The radii of curvature of the faces of a doub | 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:$f = 12 \ cm$,$R_1 = 10 \ cm$,$R_2 = -1

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Given:$f = 12 \ cm$,$R_1 = 10 \ cm$,$R_2 = -15 \ cm$ (for a double convex lens,the second surface has a negative radius of curvature in the sign convention).Substituting the values:$\frac{1}{12} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{-15} ight)$$\frac{1}{12} = (\mu - 1) \left( \frac{1}{10} + \frac{1}{15} ight)$$\frac{1}{12} = (\mu - 1) \left( \frac{3 + 2}{30} ight)$$\frac{1}{12} = (\mu - 1) 	imes \frac{5}{30}$$\frac{1}{12} = (\mu - 1) 	imes \frac{1}{6}$$\frac{6}{12} = \mu - 1$$0.5 = \mu - 1$$\mu = 1.5$Thus,the refractive index of the material of the lens is $1.5$.

The radii of curvature of the faces of a double convex lens are $10 \ cm$ and $15 \ cm$. Its focal length is $12 \ cm$. What is the refractive index of the material of the lens?

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