Question

Define magnifying power of telescope. Write its expression.
A small telescope has an objective lens of focal length 150cm and an eye piece of focal length 5cm. if this telescope is used to view 100m high towers 3km away. Find the height of the final image when it is formed 25cm away from the eye piece.

Answer 1

Hint: Magnification is describe as simply by dividing the focal length of the telescope by focal length of the eyepiece, which means that the smaller number on an eyepiece gives a higher magnification such as the 10 mm of eyepiece will give the magnification twice the exit eyepiece i-e 20mm eyepiece.

Step-By-Step answer:
The Magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the image formed at least distance of distinct vision to angle subtended at the eye by the object lying in infinity.
$M = \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)$
fo= focal length of the object=150cm
fe= focal length of the eyepiece=5cm
D= least distance of the distinct vision=25cm
$
  M = \dfrac{{ - 150}}{5}\left( {1 + \dfrac{5}{{25}}} \right) = - 36 \\
  M = \dfrac{\beta }{\alpha } = \dfrac{{\tan \beta }}{{\tan \alpha }}(as{\text{ we know angles are small)}} \\
  {\text{tan}}\alpha {\text{ = }}\dfrac{H}{u} = \dfrac{{100}}{{3000}} = \dfrac{1}{{30}} \\
 $
Where H= height of the object and u= distance of the object from the objective.
$M = \dfrac{{\tan \beta }}{{\left( {\dfrac{1}{{30}}} \right)}} = \dfrac{{ - 36}}{{30}} = \dfrac{{{H^ / }}}{D}$
${H^ / }$=the height of the image and D= distance of the image formation
Thus,
${H^ / } = \dfrac{{ - 36 \times 25}}{{30}} = - 30cm$
Therefore, the negative sign indicates that we get an inverted image.

Note: The magnification power is the amount that a telescope enlarges with its subject. Moreover, it is equal to the telescope’s focal length divided by the eyepiece’s focal length. As a thumb rule, the telescope's maximum useful magnification is 50 times its aperture in inches.