Question

The lateral magnification of the lens with an object located at two different positions $u_1$ and $u_2$ are $m_1$ and $m_2$, respectively. Then the focal length of the lens is:
A. $f=\sqrt{{m}_{1}{m}_{2}}({u}_{2}-{u}_{1})$
B. $\dfrac{{m}_{2}{u}_{2}–{m}_{1}{u}_{1}}{{m}_{2}–{m}_{1}}$
C. $\dfrac{({u}_{2}–{u}_{1})}{\sqrt{{m}_{2}{m}_{1}}}$
D. $\dfrac{({u}_{2}–{u}_{1})}{{{m}_{2}}^{-1} – {{m}_{1}}^{-1}}$

Answer 1

Hint: By using the lens formula, we will first find out the ratio of the image and object distance which will give us the expression for lateral magnification. Then we will find the lateral magnification for the object at two given locations and upon equation both, we will get the required result.

Formulae used:
Lens formula, $\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$
Lateral magnification, $m=\dfrac{h_i}{h_o}=\dfrac{v}{u}=\dfrac{f}{u+f}$

Complete step by step solution:
Since, for a lens with focal length $f$ we know that, $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$, where $u$ is the object distance from the optical centre and $v$ is the image distance.
$\implies \dfrac{u}{f}=\dfrac{u}{v}-1 \implies \dfrac{v}{u}=\dfrac{f}{u+f}$
Now, lateral magnification for a lens is given by, $m=\dfrac{v}{u}=\dfrac{f}{u+f}$
Now, for the object at location $u_1$, lateral magnification $m_1$ is given by,
$m_1=\dfrac{f}{f+u_1}\implies \dfrac{1}{m_1}=\dfrac{u_1 +f}{f}$ ………. (i)
Similarly, for the object at distance $u_2$, lateral magnification $m_2$ is given by
$m_2 =\dfrac{f}{u_2 +f}\implies \dfrac{1}{m_2}=\dfrac{u_2 +f}{f}$ ………. (ii)
Noe, subtracting equation (i) from equation (ii), we get
$\dfrac{1}{m_2}-\dfrac{1}{m_1}=\dfrac{u_2 -u_1}{f}\implies f=\dfrac{u_2 – u_1}{(m_2)^{-1}-(m_1)^{-1}}$
Hence, option d is the correct answer.

Note: The lateral magnification is given by the ratio of height of image to the height of object. In a lens, the triangle formed by the optical centre, image distance and the height of image is similar to that of the triangle formed by the optical centre, object distance and object height. And for similar triangles, the ratio of corresponding sides is equal and that is what we have used to solve the given problem.