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 (Lens maker's formula): The focal length of a lens placed in a medium of refractive index \( n_m \) is given by
\[ \frac{1}{f} = \left(\frac{n_l}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
and the power is \( P = \dfrac{1}{f} \), so \( P \propto \left(\dfrac{n_l}{n_m} - 1\right) \).
Step 2 (Compare the two media): For air \( n_m = 1 \); for water \( n_m = 1.33 \). The lens (glass) has \( n_l = 1.5 \).
In air: \( \dfrac{n_l}{n_m} - 1 = \dfrac{1.5}{1} - 1 = 0.5 \).
In water: \( \dfrac{n_l}{n_m} - 1 = \dfrac{1.5}{1.33} - 1 = 0.13 \).
Step 3 (Conclusion): Since the bracket, and hence the power, is much larger in air, the lens bends light more strongly and has greater power in air. In water its power falls (focal length increases).
\[\boxed{\text{The power of the lens is larger in air.}}\]