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 is 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

Hint: In this question, we have to assume the object distance as $x$ and the image distance will become $D-x$ as the lens of focal length $f$ is placed in between the object and a screen. After applying the lens formula we get the required answer.

Formula used:
Lens equation is used.
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
Where f is the focal length of the lens, u is the object distance and v is the image distance.

Complete step by step solution:
It is given that the luminous object and screen are placed D distance apart. A converging lens of focal length f is placed between object and screen. Let the object distance be x. Then, the image distance is D-x. Let d be the separation of distance of two lens positions for which real image is formed.

Now by applying lens equation we get:
$\dfrac{1}{x}+\dfrac{1}{D-x}=\dfrac{1}{f}$
On doing further simplification, we get distance D as
${{x}^{2}}-Dx+Df=0$
On solving this quadratic equation, we get:
$x=\dfrac{D-\sqrt{D(D-4f)}}{2} \\ $
$\Rightarrow x'=\dfrac{D+\sqrt{D(D-4f)}}{2}$
Therefore, the distance between the two object positions is
$\therefore d=x'-x=\sqrt{D(D-4f)}$
That is, a real image of the object is formed on the screen for two lens positions if they are separated by a distance d equal to $\sqrt{D(D-4f)}$.

Therefore, the correct answer is option B.

Notes: We have to apply sign conversion to the lens equation. Remember we have to find the distance of separation not the distance of formation of real image. We also have to note that converging lenses are used. Note that mirror formula, lens formula and lens maker’s formula are different.