The figure shows a thin lens with centres of | vedclass

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(D) From the figure,the radii of curvature are $R_1 = 10 \ cm$ and $R_2 = 20 \ cm$. Both centers of curvature lie to the right of the lens,so by sign

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$40$

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(D) From the figure,the radii of curvature are $R_1 = 10 \ cm$ and $R_2 = 20 \ cm$. Both centers of curvature lie to the right of the lens,so by sign convention,both $R_1$ and $R_2$ are positive.Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Substitute the given values: $\mu = 1.5$,$R_1 = 10 \ cm$,$R_2 = 20 \ cm$.$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{10} - \frac{1}{20} ight)$$\frac{1}{f} = (0.5) \left( \frac{2 - 1}{20} ight)$$\frac{1}{f} = 0.5 	imes \frac{1}{20} = \frac{0.5}{20} = \frac{1}{40}$Therefore,$f = 40 \ cm$.

The figure shows a thin lens with centres of curvature $C_1$ and $C_2$. Find its focal length in $cm$. (Take refractive index $\mu = 1.5$)

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