The focal length of an equi-convex lens is gr | vedclass

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For an equi-convex lens,$R_1 = R$ and

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greater than zero but less than $1.5$

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(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For an equi-convex lens,$R_1 = R$ and $R_2 = -R$.Substituting these values,we get $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$.Thus,$f = \frac{R}{2(\mu - 1)}$.According to the problem,the focal length $f$ is greater than the radius of curvature $R$,so $f > R$.Substituting the expression for $f$,we get $\frac{R}{2(\mu - 1)} > R$.This simplifies to $\frac{1}{2(\mu - 1)} > 1$,which implies $2(\mu - 1) Therefore,$\mu - 1 Since the refractive index $\mu$ must be greater than $1$ for a converging lens in air,the range is $1 < \mu < 1.5$. However,based on the provided options,the correct range is greater than zero but less than $1.5$.

The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is

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