If in a plano-convex lens,the radius of curva | vedclass

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(C) For a plano-convex lens,the lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given that the

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$1.33$

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(C) For a plano-convex lens,the lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given that the lens is plano-convex,$R_1 = 10 \ cm$ and $R_2 = \infty$.The focal length $f = 30 \ cm$.Substituting these values into the formula:$\frac{1}{30} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{\infty} \right)$Since $\frac{1}{\infty} = 0$,we have:$\frac{1}{30} = (\mu - 1) \left( \frac{1}{10} \right)$$\frac{10}{30} = \mu - 1$$\frac{1}{3} = \mu - 1$$\mu = 1 + 0.333 = 1.33$.

If in a plano-convex lens,the radius of curvature of the convex surface is $10 \ cm$ and the focal length of the lens is $30 \ cm,$ then the refractive index of the material of the lens will be

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