A convex lens of refractive index frac{3}{2} | vedclass

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(D) Using the Lens Maker's formula, the power $P$ of a lens in air is given by:$P = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 2.5 \ D$W

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$-1.25$

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(D) Using the Lens Maker's formula, the power $P$ of a lens in air is given by:$P = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = 2.5 \ D$When the lens is placed in a liquid of refractive index $\mu_l$, the effective refractive index of the lens relative to the liquid is $\mu' = \frac{\mu}{\mu_l}$.The new power $P'$ is given by:$P' = (\mu' - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Taking the ratio of the two equations:$\frac{P'}{P} = \frac{\mu' - 1}{\mu - 1}$Given $\mu = 1.5$ and $\mu_l = 2$, we have $\mu' = \frac{1.5}{2} = 0.75$.Substituting the values:$\frac{P'}{2.5} = \frac{0.75 - 1}{1.5 - 1} = \frac{-0.25}{0.5} = -0.5$$P' = -0.5 	imes 2.5 = -1.25 \ D$

$A$ convex lens of refractive index $\frac{3}{2}$ has a power of $2.5 \ D$. If it is placed in a liquid of refractive index $2$, the new power of the lens is: (in $D$)

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