Assertion: A double convex lens (mu = 1.5) ha | vedclass

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(A) The Lens Maker's Formula is given by $\frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,which can

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If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

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(A) The Lens Maker's Formula is given by $\frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$,which can be written as $\frac{1}{f} = \left( \frac{\mu_l - \mu_m}{\mu_m} ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$. Thus,the Reason is correct.For the lens in air $(\mu_m = 1)$:$\frac{1}{10} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) \implies \frac{1}{10} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} ight) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = \frac{1}{5} \dots (i)$For the lens in water $(\mu_m = 4/3)$:$\frac{1}{f'} = \left( \frac{1.5 - 4/3}{4/3} ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$$\frac{1}{f'} = \left( \frac{4.5/3 - 4/3}{4/3} ight) \left( \frac{1}{5} ight) = \left( \frac{0.5/3}{4/3} ight) \left( \frac{1}{5} ight) = \left( \frac{0.5}{4} ight) \left( \frac{1}{5} ight) = \frac{1}{8} 	imes \frac{1}{5} = \frac{1}{40}$Therefore,$f' = 40 \, cm$. The Assertion is also correct and the Reason explains it.

Assertion: $A$ double convex lens $(\mu = 1.5)$ has a focal length of $10 \, cm$. When the lens is immersed in water $(\mu = 4/3)$,its focal length becomes $40 \, cm$.
Reason: $\frac{1}{f} = \left( \frac{\mu_l - \mu_m}{\mu_m} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

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Assertion: $A$ double convex lens $(\mu = 1.5)$ has a focal length of $10 \, cm$. When the lens is immersed in water $(\mu = 4/3)$,its focal length becomes $40 \, cm$.Reason: $\frac{1}{f} = \left( \frac{\mu_l - \mu_m}{\mu_m} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$