Ray Optics

JEE Main level, single correct, medium

150 of 295

Question 150 of 295

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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Answer: (B) 0.040.04

P=1f So P1=1F1⇒F1=12.5 Next F2=12.6∣ΔFF1∣=∣F2−F1∣F1=∣F2F1−1∣⇒∣ΔFF1∣=∣2.52.6−1∣=0.12.6≈0.04\begin{aligned} & P=\frac{1}{f} \\ & \text { So } P_1=\frac{1}{F_1} \Rightarrow F_1=\frac{1}{2.5} \\ & \text { Next } F_2=\frac{1}{2.6} \\ & \left|\frac{\Delta F}{F_1}\right|=\frac{\left|F_2-F_1\right|}{F_1}=\left|\frac{F_2}{F_1}-1\right| \\ & \Rightarrow\left|\frac{\Delta F}{F_1}\right|=\left|\frac{2.5}{2.6}-1\right|=\frac{0.1}{2.6} \approx 0.04\end{aligned}

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