Question

In which the power of a lens will be large, in air or water?

Hint:

Use \( \dfrac{1}{f} = \left(\dfrac{n_l}{n_m}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right) \); the smaller the difference between lens and medium index, the smaller the power.

Answer 1

Step 1 (Why the medium matters): A lens works only because light bends at its surfaces, and bending depends on how different the lens's refractive index is from the surrounding medium. The closer the medium's index is to the lens's index, the weaker the bending.
Step 2 (Air versus water): Water (\(n = 1.33\)) is optically much closer to glass (\(n = 1.5\)) than air (\(n = 1\)) is. So when the lens sits in water, the relative index \(n_l/n_m\) drops from \(1.5\) to about \(1.13\), and the refracting power of each surface shrinks.
Step 3 (Effect on power): Because \(P = 1/f\) and a weaker bend means a longer focal length, the focal length grows in water and the power falls. In air the contrast in indices is largest, so the focal length is shortest and the power is greatest.
\[\boxed{\text{Power is greater in air than in water.}}\]