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A lens is made of flint glass (refractive index $=1.5$ ). When the lens is immersed in a liquid of refractive index 1.25 , the focal length
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Verified Answer
The correct answer is:
increases by a factor of 2.5
Let $f$ be focal length of lens.
$$
\begin{aligned}
& \frac{1}{f}=\left({ }^a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \text { or, } \frac{1}{f}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \text { or, } \frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{0.5 f} . .
\end{aligned}
$$
Let $f^{\prime}$ be focal length of the lens when immersed in a liquid.
$$
\begin{aligned}
& \frac{1}{f^{\prime}}=\left({ }^L \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{1.5}{1.25}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& =\left(\frac{0.25}{1.25}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{0.25}{1.25} \times \frac{1}{0.5 f} \text { (using (i)) } \\
& \frac{1}{f^{\prime}}=\frac{1}{2.5 f} \text { or, } f^{\prime}=2.5 f .
\end{aligned}
$$
i.e. focal length of lens when immersed in a liquid is increased by a factor of 2.5 .
$$
\begin{aligned}
& \frac{1}{f}=\left({ }^a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \text { or, } \frac{1}{f}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \text { or, } \frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{0.5 f} . .
\end{aligned}
$$
Let $f^{\prime}$ be focal length of the lens when immersed in a liquid.
$$
\begin{aligned}
& \frac{1}{f^{\prime}}=\left({ }^L \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{1.5}{1.25}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& =\left(\frac{0.25}{1.25}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{0.25}{1.25} \times \frac{1}{0.5 f} \text { (using (i)) } \\
& \frac{1}{f^{\prime}}=\frac{1}{2.5 f} \text { or, } f^{\prime}=2.5 f .
\end{aligned}
$$
i.e. focal length of lens when immersed in a liquid is increased by a factor of 2.5 .
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