Magnification By Lenses

Question

A 2 cm tall arrow is positioned 20 cm in front of a thin convex lens of focal length 15 cm. What is the height of the real image formed on the far side of the lens?

RankGuru · Accepted Answer

Image height depends on the linear magnification $m=\frac{v}{u}$ for a lens, where $v$ and $u$ carry their Cartesian signs and image height $h'=m\,h$. First locate the image using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $f=+15$ cm and $u=-20$ cm: $\frac{1}{v}=\frac{1}{15}+\frac{1}{-20}=\frac{4-3}{60}=\frac{1}{60}$, so $v=+60$ cm. Then $m=\frac{v}{u}=\frac{60}{-20}=-3$, and the image height is $h'=m\,h=-3	imes2=-6$ cm, meaning a 6 cm inverted image. The sign of the magnification carries genuine physical meaning: a negative value tells us the image is inverted relative to the object, a direct consequence of rays crossing the principal axis after passing through a converging lens to form a real image. Whenever a real object lies between the focus and twice the focal length of a convex lens, the image is necessarily real, inverted and enlarged, which is the regime exploited in a projector. The value 2 cm is wrong because it assumes unit magnification, ignoring the object's position. The value 4 cm is wrong as it would need a magnification of 2, corresponding to a different geometry. The value 1.5 cm is wrong since it inverts the magnification ratio. This NCERT relation links image size to object and image distances. A magnitude check confirms the image is enlarged and inverted, consistent with an object placed between $f$ and $2f$.

Magnification By Lenses

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