MCQ
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 dioptre]
- ✓0.04
- B0.40
- C0.1
- D0.01
✓
Answer
Correct option: A.
0.04
(A)
Sol. When $\mathrm{P}=2.5 \mathrm{D}$
$
\mathrm{F}=\frac{1}{\mathrm{P}}=\frac{1}{2.5}
$
When $\mathrm{P}^{\prime}=2.6 \mathrm{D}$
$\mathrm{F}^{\prime}=\frac{1}{\mathrm{P}^{\prime}}=\frac{1}{2.6}$
Relative decrease in focal length
$
\frac{\mathrm{F}-\mathrm{F}^{\prime}}{\mathrm{F}}=\frac{\frac{2}{5}-\frac{5}{13}}{\frac{2}{5}}=1-\frac{25}{26}=\frac{1}{26}=0.04
$
Sol. When $\mathrm{P}=2.5 \mathrm{D}$
$
\mathrm{F}=\frac{1}{\mathrm{P}}=\frac{1}{2.5}
$
When $\mathrm{P}^{\prime}=2.6 \mathrm{D}$
$\mathrm{F}^{\prime}=\frac{1}{\mathrm{P}^{\prime}}=\frac{1}{2.6}$
Relative decrease in focal length
$
\frac{\mathrm{F}-\mathrm{F}^{\prime}}{\mathrm{F}}=\frac{\frac{2}{5}-\frac{5}{13}}{\frac{2}{5}}=1-\frac{25}{26}=\frac{1}{26}=0.04
$
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