Question

What is the difference between the construction of an astronomical telescope and a compound microscope? The focal lengths of the objective and eyepiece of a compound microscope are $1.25cm$ and $5.0cm$, respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of $30$when the final image is formed at the near point.

Answer 1

Hint We know that both telescopes and microscopes are used for observation purposes. Though they both do similar kinds of functions the working and usage of a telescope and microscope are entirely different. Here we are also given the specifications of the eyepiece and the objective of a compound microscope. We have to find a position where we should place the object with respect to the objective in order to get a magnification of $30$ when the final image is formed.

Step By Step Solution
A telescope is used to view distant objects clearly and closer whereas a microscope is used to view very small objects magnified. Since the purposes of both are different, the construction of a telescope and microscope is also different.
In a telescope, the objective is made in such a way that it has a long focal length. Meanwhile, in a microscope, the objective is required to have a shorter focal length.
The diameters of the lenses are also substantially different in a telescope and a microscope.
Now, for the calculation part
It is given that,
The focal length of the objective ${f_0} = 1.25cm$
The focal length of the eyepiece ${f_e} = 5cm$
The angular magnification that we require $= 30$
For a microscope in normal use,
The distance of vision will be, $D = 25cm$
From this, we can find the magnification of the eyepiece as,
${m_e} = \left( {1 + \dfrac{D}{{{f_e}}}} \right) = \left( {1 + \dfrac{{25}}{5}} \right) = 1 + 5 = 6$
The total magnification will be the product of the magnification of the object and the eyepiece.
The total magnification can be written as,
$m = {m_o} \times {m_e}$
From this, we get
${m_o} = \dfrac{m}{{{m_e}}}$
Substituting the values, we get
${m_o} = \dfrac{m}{{{m_e}}} = \dfrac{{30}}{6} = 5$
The real image will be formed by the objective lens.
Therefore, the magnification of the objective lens can be written as,
${m_o} = \dfrac{v}{u} = - 5$
Where $v$ is the distance of the image from the objective and $u$ is the distance of the object from the objective.
From this, we can write,
$\dfrac{v}{u} = 5 \Rightarrow v = - 5u$
We know that,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{{{f_o}}}$
Substituting the values of $u,v$and ${f_o}$, we get
$\dfrac{1}{{ - 5u}} - \dfrac{1}{u} = \dfrac{1}{{1.25}}$
Solving the equation we get
$\dfrac{{u + 5u}}{{5{u^2}}} = \dfrac{1}{{1.25}}$
$\Rightarrow \dfrac{{6u}}{{5{u^2}}} = \dfrac{1}{{1.25}}$
From this, we get $u$as
$\Rightarrow \dfrac{6}{{5u}} = \dfrac{1}{{1.25}}$
$\Rightarrow u = \dfrac{{6 \times 1.25}}{5} = 1.5cm$
The object should be placed $1.5cm$ away from the objective to get a magnification of $30$

The answer is: $1.5cm$

Note
The lens that is situated closer to the object is called the objective lens. The eyepiece is the lens of the microscope which is close to our eye, hence it is called the eyepiece. In a compound microscope, the object should be placed at a distance slightly larger than the focal length of the objective lens. The image formed by the objective will be real, inverted, and magnified. This will again be magnified by the eyepiece. It should be noted that in the normal case the distance of vision of a microscope will be $25cm.$