A biconvex lens has a focal length 2/3 times | vedclass

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(A) For a biconvex lens, the radii of curvature are $R_1 = R$ and $R_2 = -R$. According to the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \

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$1.75$

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(A) For a biconvex lens, the radii of curvature are $R_1 = R$ and $R_2 = -R$. According to the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Substituting the values, we get: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$. Given that the focal length $f = \frac{2}{3}R$, we substitute this into the equation: $\frac{1}{(2/3)R} = (\mu - 1) \left( \frac{2}{R} \right)$. $\frac{3}{2R} = (\mu - 1) \frac{2}{R}$. Multiplying both sides by $R$, we get: $\frac{3}{2} = 2(\mu - 1)$. $0.75 = \mu - 1$. $\mu = 1.75$.

$A$ biconvex lens has a focal length $2/3$ times the radius of curvature of either surface. The refractive index of the material is:

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