Ray Optics

JEE Main level, single correct, medium

22 of 295

Question 22 of 295

A mirror is used to produce an image with magnification of 14\frac{1}{4}. If the distance between object and its image is 40 cm , then the focal length of the mirror is ________

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Answer: (C) 10.7 cm

m=−vu=−(v−u)=vu14=vu⇒u=4vv+u=405v=40v=8 cmu=32 cm1v+vu=1f18−132=1f4−132=1ff=323=10.7 cm\begin{aligned} & \mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=-\left(\frac{\mathrm{v}}{-\mathrm{u}}\right)=\frac{\mathrm{v}}{\mathrm{u}} \\ & \frac{1}{4}=\frac{\mathrm{v}}{\mathrm{u}} \Rightarrow \mathrm{u}=4 \mathrm{v} \\ & \mathrm{v}+\mathrm{u}=40 \\ & 5 \mathrm{v}=40 \\ & \mathrm{v}=8 \mathrm{~cm} \\ & \mathrm{u}=32 \mathrm{~cm} \\ & \frac{1}{\mathrm{v}}+\frac{\mathrm{v}}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{8}-\frac{1}{32}=\frac{1}{\mathrm{f}} \\ & \frac{4-1}{32}=\frac{1}{\mathrm{f}} \\ & \mathrm{f}=\frac{32}{3}=10.7 \mathrm{~cm}\end{aligned}

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