Question

How is the working of a telescope different from that of a microscope? The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively, Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.

Answer 1

Hint: As we know that the telescope and microscope both use lenses for the magnification of the object. But we can differentiate the working of both the instruments with the help information regarding image formation and focal length.

Complete step by step answer:
The main difference between the telescope and microscope is that the telescope is used to see the distant object with the help of magnification but the image is apparent but the microscope is used to see the small object with the help of magnification but the image is apparent. The other difference is that the objective of a microscope has small focal length and small aperture but the objective of the telescope has large focal length and large aperture. The image size formed from the telescope is smaller than the original size of object but in the microscope, the image size is quite larger than the original size of object.
It is given that focal length of the objective of a microscope is 1.25 cm and focal length of the eyepiece is 5 cm.
Let the distance of the object from the objective is ${u_o}$.
Magnification of 30 is given, so we can use the expression of the magnification $m = \dfrac{{{v_o}}}{{{u_o}}}\left( {1 + \dfrac{D}{{{f_e}}}} \right)$ for the relation of ${v_o}$ and ${u_o}$.
Here, $m$ is the magnification, ${v_o}$ is the distance of the image from objective and ${u_o}$ is the distance of the object from the objective. $D$ is the distance of the least distinct vision and ${f_e}$ is the focal length of the eyepiece.
Substitute the respective values in the expression of the magnification.
Therefore, we get
$\begin{array}{l}
m = \dfrac{{{v_o}}}{{{u_o}}}\left( {1 + \dfrac{D}{{{f_e}}}} \right)\\
 - 30 = \dfrac{{{v_o}}}{{{u_o}}}\left( {1 + \dfrac{{25\;{\rm{cm}}}}{{5\;{\rm{cm}}}}} \right)\\
 - 30 = \dfrac{{{v_o}}}{{{u_o}}}\left( 6 \right)\\
{v_o} = - 5{u_o}
\end{array}$…… (1)
Now after obtaining the relation between ${v_o}$ and ${u_o}$, we can use the lens formula for the calculation of object distance from the objective. So,
$\dfrac{1}{{{f_o}}} = \dfrac{1}{{{v_o}}} - \dfrac{1}{{{u_o}}}$
Substitute the values in the above equation.
Therefore, we get
$\begin{array}{l}
\dfrac{1}{{1.25\,{\rm{cm}}}} = \dfrac{1}{{ - 5{u_o}}} - \dfrac{1}{{{u_o}}}\\
\dfrac{1}{{1.25\;{\rm{cm}}}} = \dfrac{{ - 1 - 5}}{{ - 5{u_o}}}\\
{u_o} = \dfrac{{1.25\;{\rm{cm}} \times \left( { - 6} \right)}}{5}\\
{u_o} = - 1.5\;{\rm{cm}}
\end{array}$

Hence, the distance of the object relative to the objective is $1.5\;{\rm{cm}}$.

Note: Try to remember the expression of the magnification and the lens formula. Generally in the problems of telescopes these formulas are very important because telescopes work with the help of magnification. If the different unit of distances is given then convert into the same measurable units and put in the formula for the correct calculation.