Question

A convex lens of focal length $20cm$ produces image of the same magnitude $2$ when an object is kept at two distances ${{x}_{1}}$ and ${{x}_{2}}({{x}_{1}}>{{x}_{2}})$ from the lens. The ratio of ${{x}_{i}}$ and ${{x}_{2}}$ is?
$A.5:3$
$B.2:1$
$C.4:3$
$D.3:1$

Answer 1

Magnification of a lens is defined as the ratio of height of image to the height of object. We have to apply the relation for magnification of a lens with its focal length. Convex lens produces real images at all possible positions of objects except when the object is placed between Focus and Pole. As ${{x}_{1}}>{{x}_{2}}$, ${{x}_{1}}$ is object distance for real image and ${{x}_{2}}$ for virtual image respectively. Lens formula and formula for magnification of lenses will be used to solve this problem.

Formula used:
In solving this particular problem, we first apply lens formula and then formula for linear magnification. Lens formula and formula of magnification are given below respectively:-
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ and $m=\dfrac{v}{u}$.

Complete answer:
Magnification is defined as the ratio of height of image to the height of object. If $m=$magnification, $u=$object distance and $v=$image distance, then
$m=\left| \dfrac{v}{u} \right|$.
$2=\left| \dfrac{v}{u} \right|$.
$v=2u$.
For real images in convex lenses we have Object distance as negative and image distance as positive. From the sign convention we have
$u=-{{x}_{1}}$and $v=+2{{x}_{1}}$
Now, applying lens formula we get
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{2{{x}_{1}}}-\dfrac{1}{-{{x}_{1}}}=\dfrac{1}{f}$
$\dfrac{1}{2{{x}_{1}}}+\dfrac{1}{{{x}_{1}}}=\dfrac{1}{f}$
$\dfrac{3}{2{{x}_{1}}}=\dfrac{1}{f}$
${{x}_{1}}=\dfrac{3f}{2}$…………….. $(i)$
For a virtual image in a convex lens we have object distance as negative and image distance is negative also. From the sign convention we have
$u=-{{x}_{2}}$and $v=-2{{x}_{2}}$
Now, applying lens formula we get
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{-2{{x}_{2}}}-\dfrac{1}{-{{x}_{2}}}=\dfrac{1}{f}$
$\dfrac{1}{-2{{x}_{2}}}+\dfrac{1}{{{x}_{2}}}=\dfrac{1}{f}$
$\dfrac{1}{2{{x}_{2}}}=\dfrac{1}{f}$
${{x}_{2}}=\dfrac{f}{2}$…………… $(ii)$
On dividing $(i)$by $(ii)$we get
$\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{\dfrac{3f}{2}}{\dfrac{f}{2}}$
$\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{3f\times 2}{2\times f}$
On simplification, we get
$\dfrac{{{x}_{1}}}{{{x}_{2}}}=\dfrac{3}{1}$
So, ${{x}_{1}}:{{x}_{2}}=3:1$

So, the correct answer is “Option D”.

Note:
We should take care of sign conventions for convex and concave lenses respectively. We should also take care of the fact that lens formula should be used for lenses and mirror formula for mirrors respectively. We should take care that linear magnification and angular magnification are different terms.