Question

A lens has a power of +5 diopter in the air. What will be its power if completely immersed in water? The refractive index of the lens in air is 1.5 $({\mu }_w=\frac{4}{3})$

• A. $+\frac{5}{4}D$
• B. $+\frac{21}{4}D$
• C. $+\frac{10}{3}D$
• D. $+5D$

Answer 1

The correct option is A $+\frac{5}{4}D$

from lens makers formula

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})=Power$

$+5=\left(\mu 005\right)(\frac{1}{R_1}-\frac{1}{R_2})$

$P_7=(\frac{3/2}{4/3}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{5}{P}=\frac{0.5}{1/8}=\frac{8}{2}$

$P=\frac{5}{4}D=+1.25D$