Question

When an object is moved along the principal axis of a thin converging lens of focal length 12 cm, two images, three times the size of the object are obtained when the object is at $x_1$ and $x_2$ distance from the lens. Which one of the following could be the value of the ratio $\frac{x_1}{x_2}$ ?

• A. 3
• B. 6
• C. 4
• D. 2

Answer 1

Answer: (D)

2

Using lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

We have,

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

$\frac{1}{v}=\frac{u+f}{fu}$

$\Rightarrow v=\frac{uf}{u+f}$

We know,

$m=\frac{v}{u}$

$\Rightarrow m=\frac{(\frac{uf}{u+f})}{u}=\frac{f}{u+f}$

Given, $f=12cm$

Also, $m=\pm 3$, because image can be 3 times the size of object but can be real and inverted as well as virtual and erect.

$\Rightarrow \pm 3=\frac{f}{u+f}$

CASE 1: Let $u=x_1$ and image be real and inverted

$-3=\frac{12}{x_1+12}$

$-3x_1-36=12\Rightarrow -3x_1=48\Rightarrow x_1=-16$

$\Rightarrow x_1=-16cm$

CASE 2: Let $u=x_2$ and image be virtual and erect

$3=\frac{12}{x_2+12}$

$3x_2+36=12\Rightarrow 3x_2=-24\Rightarrow x_2=-8$

$\Rightarrow x_2=-8cm$

Now,

$\frac{x_1}{x_2}=\frac{-16}{-8}=2$

Hence, the correct answer is OPTION D.