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]
Solution
<p>$P_1 = 2.5\,D,\quad f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4\,m$</p>
<p>After an increase in optical power by 0.1 D:</p>
<p>$$ P_2 = 2.5\,D + 0.1\,D = 2.6\,D,\quad f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846\,m $$</p>
<p>The change in focal length is:</p>
<p>$\Delta f = f_1 - f_2 = 0.4\,m - 0.3846\,m \approx 0.0154\,m$</p>
<p>The relative decrease in focal length is then calculated as:</p>
<p>$$ \text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} \approx 0.0385 \approx 0.04 $$</p>
<p>Thus, the relative decrease in focal length is approximately <strong>0.04</strong>.</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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