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(B) The power of a lens is given by $P = \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a bi-convex lens,$R_1 = R$ and $R_2

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$1/5 \ cm$ and $1/7 \ cm$

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(B) The power of a lens is given by $P = \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a bi-convex lens,$R_1 = R$ and $R_2 = -R$,so $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} ight) = (\mu - 1) \frac{2}{R}$.Given $R = 1/6 \ cm$,the power is proportional to $\frac{2}{R} = 2 	imes 6 = 12 \ cm^{-1}$.For the new lens,we need $\frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} = 12$.Checking the options:For option $C$: $R_1 = 1/3 \ cm$ and $R_2 = 1/7 \ cm$,we get $\frac{1}{R_1} + \frac{1}{R_2} = 3 + 7 = 10 
eq 12$.Wait,let's re-evaluate: $\frac{1}{R_1} + \frac{1}{R_2} = 12$.For option $B$: $R_1 = 1/5 \ cm$ and $R_2 = 1/7 \ cm$,we get $5 + 7 = 12$.Thus,the correct combination is $R_1 = 1/5 \ cm$ and $R_2 = 1/7 \ cm$.

$A$ bi-convex lens has a radius of curvature for both surfaces equal to $1/6 \ cm$. If this lens is to be replaced by another convex lens having different radii of curvature on both sides $(R_1 \neq R_2)$,without any change in the lens power,then the possible combination of $R_1$ and $R_2$ is:

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