Question

A thin glass ($\mu$=1.5) lens has optical power of -10D in air. Its optical power in a liquid medium with refractive index 1.6 will be

• A. +1D
• B. +2D
• C. +1.25 D
• D. +2.5 D

Answer 1

Answer: (C)

+1.25 D

Answer 2

The optical power $P$ of a lens is given by $P = \frac{1}{f} = (\frac{\mu_{lens}}{\mu_{medium}} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$. Let $C = (\frac{1}{R_1} - \frac{1}{R_2})$ be the curvature term.

Power in air ($P_{air}$): $P_{air} = (\frac{\mu_g}{\mu_{air}} - 1) C$ Given $\mu_g = 1.5$, $\mu_{air} = 1$, and $P_{air} = -10$ D. $-10 = (\frac{1.5}{1} - 1) C = 0.5 C$ $C = \frac{-10}{0.5} = -20$ m$^{-1}$.

Power in liquid ($P_{liquid}$): $P_{liquid} = (\frac{\mu_g}{\mu_l} - 1) C$ Given $\mu_g = 1.5$ and $\mu_l = 1.6$. $P_{liquid} = (\frac{1.5}{1.6} - 1) (-20)$ $P_{liquid} = (\frac{15}{16} - 1) (-20)$ $P_{liquid} = (\frac{-1}{16}) (-20)$ $P_{liquid} = \frac{20}{16} = 1.25$ D.

Alternatively, using the ratio of powers: $\frac{P_{liquid}}{P_{air}} = \frac{(\frac{\mu_g}{\mu_l} - 1)}{(\frac{\mu_g}{\mu_{air}} - 1)}$ $\frac{P_{liquid}}{-10} = \frac{(\frac{1.5}{1.6} - 1)}{(\frac{1.5}{1} - 1)} = \frac{(\frac{-1}{16})}{(0.5)} = \frac{-1}{8}$ $P_{liquid} = -10 \times (\frac{-1}{8}) = \frac{10}{8} = 1.25$ D.