Question

The magnification produced by a convex lens for two different positions of an object are ${m_1}$ and ${m_2}$ respectively (${m_1} > {m_2}$). If $'d'$ is the separation between the two positions of the object then the focal length of the lens is
A. $\sqrt {{m_1}{m_2}} $
B. $\dfrac{1}{{\sqrt {{m_1} - {m_2}} }}$
C. $\dfrac{{d\,{m_1}{m_2}}}{{{m_1} - {m_2}}}$
D. $\dfrac{d}{{{m_1} - {m_2}}}$

Answer 1

Hint: Here, we will use the formula of magnification of the lens and the lens formula to calculate the focal length of the lens. Focal length is defined as the distance of the lens from the focal point of the lens.

Complete step by step answer:
Here, we will consider a convex lens placed in between the object and the screen

Where, $u$ is the distance of the object from the lens and $v$ is the distance of the image from the lens.
Here, in this case, magnification is given by
${m_1} = \dfrac{v}{u}$
$v = {m_1}u$
Now, let the object be displaced from its original position at a distance $d$ from its original position as shown below.

Where, $d$ is the distance of the lens from its original position.
Here, in this case, the magnification is given by
${m_2} = \dfrac{u}{v}$
$u = {m_2}v$
Now, multiplying both the magnifications, we get
${m_1} \times {m_2} = \dfrac{v}{u} \times \dfrac{u}{v}$
$ \Rightarrow \,{m_1} \times {m_2} = 1$

Now, using lens formula,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Now, putting the value of $v$ in the above equation, we get
$\dfrac{1}{f} = \dfrac{1}{{u{m_1}}} + \dfrac{1}{u}$
$ \Rightarrow \,\dfrac{1}{f} = \dfrac{{1 + {m_1}}}{{u{m_1}}}$
From here, we can get the value of $u$ , as shown below
$u = \dfrac{{f\left( {1 + {m_1}} \right)}}{{{m_1}}}$
Also, by putting the value of $v$ by using lens formula as
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
$ \Rightarrow \,\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{{{m_2}v}}$
$ \Rightarrow \,\dfrac{1}{f} = \dfrac{{{m_2} + 1}}{{{m_2}v}}$
From here, we can get the value of $v$ as
$v = \dfrac{{f\left( {{m_2} + 1} \right)}}{{{m_2}}}$

Now, when the object is displaced from its original position the distance between the original position and the displaced of the object is given by
$d = v - u$
Now, putting the values of $v$ and $u$ , we get
$d = \dfrac{{f\left( {{m_2} + 1} \right)}}{{{m_2}}} - \dfrac{{f\left( {{m_1} + 1} \right)}}{{{m_1}}}$
$ \Rightarrow \,d = \dfrac{{{m_1}f\left( {{m_2} + 1} \right) - {m_2}f\left( {{m_1} + 1} \right)}}{{{m_1}{m_2}}}$
Now taking $f$ common, we get
$d = f\dfrac{{{m_1}\left( {{m_2} + 1} \right) - {m_2}\left( {{m_1} + 1} \right)}}{{{m_1}{m_2}}}$

Now, we will get the value of $f$ as
$f = \dfrac{{d{m_1}{m_2}}}{{{m_1}\left( {{m_2} + 1} \right) - {m_2}\left( {{m_1} + 1} \right)}}$
$ \Rightarrow \,f = \dfrac{{d{m_1}{m_2}}}{{{m_1}{m_2} + {m_1} - {m_1}{m_2} - {m_2}}}$
$ \Rightarrow \,f = \dfrac{{d{m_1}{m_2}}}{{{m_1} - {m_2}}}$
But, $\,{m_1} \times {m_2} = 1$
Therefore, we get
$\therefore f = \dfrac{d}{{{m_1} - {m_2}}}$
Therefore, the focal length of the lens is $f = \dfrac{d}{{{m_1} - {m_2}}}$

Hence, option D is the correct option.

Note:Magnifying is the process in which the object is magnified or enlarged so that one must have a clear vision of the distant object. In some cases, we can also magnify an image. It is also defined as the ratio of the height of an image to the height of an object.