Lens Displacement Method MCQ - Practice Questions with Answers

Displacement Method-

This method helps us to find the another position of the object which cause the sharp image on the screen. 

For this let us consider a convex lens L placed between an object O and a screen S. If the distance between the object and the screen is D and the positions of the object and the screen are held fixed. The lens is allowed to move along the axis of the system and at a position 'I' a sharp image will be formed on the screen. Interestingly, there is another position on the same axis where a sharp image will once again be obtained on the screen. See in the figure given below that the second position is 'II'

In the figure, let the distance of position I from the object be $x_1$. Then, the distance of the screen from the lens is $D-x_1$. Therefore, $u=-x_1$ and $v=+\left(D-x_1\right)$.

Substituting in the lens equation -

$$
\begin{gathered}
\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
\Rightarrow \frac{1}{D-x_1}+\frac{1}{x_1}=\frac{1}{f} \ldots
\end{gathered}
$$

At position II, let the distance of the lens from the screen be $x_2$. Then, the distance of the lens from the object is $D-x_2$. Therefore, $u=-x_2$ and $v=+\left(D-x_2\right)$.

Now substituting in lens formula -

$$
\frac{1}{D-x_2}+\frac{1}{x_2}=\frac{1}{f} \ldots
$$

Comparing (1) and (2), we get -
1. $x_1=x_2$; or
2. $D-x_1=x_2$ and $D-x_2=x_1$

The first solution is trivial. Therefore, if the first position of the lens, for a sharp image, is $x_1$ from the object, the second position is at $D-x_1$ from the object. Let the distance between the two positions I and If be d . From the diagram, it is clear that

$$
D=x_1+x_2 \quad \text { and } \quad d=x_2-x_1
$$

Solving the two equations in (iii), we obtain

$$
x_1=\frac{D-d}{2} \text { and } D-x_1=\frac{D+d}{2} \ldots
$$

Substituting Eq. (iv) in Eq. (i), we get

$$
\begin{gathered}
\frac{1}{f}=\frac{2}{D-d}+\frac{2}{D+d} \\
f=\frac{D^2-d^2}{4 D} \ldots
\end{gathered}
$$

Also, $d=\sqrt{D^2-4 D f}=\sqrt{D(D-4 f)} \ldots$

                                          We notice from Eq. (v) that a solution for $d$ is possible only when $D \geq 4 f$.
When $D<4 f$, there is no position for which a sharp image can be formed.
When $D=2 f$, there is only one position where a sharp image is formed.
When $D>2 f$, there are two positions where a sharp image is formed.

Note - This method is applicable to convex lens, not for concave lens