Question 104: Ray Optics

Question

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 $\left(R_1 
eq R_2ight)$, without any change in lens power then possible combination of $R_1$ and $R_2$ is :

Accepted Answer

To replace a bi-convex lens with another convex lens that has different radii of curvature on each side (i.e., $ R_1 
eq 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} \, 	ext{cm} $. The lens formula for power is tied to the radii through:
$ \frac{1}{f_1} = (\mu - 1) \left( \frac{2}{R} ight) $
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} ight) $
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} ight) = (\mu - 1) \left( \frac{2}{R} ight) $
This simplifies to:
$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} $
Given $ R = \frac{1}{6} \, 	ext{cm} $, it follows that:
$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{\left(\frac{1}{6}ight)} $
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.

Answer

$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$

Answer

$\frac{1}{5} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$

Answer

$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{3} \mathrm{~cm}$

Answer

$\frac{1}{6} \mathrm{~cm}$ and $\frac{1}{9} \mathrm{~cm}$

Ray Optics

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