A biconvex lens (R_{1}=R_{2}=20 cm) has a fo | vedclass

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(A) For a biconvex lens,the lens maker's formula is given by: $\frac{1}{f} = (n-1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.Given $n = 1.5$,$

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$-40$

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(A) For a biconvex lens,the lens maker's formula is given by: $\frac{1}{f} = (n-1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$.Given $n = 1.5$,$R_{1} = 20 \ cm$,and $R_{2} = -20 \ cm$.Substituting these values: $\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} ight) = 0.5 \left( \frac{1}{20} + \frac{1}{20} ight) = 0.5 \left( \frac{2}{20} ight) = 0.5 	imes 0.1 = 0.05$.Thus,$f = \frac{1}{0.05} = 20 \ cm$.The focal length of the concave mirror is equal to the focal length of the lens,so $f_{mirror} = 20 \ cm$.For a concave mirror,the focal length is negative,so $f = -20 \ cm$.The radius of curvature $R$ is given by $R = 2f = 2(-20 \ cm) = -40 \ cm$.

$A$ biconvex lens $(R_{1}=R_{2}=20 \ cm)$ has a focal length equal to the focal length of a concave mirror. The radius of curvature of the concave mirror is (Refractive Index of glass lens $= 1.5$) (in $cm$)

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