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 of they are separated by a distance $$d$$ equal to
A.
$$\sqrt {D\left( {D + 4f} \right)} $$
B.
$$\sqrt {D\left( {D - 4f} \right)} $$
C.
$$\sqrt {2D\left( {D - 4f} \right)} $$
D.
$$\sqrt {{D^2} + 4f} $$
Answer :
$$\sqrt {D\left( {D - 4f} \right)} $$
Solution :
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
$$x2 - Dx + Df = 0$$
On solving for $$x,$$ we get
$$\eqalign{
& {x_1} = \frac{{D - \sqrt {D\left( {D - 4f} \right)} }}{2} \cr
& {x_2} = \frac{{D + \sqrt {D\left( {D - 4f} \right)} }}{2} \cr} $$
The distance between the two object positions is
$$d = {x_2} - {x_1} = \sqrt {D\left( {D - 4f} \right)} $$