A thin glass (refractive index 1.5) lens has | vedclass

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(B) The power of a lens in air is given by the Lens Maker's Formula: $\frac{1}{f_{air}} = P_{air} = (\mu_g - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \

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$5/8\,D$

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(B) The power of a lens in air is given by the Lens Maker's Formula: $\frac{1}{f_{air}} = P_{air} = (\mu_g - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.Given $P_{air} = -5\,D$ and $\mu_g = 1.5$,we have: $-5 = (1.5 - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right] = 0.5 \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.Thus,$\left[ \frac{1}{R_1} - \frac{1}{R_2} \right] = \frac{-5}{0.5} = -10$.Now,the power of the same lens in a liquid with refractive index $\mu_l = 1.6$ is given by: $P_{liquid} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.Substituting the values: $P_{liquid} = \left( \frac{1.5}{1.6} - 1 \right) (-10)$.$P_{liquid} = \left( \frac{1.5 - 1.6}{1.6} \right) (-10) = \left( \frac{-0.1}{1.6} \right) (-10) = \left( \frac{-1}{16} \right) (-10) = \frac{10}{16} = \frac{5}{8}\,D$.

$A$ thin glass (refractive index $1.5$) lens has an optical power of $-5\,D$ in air. Its optical power in a liquid medium with a refractive index of $1.6$ will be:

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