Question

A luminous object and a screen are at a fixed distance $D$ apart. A converging lens of focal length $f$ is placed between the object and screen. A real image of the object in formed on the screen for two lens positions if they are separated by a distance $d$ equal to

• A. $\sqrt{D(D+4f)}$
• B. $\sqrt{D(D-4f)}$
• C. $\sqrt{2D(D-4f)}$
• D. $\sqrt{D^2+4f}$

Answer 1

Answer: (B)

$\sqrt{D(D-4f)}$

Let the object distance be $x$. Then, the image distance is $D-x$.
From lens equation,
$\frac{1}{x}+\frac{1}{D-x}=\frac{1}{f}$
On algebraic rearrangement, we get
$x^2-Dx+Df=0$
On solving for $x$, we get
$x_1=\frac{D-\sqrt{D(D-4f)}}{2}$
$x_2=\frac{D+\sqrt{D(D-4f)}}{2}$
The distance between the two object positions is
$d=x_2-x_1=\sqrt{D(D-4f)}$.