A convex lens of radii of curvature 6 cm and | vedclass

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(C) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Here, $\mu_{rel} = \frac{\mu_{le

Vedclass · Accepted Answer

$26$

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(C) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Here, $\mu_{rel} = \frac{\mu_{lens}}{\mu_{liquid}} = \frac{1.5}{1.3} = \frac{15}{13}$.For a convex lens, $R_1 = +6 \ cm$ and $R_2 = -12 \ cm$.Substituting these values into the formula:$\frac{1}{f} = \left( \frac{15}{13} - 1 \right) \left( \frac{1}{6} - \frac{1}{-12} \right)$$\frac{1}{f} = \left( \frac{2}{13} \right) \left( \frac{1}{6} + \frac{1}{12} \right)$$\frac{1}{f} = \left( \frac{2}{13} \right) \left( \frac{2+1}{12} \right) = \left( \frac{2}{13} \right) \left( \frac{3}{12} \right) = \left( \frac{2}{13} \right) \left( \frac{1}{4} \right) = \frac{2}{52} = \frac{1}{26}$.Therefore, $f = 26 \ cm$.

$A$ convex lens of radii of curvature $6 \ cm$ and $12 \ cm$ is immersed in a liquid of refractive index $1.3$. If the refractive index of the material of the lens is $1.5$, then the focal length of the lens when immersed in the liquid is (in $cm$)

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