Ray Optics

JEE Main level, single correct, medium

25 of 295

Question 25 of 295

When an object is placed 40 cm away from a spherical mirror an image of magnification 12\frac{1}{2} is produced. To obtain an image with magnification of 13\frac{1}{3}, the object is to be moved :

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Answer: (A) 40 cm away from the mirror.

m=12=ff−u12=ff−(−40)f+40=2f⇒f=40 cm now m=13=4040−u40−u=120⇒u=−80\begin{aligned} & \mathrm{m}=\frac{1}{2}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}} \\ & \frac{1}{2}=\frac{\mathrm{f}}{\mathrm{f}-(-40)} \\ & \mathrm{f}+40=2 \mathrm{f} \Rightarrow \mathrm{f}=40 \mathrm{~cm} \\ & \text { now } \mathrm{m}=\frac{1}{3}=\frac{40}{40-\mathrm{u}} \\ & 40-\mathrm{u}=120 \Rightarrow \mathrm{u}=-80\end{aligned}

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