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(B) According to 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

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$4/3$

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(B) According to 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 curvature of the plane surface is $R_2 = \infty$ and the radius of curvature of the convex surface is $R_1 = 10 \ cm$.Given the focal length $f = 30 \ cm$,we substitute the values into the formula:$\frac{1}{30} = (\mu - 1) \left[ \frac{1}{10} - \frac{1}{\infty} \right]$Since $\frac{1}{\infty} = 0$,the equation becomes:$\frac{1}{30} = (\mu - 1) \left( \frac{1}{10} \right)$Multiplying both sides by $10$:$\frac{10}{30} = \mu - 1$$\frac{1}{3} = \mu - 1$Therefore,$\mu = 1 + \frac{1}{3} = \frac{4}{3}$.

If the radius of curvature of the convex surface of a plano-convex lens is $10 \ cm$ and its focal length is $30 \ cm$,what is the refractive index of the material of the lens?

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