What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5D? ['D' stands…

The focal length F is the reciprocal of optical power P :

\(F=\frac{1}{P}\)

When P =2.5 D,

\(F=\frac{1}{2.5}=\frac{2}{5}\)

When P increases to 2.6 D ,

\({F}^{'}=\frac{1}{2.6}=\frac{5}{13}\)

Relative decrease in focal length:

\(\frac{F-{F}^{'}}{F}=\frac{\frac{2}{5}-\frac{5}{13}}{\frac{2}{5}}\)

Simplifying:

\(=\frac{26-25}{26}=\frac{1}{26}\approx 0.04\)

the correct answer is 0.04 .