The power of a thin convex lens placed in air | vedclass

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(C) For a thin convex lens, $P = +4 \ D$, $\mu_{\ell} = \frac{3}{2}$, $\mu_m = \frac{5}{3}$.Since $P = \frac{1}{f_a}$, we have $f_a = \frac{1}{4} \ m

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it behaves like a concave lens of focal length $125 \ cm$

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(C) For a thin convex lens, $P = +4 \ D$, $\mu_{\ell} = \frac{3}{2}$, $\mu_m = \frac{5}{3}$.Since $P = \frac{1}{f_a}$, we have $f_a = \frac{1}{4} \ m = 25 \ cm$.In air, the lens maker's formula is $\frac{1}{f_a} = (\mu_{\ell} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$ ... $(i)$.In a liquid medium, the focal length $f_m$ is given by $\frac{1}{f_m} = \left(\frac{\mu_{\ell}}{\mu_m} - 1ight) \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$ ... (ii).Dividing equation (ii) by $(i)$, we get $\frac{f_a}{f_m} = \frac{(\mu_{\ell} - 1)}{(\frac{\mu_{\ell}}{\mu_m} - 1)}$.Substituting the values: $\frac{f_a}{f_m} = \frac{(\frac{3}{2} - 1)}{(\frac{3/2}{5/3} - 1)} = \frac{1/2}{(9/10 - 1)} = \frac{1/2}{-1/10} = -5$.Therefore, $f_m = -5 	imes f_a = -5 	imes 25 \ cm = -125 \ cm$.The negative sign indicates that the lens behaves like a concave lens with a focal length of $125 \ cm$.

The power of a thin convex lens placed in air is $+4 \ D$. The refractive index of the material of the convex lens is $\frac{3}{2}$. If this convex lens is immersed in a liquid of refractive index $\frac{5}{3}$, then

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