The image formed by a convex lens is erect and its size is four times the size of the object. If the focal length of the lens is \( 20 \) cm, the object and image distances are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -15 \) cm and \( -60 \) cm, respectively
A convex lens forms an erect, magnified image only when the object is inside the focus; the image is then virtual and on the same side as the object. With the Cartesian sign convention the lens magnification is \( m = v/u \), positive for an erect image.
\( m = v/u = +4 \Rightarrow v = 4u \)
\( f = +20\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{4u} - \frac{1}{u} = \frac{1}{20} \)
\( \displaystyle -\frac{3}{4u} = \frac{1}{20} \Rightarrow u = -15\,\mathrm{cm} \)
\( v = 4u = -60\,\mathrm{cm} \)
Both distances are negative: object and image are on the same side of the lens, as expected for a virtual image.
Shortcut: An erect image by a lens needs \( u \) and \( v \) of the same sign with \( v/u = 4 \). Only Options (1) and (3) have the same sign, and only Option (3) has \( v/u = 60/15 = 4 \).