Question

The focal lengths of the objective and eye lens of an astronomical telescope are respectively 2 meter and 100 cm. Final image is formed at
(i) Least distance of distinct vision
(ii) Infinity
Magnifying power in two cases will be:
(A) $-48,-40$
(B) $-40,-48$
(C) $-40,+48$
(D) $-48+40$

Answer 1

An astronomical telescope is an optical instrument which is used for observing distinct images of heavenly bodies like stars, planets etc. It consists of two lenses:
-Objective lens, which is of large focal length and large aperture
-Eye for piece, which has a small focal length and small aperture
In case of final image is formed at infinity
Magnifying power, we can use $\text{m}=\dfrac{{{\text{f}}_{\text{o}}}}{{{\text{f}}_{\text{e}}}}$
Where ${{\text{f}}_{\text{o}}}$ =focal length of objective lens
$-{{\text{f}}_{\text{e}}}$ =focal length of eye lens.
In case of final image is formed at least distance of distinct vision, we use
$\text{m}=\dfrac{-{{\text{f}}_{\text{o}}}}{{{\text{f}}_{\text{e}}}}\left( 1+\dfrac{{{\text{f}}_{\text{e}}}}{\text{d}} \right)$

Complete step by step solution:
In case of image is formed at infinity
To find out the magnifying power,
$\text{m}=\dfrac{{{\text{f}}_{\text{o}}}}{-{{\text{f}}_{\text{e}}}}$ …….. (1)
Here,
$\begin{aligned} & {{\text{f}}_{\text{o}}}=2\text{ m} \\\ & \text{ }=2\times 100\text{ cm} \\\ & \text{ }=200\text{ cm} \\\ \end{aligned}$
${{\text{f}}_{\text{e}}}=\text{5 cm}$
Put the values in equation (1)
$\begin{aligned} & =\dfrac{200}{5} \\\ & =40 \\\ & \text{m}=-40 \\\ \end{aligned}$
In case of image is formed at least of distinct vision
$\text{m}=\dfrac{-{{\text{f}}_{\text{o}}}}{{{\text{f}}_{\text{e}}}}\left( 1+\dfrac{{{\text{f}}_{\text{e}}}}{\text{d}} \right)$ ….. (2)
We know, least distance of distinct vision $\text{d}=25\text{ cm}$
$\begin{aligned} & {{\text{f}}_{\text{o}}}=200\text{ cm} \\\ & {{\text{f}}_{\text{e}}}=5\text{ cm} \\\ \end{aligned}$
Put the values in equation (2)
$\begin{aligned} & \text{m}=\dfrac{-200}{5}\left( 1+\dfrac{5}{25} \right) \\\ & \text{ }=\dfrac{-200}{5}\left[ \dfrac{25+5}{25} \right] \\\ & \text{ }=\dfrac{-200}{5}\times \dfrac{30}{25} \\\ & \text{ }=\dfrac{-1200}{25} \\\ & \text{ }=-48 \\\ \end{aligned}$
$\therefore$ Final image is formed at
-Least distance of distinct vision $\text{m}=-48$
-Infinity, $\text{m}=-40$
So the option (A) is correct.

Note
Remember that there are different methods to solve magnifying power in the case of astronomical telescopes. As the magnifying power is negative, the final image in an astronomical telescope is inverted i.e. upside down and left turned right.