Radii of curvature of a converging lens are i | vedclass

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(D) Given: Ratio of radii of curvature $\frac{R_{1}}{R_{2}} = \frac{1}{2}$, focal length $f = 6 \,cm$, and refractive index $\mu = 1.5$.Let $R_{1} = R

Vedclass · Accepted Answer

$4.5 \,cm$ and $9 \,cm$

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(D) Given: Ratio of radii of curvature $\frac{R_{1}}{R_{2}} = \frac{1}{2}$, focal length $f = 6 \,cm$, and refractive index $\mu = 1.5$.Let $R_{1} = R$ and $R_{2} = 2R$.For a converging lens, the lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_{1}} + \frac{1}{R_{2}} ight)$.Substituting the values: $\frac{1}{6} = (1.5 - 1) \left( \frac{1}{R} + \frac{1}{2R} ight)$.$\frac{1}{6} = 0.5 \left( \frac{2 + 1}{2R} ight) = 0.5 \left( \frac{3}{2R} ight) = \frac{1.5}{2R} = \frac{3}{4R}$.Solving for $R$: $4R = 18$, so $R = 4.5 \,cm$.Therefore, $R_{1} = 4.5 \,cm$ and $R_{2} = 2 	imes 4.5 = 9 \,cm$.

Radii of curvature of a converging lens are in the ratio $1 : 2$. Its focal length is $6 \,cm$ and refractive index is $1.5$. Then its radii of curvature are ......... respectively.

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