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,$f > R$.Therefore,$\frac{R}{2(\mu - 1)} > R$,which implies $\frac{1}{2(\mu - 1)} > 1$.This simplifies to $2(\mu - 1) So,$\mu Since the lens must be made of a material with a refractive index greater than $1$ (for it to act as a lens in air),the refractive index $\mu$ must be greater than $1$ 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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