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

Initial optical power, $P_1 = 2.5D$

Final optical power, $P_2 = 2.5D + 0.1D = 2.6D$

Step 1: Relation Between Focal Length and Power

The focal length $f$ (in meters) of a lens is related to its optical power $P$ by:

$$P = \frac{1}{f}$$

So,

$$f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \text{ m}$$

 $$f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \text{ m}$$

Step 2: Relative Decrease in Focal Length

The relative decrease in focal length is given by:

$$\frac{\Delta f}{f_1} = \frac{f_1 - f_2}{f_1}$$

Substituting the values:

$$\frac{\Delta f}{f_1} = \frac{0.4 - 0.3846}{0.4}$$

 $$= \frac{0.0154}{0.4}$$=0.0385 which is approx 0.04

Final Answer: 0.04