Ray Optics

JEE Main level, single correct, medium

30 of 295

Question 30 of 295

A convex mirror of radius of curvature 30cm30cm forms an image that is half the size of the object. The object distance is :

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Answer: (C) −15cm-15cm

Given R=30cmR=30cm
So, f=R2=+15cmf=\dfrac{R}{2}=+15cm
For convex mirror, virtual image is formed for real object. Thus, magnification (m)=12\left(m\right)=\dfrac{1}{2}
The magnification can be written as
m=−vu...(1)m=-\dfrac{v}{u}...\left(1\right)
Equation (1) implies that
12=−vu⇒v=−u2\dfrac{1}{2}=-\dfrac{v}{u} \Rightarrow v=-\dfrac{u}{2}
The relation between the object distance, image distance and the focal length of the mirror is given by
1f=1v+1u...(2)\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}...\left(2\right)
Equation (2) implies that
115=1−u2+1u=−1u⇒u=−15cm\dfrac{1}{15}=\dfrac{1}{-\dfrac{u}{2}}+\dfrac{1}{u} =-\dfrac{1}{u} \Rightarrow u=-15cm

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