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The power of a thin convex lens $\left(\mathrm{a}_{\mathrm{g}}=1.5\right)$ is $5.0 \mathrm{D} .$ When it is placed in a liquid of refractive index $_{a} n_{b}$, then it behaves as a concave lens of focal length $100 \mathrm{~cm}$. The refractive index of the liquid $\mathrm{a}_{l}$ will be
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Verified Answer
The correct answer is:
$5 \sqrt{3}$
By using lens maker's formula,
$$
\begin{array}{l}
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\
\Rightarrow 5=(1.5-1)\left(\frac{2}{R}\right)...(i)
\end{array}
$$
If a lens of refractive index $\mu_{g}$ is immersed in a liquid of refractive index $\mu_{1}$, then its focal length in liquid
$$
\begin{array}{l}
\frac{1}{f_{l}}=\left(_{\ell} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\
\Rightarrow-1=\left(\frac{1.5}{n}-1\right)\left(\frac{2}{R}\right) ...(ii)\\
\text { Dividing, (i) by (ii) }-5=\frac{0.5 n}{1.5-n} \\
\Rightarrow-7.5+5 n=0.5 n \Rightarrow-7.5=-4.5 n \\
\Rightarrow n=\frac{75}{45}=\frac{5}{3}
\end{array}
$$
$$
\begin{array}{l}
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\
\Rightarrow 5=(1.5-1)\left(\frac{2}{R}\right)...(i)
\end{array}
$$
If a lens of refractive index $\mu_{g}$ is immersed in a liquid of refractive index $\mu_{1}$, then its focal length in liquid
$$
\begin{array}{l}
\frac{1}{f_{l}}=\left(_{\ell} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\
\Rightarrow-1=\left(\frac{1.5}{n}-1\right)\left(\frac{2}{R}\right) ...(ii)\\
\text { Dividing, (i) by (ii) }-5=\frac{0.5 n}{1.5-n} \\
\Rightarrow-7.5+5 n=0.5 n \Rightarrow-7.5=-4.5 n \\
\Rightarrow n=\frac{75}{45}=\frac{5}{3}
\end{array}
$$
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