Question

The focal length of the objective and the eyepiece of a telescope are 50cm and 5cm respectively. If the telescope is focused for distinct vision on a distant scale 2m from its objective, then its magnifying power for near point will be:
(A) – 2
(B) – 4
(C) 8
(D) – 8

Answer 1

Hint Use the lens formula and calculate the image distance of objective. Similarly calculate the object distance of the eyepiece. Now, calculate the magnification for both lenses by $m = \dfrac{v}{u}$. Total magnification of a telescope is given by the product of both objective and eyepiece magnification.

Complete step-by-step answer
We know the lens formula relating object distance $\;u$, image distance $v$ and the focal length $f$.
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Let${f_o}$ and ${f_e}$be the focal length of the objective and eyepiece respectively.
It is given that
${f_o} = 50cm$
${f_e} = 5cm$
${u_o} = - 2m = - 200cm$
Substitute the${f_o}$ and ${u_o}$in the lens formula.
$\dfrac{1}{v} = \dfrac{1}{{50}} + \dfrac{1}{{200}}$
$\dfrac{1}{v} = \dfrac{{4 - 1}}{{200}}$
$v = + \dfrac{{200}}{3}cm$
The image acts as the object for the eyepiece, ${v_e} = - 25cm$
Now, substitute the ${f_e}$ and ${v_e}$ again in the lens formula for the eyepiece
$\dfrac{1}{u} = \dfrac{1}{{ - 25}} - \dfrac{1}{5}$
$\dfrac{1}{u} = - \dfrac{{1 + 5}}{{25}}$
$u = - \dfrac{{25}}{6}cm$
Magnification is the ratio of image distance to the object distance. It is given by,
$m = \dfrac{v}{u}$
${m_o} = - \dfrac{{\dfrac{{200}}{3}}}{{200}} = - \dfrac{1}{3}$
${m_e} = \dfrac{{ - 25}}{{\dfrac{{ - 25}}{6}}} = 6$
The magnification produced by the telescope is the product of the individual magnification produced by objective and eyepiece.
$m = {m_o} \times {m_e}$
$m = - \dfrac{1}{3} \times 6$
$m = - 2$

Hence, magnification produced is $- 2$ and the correct option is A.

Note There are different types of magnification like axial, lateral and areal magnification. From the magnification we can know about the nature of the image as well. If the magnification is negative the image is real and inverted. If it is positive then the image is virtual and erect.