Question

When an object is at distances x and y from a lens, a real image and a virtual image is formed respectively having same magnification. The focal length of the lens is

• A. $\frac{x+y}{2}$
• B. $x-y$
• C. $\sqrt{xy}$
• D. $x+y$

Answer 1

The correct option is A $\frac{x+y}{2}$

The given lens is a convex lens. Let the magnification be m, then for real image
$\frac{1}{mx}+\frac{1}{x}=\frac{1}{f}$
$\Rightarrow \frac{1}{x}(\frac{1}{m}+1)=\frac{1}{f}$
$\Rightarrow \frac{1}{m}=(\frac{x}{f}-1)$ ----- (1)
And for virtual image $\frac{1}{-my}+\frac{1}{y}=\frac{1}{f}$
putting $\frac{1}{m}=(\frac{x}{f}-1)$ in the above equation,
$-(\frac{x}{f}-1)\frac{1}{y}+\frac{1}{y}=\frac{1}{f}$
taking $\frac{1}{y}$ out as common, we have
$\frac{1}{y}(1-\frac{x}{f}+1)=\frac{1}{f}$
$2-\frac{x}{f}=\frac{y}{f}$
$2=\frac{y+x}{f}$ or,
$f=\frac{x+y}{2}$