Question

Two thin converging lenses are placed on a common axis, so that the center of one of them coincides with the focus of the other. An object is placed at a distance twice the focal length from the left-hand lens. Where will its image be? What is the lateral magnification? The focal length of each lens is $f$.

• A. $0.25$
• B. $-0.5$
• C. $0.75$
• D. $-1$

Answer 1

The correct option is B $-0.5$


The image formed by first lens will be at a distance $2f$ with lateral magnification $m_1=-1$.
For the second lens this image will behave as a virtual object.
Using the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ we have, for $u=f$
$\frac{1}{v}-\frac{1}{f}=\frac{1}{f}\therefore v=\frac{f}{2}$
$m_2=\frac{v_2}{u_2}=\frac{\left(f/2\right)}{f}=\frac{1}{2}$
Therefore, final image is formed at a distance $\frac{f}{2}$ from the second lens with total lateral magnification,
$m=m_1\times m_2=(-1)\times \left(\frac{1}{2}\right)=-\frac{1}{2}$