Question

A 16cm long image of an object is formed by a convex lens on a screen. On moving the lens towards the screen, a $9{\text{cm}}$ long image is formed again on the screen. Find the size of the object.
A) $9{\text{cm}}$
B) ${\text{11cm}}$
C) ${\text{12cm}}$
D) ${\text{13cm}}$

Answer 1

Hint: Here as the lens is moved towards the screen, the height of the image formed decreases. According to the displacement method, the height of the object and the height of the two images formed are related to each other. The displacement method gives the size of the object as the square root of the product of the heights of the images formed before and after the lens is moved.

Formula used:
The height of the object is given by,
${h_o} = \sqrt {{h_1}{h_2}} $, where ${h_1}$ is the height of the image before the position of the lens is changed and ${h_2}$ is the height of the image after the position of the lens is changed.

Complete step by step answer:
Step 1: List the parameters given in the question.
The problem at hand discusses the scenario where a change in the position of the convex lens causes a corresponding change in the size of the images formed on the screen by the given lens.
The size of the image formed on the screen before moving the lens is given to be ${h_1} = 16{\text{cm}}$.
The size of the image formed on the screen after moving the lens is given to be ${h_2} = 9{\text{cm}}$.
We have to determine the size of the object ${h_o}$ whose images are formed on the screen for two different positions from the screen.
Step 2: Express the relation for the height of the object based on the displacement method.
The displacement method gives the height of the object as
${h_o} = \sqrt {{h_1}{h_2}} $ ---------- (1)
Substituting for ${h_1} = 16{\text{cm}}$ and ${h_2} = 9{\text{cm}}$ in equation (1) we get,
$\Rightarrow {h_o} = \sqrt {16 \times 9} = 12{\text{cm}}$ .
Thus the height of the object is obtained as ${h_o} = 12{\text{cm}}$.

Therefore, the correct option is C.

Note:
In the displacement method, two positions exist for an object where a clear and sharp image of the object is obtained on a screen. The convex lens is a converging lens and it forms magnified images. The product of the magnification of the two images is found to be one i.e., ${m_1}{m_2} = 1$, where ${m_1} = \dfrac{{{h_1}}}{{{h_o}}}$ is the magnification of the image for the first position and ${m_2} = \dfrac{{{h_2}}}{{{h_o}}}$ is the magnification of the image formed for the second position of the lens. This magnifying property of the convex lens is used to obtain the relation ${h_o} = \sqrt {{h_1}{h_2}} $.