A double convex lens has a focal length of 25 | vedclass

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(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given: $f = 25\; cm$,$\mu = 1.5$. For a dou

Vedclass · Accepted Answer

$18.75\; cm, 37.5\; cm$

Vedclass · Answer

(C) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Given: $f = 25\; cm$,$\mu = 1.5$. For a double convex lens,$R_1 = R$ and $R_2 = -2R$.Substituting the values: $\frac{1}{25} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-2R} ight)$.$\frac{1}{25} = 0.5 \left( \frac{1}{R} + \frac{1}{2R} ight) = 0.5 \left( \frac{3}{2R} ight) = \frac{1.5}{2R} = \frac{3}{4R}$.$4R = 25 	imes 3 = 75$.$R = \frac{75}{4} = 18.75\; cm$.Therefore,the radii are $R_1 = 18.75\; cm$ and $R_2 = 2R = 37.5\; cm$.

$A$ double convex lens has a focal length of $25\; cm$. The radius of curvature of one of the surfaces is double the other. Find the radii if the refractive index of the material of the lens is $1.5$.

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