Question

A tower $100\,\,m$ tall at a distance of $3\,\,Km$ is seen through a telescope having objective of focal length $140\,\,cm$ and eyepiece of focal length $5\,\,cm$. Then the size of the final image if it is at $25\,\,cm$ from the eye?
(A) $14\,\,cm$
(B) $28\,\,cm$
(C) $42\,\,cm$
(D) $56\,\,cm$

Answer 1

Hint: The final size of the image that is the height of the tower can be found with the help of the formula for the magnification of the lens of the eyepiece and the height of the image of the tower formed by the objective lens.

Formulae Used:
The magnification of the eyepiece is given by:
${m_e} = \left( {1 + \dfrac{d}{{{f_e}}}} \right)$
Where, ${m_e}$ denotes the magnification of the eyepiece lens, $d$ denotes the distance of the image from the eyepiece, ${f_e}$ denotes the focal length of the eyepiece.

Complete step-by-step solution:
The data given in the problem are;
Height of the tower,${h_o} = 100\,\,m$,
Distance of the object, $
  u = 3\,\,Km \\
  u = 3000\,\,m \\
 $,
Focal length of the objective lens, ${f_o} = 140\,\,cm$
Focal length of the eyepiece lens, ${f_e} = 5\,\,cm$
Distance of the image, $
  v = 25\,\,cm \\
  v = 0.25\,\,m \\
 $
$f = {f_o} + {f_e}$
Substitute the values;
$
  f = 140 + 5 \\
  f = 145\,\,m \\
 $
Height of the image of the tower formed by the objective lens;
$
  \alpha = \dfrac{{{h_0}}}{u} \\
  \alpha = \dfrac{{100}}{{3000}} \\
  \alpha = \dfrac{1}{{30}}\,\,rad\,\,..........\left( 1 \right) \\
 $
$
  \alpha = \dfrac{h}{{{f_o}}} \\
  \Rightarrow \alpha = \dfrac{h}{{140}}\,\,rad\,\,..........\left( 2 \right) \\
 $
By equating the equation (1) and the equation (2) we get;
$
  \dfrac{h}{{140}} = \dfrac{1}{{30}} \\
  \Rightarrow h = \dfrac{{140}}{{30}} \\
  \Rightarrow h = 4.7\,\,cm \\
 $
The magnification of the eyepiece is given by:
$
  {m_e} = \left( {1 + \dfrac{d}{{{f_e}}}} \right) \\
  \Rightarrow {m_e} = \left( {1 + \dfrac{{25}}{5}} \right) \\
  \Rightarrow {m_e} = \left( {\dfrac{{5 + 25}}{5}} \right) \\
  \Rightarrow \left( {\dfrac{{30}}{5}} \right) \\
  \Rightarrow {m_e} = 6\,\,cm \\
 $
Height of the final image is;
$
  {h_i} = {m_e} \times h \\
  \Rightarrow {h_i} = 6 \times 4.7 \\
  \Rightarrow {h_i} = 28\,\,cm \\
 $
Therefore, the height of the final image of the tower is ${h_i} = 28\,\,cm$.
Hence, the option (B) ${h_i} = 28\,\,cm$ is the correct answer.

Note:- The enlargement that is obtained by the lens is interpreted as the ratio of the height of the image to the height of the object. The enlargement obtained by a lens is the same as the ratio of distance of an image to the distance of the object.