A bi-convex lens has radius of curvature of both the surfaces same as $1 / 6 \mathrm{~cm}$. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides $\left(R_1 \neq R_2\right)$, without any change in lens power then possible combination of $R_1$ and $R_2$ is :

<p>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. </p> <p>For the current bi-convex lens, both radii of curvature are given as $ \frac{1}{6} \, \text{cm} $. The lens formula for power is tied to the radii through:</p> <p>$ \frac{1}{f_1} = (\mu - 1) \left( \frac{2}{R} \right) $</p> <p>When this lens is replaced by a lens with different radii ($ R_1 $ and $ R_2 $), the condition that has to be met is:</p> <p>$ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $</p> <p>For power equivalence, the expressions for $ \frac{1}{f_1} $ and $ \frac{1}{f_2} $ must be equal:</p> <p>$ (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = (\mu - 1) \left( \frac{2}{R} \right) $</p> <p>This simplifies to:</p> <p>$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} $</p> <p>Given $ R = \frac{1}{6} \, \text{cm} $, it follows that:</p> <p>$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{\left(\frac{1}{6}\right)} $</p> <p>Thus, simplifying the equation gives:</p> <p>$ \frac{1}{R_1} + \frac{1}{R_2} = 12 $</p> <p>Therefore, any valid pair of radii $ R_1 $ and $ R_2 $ must satisfy this equation.</p>