A convex lens has a power of +5.0 , D in air, | vedclass

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(A) The power of the lens in air is $P_a = +5.0 \, D$. The focal length in air is $f_a = \frac{1}{P_a} = \frac{1}{5} \, m = 20 \, cm$.Using the lens m

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$5/3$

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(A) The power of the lens in air is $P_a = +5.0 \, D$. The focal length in air is $f_a = \frac{1}{P_a} = \frac{1}{5} \, m = 20 \, cm$.Using the lens maker's formula ratio: $\frac{f_l}{f_a} = \frac{_a\mu_g - 1}{_l\mu_g - 1}$,where $_l\mu_g = \frac{_a\mu_g}{_a\mu_l} = \frac{1.5}{\mu_l}$.Given that the lens acts as a concave lens of focal length $f_l = -100 \, cm$ in the liquid.Substituting the values: $\frac{-100}{20} = \frac{1.5 - 1}{\frac{1.5}{\mu_l} - 1}$.$-5 = \frac{0.5}{\frac{1.5}{\mu_l} - 1}$.$-5 \left( \frac{1.5}{\mu_l} - 1 \right) = 0.5$.$-\frac{7.5}{\mu_l} + 5 = 0.5$.$4.5 = \frac{7.5}{\mu_l}$.$\mu_l = \frac{7.5}{4.5} = \frac{75}{45} = \frac{5}{3}$.

$A$ convex lens has a power of $+5.0 \, D$ in air,where the refractive index of the lens material is $_a\mu_g = 1.5$. In a liquid of what refractive index should it be immersed so that it acts as a concave lens of focal length $100 \, cm$?

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