The focal length of a convex lens is 10 cm a | vedclass

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(B) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given: $f = +10 \ cm$,$\mu = 1.5$,$R_1 = +7

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given: $f = +10 \ cm$,$\mu = 1.5$,$R_1 = +7.5 \ cm$.Substituting the values: $\frac{1}{10} = (1.5 - 1) \left( \frac{1}{7.5} - \frac{1}{R_2} \right)$.$0.1 = 0.5 \left( \frac{1}{7.5} - \frac{1}{R_2} \right)$.$0.2 = \frac{1}{7.5} - \frac{1}{R_2}$.$\frac{1}{R_2} = \frac{1}{7.5} - 0.2$.$\frac{1}{R_2} = \frac{1}{7.5} - \frac{1}{5} = \frac{2 - 3}{15} = -\frac{1}{15}$.Therefore,$R_2 = -15 \ cm$. The magnitude of the radius of curvature is $15 \ cm$.

The focal length of a convex lens is $10 \ cm$ and its refractive index is $1.5$. If the radius of curvature of one surface is $7.5 \ cm$,the radius of curvature of the second surface will be......$cm$.

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