The radius of curvature of the convex surface | vedclass

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(D) Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a plano-convex lens,the radius of curva

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$+4 \ D$

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(D) Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a plano-convex lens,the radius of curvature of the convex surface is $R_1 = +15 \ cm$ and the plane surface is $R_2 = \infty$.Given $\mu = 1.6$.Substituting the values: $\frac{1}{f} = (1.6 - 1) \left( \frac{1}{15} - \frac{1}{\infty} ight)$.$\frac{1}{f} = 0.6 	imes \frac{1}{15} = \frac{0.6}{15} = \frac{6}{150} = \frac{1}{25} \ cm^{-1}$.The power of the lens $P$ in Diopters $(D)$ is given by $P = \frac{100}{f(cm)}$.$P = \frac{100}{25} = +4 \ D$.

The radius of curvature of the convex surface of a thin plano-convex lens is $15 \ cm$ and the refractive index of its material is $1.6$. The power of the lens is

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