What is the focal length (in cm) of a biconve | vedclass

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(B) For a biconvex lens,the radius of curvature $R_1 = R$ and $R_2 = -R$. Given $R = 40\, cm$ and $\mu = 1.65$.Using the Lens Maker's Formula: $\frac{

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$31$

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(B) For a biconvex lens,the radius of curvature $R_1 = R$ and $R_2 = -R$. Given $R = 40\, cm$ and $\mu = 1.65$.Using the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Substituting the values: $\frac{1}{f} = (1.65 - 1) \left( \frac{1}{40} - \frac{1}{-40} ight)$.$\frac{1}{f} = (0.65) \left( \frac{1}{40} + \frac{1}{40} ight) = 0.65 	imes \frac{2}{40} = \frac{0.65}{20}$.$f = \frac{20}{0.65} \approx 30.77\, cm$.Rounding to the nearest integer,we get $f \approx 31\, cm$.

What is the focal length (in $cm$) of a biconvex lens with a radius of curvature of $40\, cm$ and a refractive index of $1.65$?

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