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A thin convergent glass lens $\left(\mu_{g}=1.5\right)$ has a power of $+5.0 \mathrm{D}$. When this lens is immersed in a liquid of refractive index $\mu,$ it acts as a divergent lens of focal length $100 \mathrm{~cm}$. The value of $\mu$ must be
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$5 / 3$
$\frac{\mathrm{P}_{\mathrm{a}}}{\mathrm{P}_{1}}=\frac{\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)}{\left(\frac{\mu_{\mathrm{g}}}{\mu_{1}}-1\right)}=\frac{+5}{-100 / 100}=-5$
$-5\left(\frac{\mu_{\mathrm{g}}}{\mu_{1}}-1\right)=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1$
$\frac{1.5}{\mu_{1}}-1=\frac{-1}{5}(1.5-1)=-0.1$
$$
\begin{array}{ll}\mu_{1}=\frac{1.5}{0.9}=\frac{5}{3}\end{array}
$$
$-5\left(\frac{\mu_{\mathrm{g}}}{\mu_{1}}-1\right)=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1$
$\frac{1.5}{\mu_{1}}-1=\frac{-1}{5}(1.5-1)=-0.1$
$$
\begin{array}{ll}\mu_{1}=\frac{1.5}{0.9}=\frac{5}{3}\end{array}
$$
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