Question

A thin Plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis. The separation between the object and image planes is \[1.8\text{ }m\]. The magnification of the image formed by one of the half lenses is 2.
Find the focal length of the lens used.
$\begin{align}
  & \text{A}\text{. }0.4m \\
 & \text{B}\text{. }0.8m \\
 & \text{C}\text{. }0.6m \\
 & \text{D}\text{. }2.0m \\
\end{align}$

Answer 1

Hint: As magnification of lens is given by the ratio of image distance to object distance from the lens, take the object distance as \[x\] and calculate the image distance. Then using the mirror formula calculate the focal length. Now the image for the first lens will be the object for the second lens and the object for the first lens will be the image for the second lens. So calculate the focal length of the second lens. Then calculate the focal lens of the combination of lens to get the focal length of the lens used.

Formula used:
The lens formula is given by $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Complete answer:
Given the magnification caused by the first lens is 2.
Magnification of a lens is defined as the ratio of image height to object height. Which can be further simplified to the ratio of image distance to the object distance from the lens.
i.e. if the image distance is $v$ and object distance is $u$ the magnification is given by,$m=\dfrac{v}{u}$
Let the object be situated at distance $x$ from the mirror. i.e. $u=O{{L}_{1}}=x$
So magnification by the first lens is
$m=\dfrac{v}{u}\Rightarrow v=mu=2x\left( \because m=2\text{ and }u=x \right)$
Given that the object plane and image plane is separated by a distance \[1.8\text{ }m\].
So

$\begin{align}
  & u+v=1.8m \\
 & \Rightarrow x+2x=1.8m \\
 & \Rightarrow 3x=1.8m \\
 & \Rightarrow x=0.6m \\
\end{align}$

The lens formula relates the focal length with object distance and image distance and is given by
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Taking sign convention, $v=2x=1.2m$, $u=-x=-0.6m$

So,
$\begin{align}
&\dfrac{1}{f}=\dfrac{1}{1.2}-\left(-\dfrac{1}{0.6}\right)=\dfrac{1}{1.2}+\dfrac{1}{0.6}=\dfrac{1+2}{1.2}=\dfrac{3}{1.2} \\
& \Rightarrow f=\dfrac{1.2}{3}=0.4m \\
\end{align}$

For the second lens image and object is interchanged. i.e. the object of the first lens will be the image of the second lens and image of first lens will be image of second lens.
So the object is situated at I at a distance ${{L}_{2}}I=x$ from the second lens. Let the separation between the lenses is $d$
Now,
\[\begin{align}
  & O{{L}_{1}}+d+{{L}_{2}}I=1.8m\Rightarrow 2x+d=1.8m \\
 & \Rightarrow d=1.8-2x=1.8m-1.2m=0.6m\left( \because x=0.6m \right) \\
\end{align}\]
For second lens taking sign convention.
$\begin{align}
  & u'=+x=+0.6m \\
 & v'=-(d+x)=-(0.6+0.6)=-1.2m \\
\end{align}$
Lens formula for second lens.
$\begin{align}
  & \dfrac{1}{f'}=\dfrac{1}{v'}-\dfrac{1}{u'} \\
 & \Rightarrow \dfrac{1}{f'}=-\dfrac{1}{1.2}-\dfrac{1}{0.6}=-\left( \dfrac{1+2}{1.2} \right)=-\dfrac{3}{1.2} \\
 & \Rightarrow f'=-\dfrac{1.2}{3}=-0.4m \\
\end{align}$
The negative sign indicates that the focal length of the second lens lies to the left of the second lens. Taking only magnitude.
$f'=0.4m$.
So the focal length of the lens will be
$F=\dfrac{f+f'}{2}=\dfrac{0.4+0.4}{2}=0.4m$

So, the correct answer is “Option A”.

Note:
In every ray optics problem the sign convention should be used in the formula. The sign convention is like the co-ordinate axis. If you go right from the lens the distance will be taken as positive and if you go left from the lens the distance will be taken as negative. Similarly moving top is positive and moving bottom will be negative.