Question

When an object is at distances $u_1$ and $u_2$ from a lens a real image and a virtual image is formed respectively having the same magnification. The focal length of the lens is

• A. $u_1+\frac{u_2}{2}$
• B. $\frac{u_1-u_2}{2}$
• C. $\frac{u_1+u_2}{2}$
• D. $u_1+u_2$

Answer 1

Answer: (C)

: For a convex lens when real image is formed, $u_1$ is negative, $v_1$ is positive, $f$ is positive. Lens formula is $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\therefore$ For real image, $\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f}$ $\therefore$ $\frac{u_1}{v_1}+\frac{u_1}{u_1}=\frac{u_1}{f}$ $\therefore$ $\frac{u_1}{v_1}=\left(\frac{u_1}{f}-1\right)$ ?.(i) When virtual image is formed, $u_2$ is negative, $v_2$ is also negative, $f$ is positive $\therefore$ For virtual image, $\frac{1}{-v_2}+\frac{1}{u_2}=\frac{1}{f}$ Or $\frac{u_2}{-v_2}+\frac{u_2}{u_2}=\frac{u_2}{f}$ or $\frac{u_2}{v_2}-1=-\frac{u_2}{f}$ Or $\frac{u_2}{v_2}=1-\frac{u_2}{f}$ ...(ii) Since magnification is same in the two cases. $\frac{v_2}{u_2}=\frac{v_2}{u_1}$ or $\frac{u_2}{v_2}=\frac{u_2}{v_1}$ .Substitute for these from (i) and (ii). Or $\left(1-\frac{u_2}{f}\right)=\left(\frac{u_1}{f}-1\right)$ or $1+1=\frac{u_1}{f}+\frac{u_2}{f}$ Or $2=\frac{(u_1+u_2)}{f}$ or $f=\frac{u_1+u_2}{2}$