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Assertion: A double convex lens \((\mu=1.5)\) has focal length \(10 \mathrm{~cm}\). When the lens is immersed in water \((\mu=4 / 3)\) its focal length becomes \(77 \mathrm{~cm}\).
Reason : \(\frac{1}{f}=\frac{\mu_g-\mu_w}{\mu_w}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
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Reason : \(\frac{1}{f}=\frac{\mu_g-\mu_w}{\mu_w}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
Solution:
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If both assertion and reason are true and reason is the correct explanation of assertion
$\begin{aligned} & \text { } \frac{1}{f}=\left({ }^a \mu_g-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\ & =\left(\frac{\mu_g}{\mu_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & \frac{1}{f}=\left(\frac{\mu_g-\mu_w}{\mu_w}\right) \cdot\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\ & =\left(\frac{1.5-1.33}{1.33}\right) \cdot\left[\frac{1}{20}+\frac{1}{20}\right] \\ & \frac{1}{f}=\frac{0.13 \times 2}{20} \Rightarrow f=\frac{20}{0.26} \\ & \therefore f=76.92 \mathrm{~cm} \text {. } \\ & \end{aligned}$
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