To replace a bi-convex lens with another convex lens that has different radii of curvature on each side (i.e., $ R_1 \neq R_2 $), while maintaining the same lens power, the radii must satisfy a specific relationship.

For the current bi-convex lens, both radii of curvature are given as $ \frac{1}{6} \, \text{cm} $. The lens formula for power is tied to the radii through:

$ \frac{1}{f_1} = (\mu - 1) \left( \frac{2}{R} \right) $

When this lens is replaced by a lens with different radii ($ R_1 $ and $ R_2 $), the condition that has to be met is:

$ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $

For power equivalence, the expressions for $ \frac{1}{f_1} $ and $ \frac{1}{f_2} $ must be equal:

$ (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = (\mu - 1) \left( \frac{2}{R} \right) $

This simplifies to:

$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} $

Given $ R = \frac{1}{6} \, \text{cm} $, it follows that:

$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{\left(\frac{1}{6}\right)} $

Thus, simplifying the equation gives:

$ \frac{1}{R_1} + \frac{1}{R_2} = 12 $

Therefore, any valid pair of radii $ R_1 $ and $ R_2 $ must satisfy this equation.