Ray Optics

JEE Main level, single correct, medium

181 of 295

Question 181 of 295

An object is placed at the focus of concave lens having focal length f.f. What is the magnification and distance of the image from the optical centre of the lens?

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Answer: (B) 12,f2\dfrac{1}{2},\dfrac{f}{2}

1F=1v−1u\dfrac{1}{F}=\dfrac{1}{v}-\dfrac{1}{u}
Put u=−fu=-f and F=−fF=-f
1−f=1v−1−f;−1f=1f−1v;−2f=1v⇒v=−f2\dfrac{1}{-f}=\dfrac{1}{v}-\dfrac{1}{-f};\dfrac{-1}{f}=\dfrac{1}{f}-\dfrac{1}{v};\dfrac{-2}{f}=\dfrac{1}{v}\Rightarrow v=-\dfrac{f}{2}
Magnification =v/u=−f2(−f)=12=v/u=\dfrac{-f}{2(-f)}=\dfrac{1}{2}

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