Question

We wish to make a microscope with the help of two positive lenses both with a focal length of $20\;{\rm{mm}}$ each and the object is positioned $25\;{\rm{mm}}$ from the objective lens. How far apart the lenses should be so that the final image is formed at infinity?
(a) $20\;{\rm{mm}}$
(b) $100\;{\rm{mm}}$
(c) $120\;{\rm{mm}}$
(d) $80\;{\rm{mm}}$

Answer 1

Hint: When two lenses are used in forming a microscope, it will be a compound microscope. We will apply the lens equation to derive the answer.

Complete step by step answer:
We want to construct a microscope with two convex lenses. In a compound microscope made with two lenses, one lens will act as an objective lens and the other acts as the eyepiece.

Let ${f_e}$ be the focal length of the eyepiece lens and ${f_o}$ be the focal length of the objective lens. It is given that the focal length of the eyepiece and the objective lens are the same.
Now, the equation for the focal length can be written using the lens equation as

$\dfrac{1}{{{f_e}}} = \dfrac{1}{{{v_o}}} - \dfrac{1}{{{u_o}}}$

Here ${u_o}$ is the distance of the object from the objective lens and ${v_o}$ is the distance of the image from the objective lens.
It is given that the object is at a distance of $25\;{\rm{mm}}$ from the objective lens. Since the object is
behind the lens, we take the distance as negative. Hence,

${u_o} = - 25\;{\rm{mm}}$

Given that the focal length of the eye piece, ${f_o} = 20\;{\rm{mm}}$
So, we can substitute the values of ${f_o}$ and ${u_o}$ in the lens equation to find the image distance. Then, we get
$
\dfrac{1}{{20}} = \dfrac{1}{{{v_o}}} - \dfrac{1}{{ - 25}}\\
\dfrac{1}{{{v_o}}} = \dfrac{1}{{20}} - \dfrac{1}{{25}}\\
= \dfrac{{25 - 20}}{{20 \times 25}}\\
 = \dfrac{5}{{500}}
$

Hence,
$\begin{array}{c}
\dfrac{1}{{{v_o}}} = \dfrac{1}{{100}}\\
{v_o} = 100\;{\rm{mm}}
\end{array}$

The image formed by the objective lens acts as the object for the eyepiece. Hence, the lens equation for the eyepiece can be written as

$\dfrac{1}{{{f_e}}} = \dfrac{1}{{{v_e}}} - \dfrac{1}{{{u_e}}}$

Here ${v_e}$ is the image distance and ${u_e}$ is the object distance for the eyepiece.
Since the final image is formed at infinity, ${v_e} = \infty $. Hence, we get

${f_e} = - {u_e}$

The negative sign implies that the object is placed behind the eye piece.
Thus, we can write the relation for the length of the microscope as

$L = {f_e} + {v_o}$

The length of the microscope is the same as the distance between the lenses of the microscope.
So, substituting the values for ${f_e}$ and ${v_o}$ in the above equation we get
$\begin{array}{c}
L = 20 + 100\\
 = 120\;{\rm{mm}}
\end{array}$

Since the distance between the lenses is obtained as $120\,{\rm{mm}}$

So, the correct answer is “Option C”.

Note:
For a compound microscope, if the final image is formed at infinity, the focal length of the eye piece will be equal to the object distance of the eye piece. Also, make sure to give correct signs to the image and object distances based on sign conventions.