Question 116: Ray Optics

Question

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 for dioptre]

Accepted Answer

$$ P_1 = 2.5\,D,\quad f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4\,m $$
After an increase in optical power by 0.1 D:
$$ P_2 = 2.5\,D + 0.1\,D = 2.6\,D,\quad f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846\,m $$
The change in focal length is:
$$ \Delta f = f_1 - f_2 = 0.4\,m - 0.3846\,m \approx 0.0154\,m $$
The relative decrease in focal length is then calculated as:
$$ 	ext{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} \approx 0.0385 \approx 0.04 $$
Thus, the relative decrease in focal length is approximately
0.04
.

Answer

0.04

Answer

0.1

Answer

0.40

Answer

0.01

Ray Optics

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 for dioptre]

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