(1: 2\)
- B \(2: 1\)
- C \(1: 4\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(D) \(4: 1\)
Step-by-step Solution
Detailed explanation
For thin prism, \(\delta=(\mu-\mathrm{i}) \mathrm{A}\)
Given, \(\mu_1=\frac{\mu_{\text {glass }}}{\mu_{\text {air }}}=\frac{3}{2}\) and \(\frac{\mu_{\text {water }}}{\mu_{\text {air }}}=\frac{4}{3}\)
\(
\begin{array}{ll}
\therefore & \mu_2=\frac{\mu_{\text {glass }}}{\mu_{\text {water }}}=\frac{\frac{3}{2}}{\frac{4}{3}}=\frac{9}{8} \\
\therefore & \frac{\delta_1}{\delta_2}=\frac{\mu_1-1}{\mu_2-1}=\frac{\frac{3}{2}-1}{\frac{9}{8}-1} \\
\therefore & \frac{\delta_1}{\delta_2}=4
\end{array}
\)
Given, \(\mu_1=\frac{\mu_{\text {glass }}}{\mu_{\text {air }}}=\frac{3}{2}\) and \(\frac{\mu_{\text {water }}}{\mu_{\text {air }}}=\frac{4}{3}\)
\(
\begin{array}{ll}
\therefore & \mu_2=\frac{\mu_{\text {glass }}}{\mu_{\text {water }}}=\frac{\frac{3}{2}}{\frac{4}{3}}=\frac{9}{8} \\
\therefore & \frac{\delta_1}{\delta_2}=\frac{\mu_1-1}{\mu_2-1}=\frac{\frac{3}{2}-1}{\frac{9}{8}-1} \\
\therefore & \frac{\delta_1}{\delta_2}=4
\end{array}
\)
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