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(C) For a plano-convex lens,the radius of curvature of the plane surface is $R_1 = \infty$ and the radius of curvature of the convex surface is $R_2 =

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$1.33$

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(C) For a plano-convex lens,the radius of curvature of the plane surface is $R_1 = \infty$ and the radius of curvature of the convex surface is $R_2 = -10 \,cm$ (using sign convention).Given focal length $f = 30 \,cm$.Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Substituting the values: $\frac{1}{30} = (\mu - 1) \left( \frac{1}{\infty} - \frac{1}{-10} \right)$.Since $\frac{1}{\infty} = 0$,we get: $\frac{1}{30} = (\mu - 1) \left( 0 + \frac{1}{10} \right)$.$\frac{1}{30} = \frac{\mu - 1}{10}$.$\mu - 1 = \frac{10}{30} = \frac{1}{3}$.$\mu = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33$.

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 $30 \,cm$,the refractive index of the material of the lens will be ..........

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