A biconvex lens with equal radii of curvature | vedclass

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(D) The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a biconvex lens with equal radii

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The lens maker's formula is given by: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a biconvex lens with equal radii of curvature,we have $R_1 = R$ and $R_2 = -R$.Substituting these values into the formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} ight) = (\mu - 1) \left( \frac{2}{R} ight)$.Given $f = 10 \ cm$ and $\mu = 1.6$,we have: $\frac{1}{10} = (1.6 - 1) \left( \frac{2}{R} ight)$.$\frac{1}{10} = (0.6) \left( \frac{2}{R} ight) = \frac{1.2}{R}$.Therefore,$R = 1.2 	imes 10 = 12 \ cm$.

$A$ biconvex lens with equal radii of curvature has a refractive index of $1.6$ and a focal length of $10 \ cm$. Its radius of curvature will be.......$cm$.

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