Question

In a compound microscope the objective and the eyepiece have focal lengths of $0.95\;{\text{cm}}$ and $5\;{\text{cm}}$ respectively, and are kept at a distance of $20\;{\text{cm}}$. The last image is formed at a distance of $25\;{\text{cm}}$ from the eyepiece. The position of object and the total magnification are
A) $\dfrac{{95}}{{94}}\;{\text{cm}}$ in front of field lens, $94$
B) $\dfrac{{80}}{{79}}\;{\text{cm}}$ in front of field lens, $93$
C) $\dfrac{{70}}{{65}}\;{\text{cm}}$ in front of field lens, $65$
D) $\dfrac{{60}}{{55}}\;{\text{cm}}$ in front of field lens, $50$

Answer 1

Hint: The length of the microscope tube is the distance between objective lens and eyepiece lens. It depends on the focal length of the eyepiece lens and position of the object. The lens maker’s formula connecting image distance, object distance and the focal length can be used.

Complete step by step solution:
Given the focal length of the objective lens is ${f_o} = 0.95\;{\text{cm}}$, focal length of eyepiece is ${f_e} = 5\;{\text{cm}}$
And the objective lens and the eyepiece lens are kept at a distance, $L = 20\;{\text{cm}}$. And image formed from the eyepiece at a distance of ${v_e} = 25\;{\text{cm}}$
The expression for the distance between the objective lens and eyepiece lens is given as,
$L = {u_0} + \dfrac{{D{f_e}}}{{D + {f_e}}}$
Where, ${u_0}$ is the distance of the object from the objective lens, $D$ is the least distance of the distinct vision and ${f_e}$ is the focal length of the eyepiece lens.
The value of $D$ is $25\;{\text{cm}}$.
Substituting the values in the above expression,
$
  20\;{\text{cm}} = {u_o} + \dfrac{{25\;{\text{cm}} \times 5\;{\text{cm}}}}{{25\;{\text{cm}} + 5\;{\text{cm}}}} \\
  20\;{\text{cm}} = {u_o} + \dfrac{{125\;}}{{30\;}}\;{\text{cm}} \\
  {u_o} = \dfrac{{95}}{6}\;{\text{cm}} \\
 $
Thus, the position of the object from the objective lens is ${u_0} = \dfrac{{95}}{6}\;{\text{cm}}$.
Now we can use the lens makers formula for the objective lens.
$\dfrac{1}{{{u_o}}} - \dfrac{1}{{{v_o}}} = \dfrac{1}{{{f_o}}}$
Substituting the values in the above expression,
$
  \dfrac{6}{{95}} - \dfrac{1}{{{v_o}}} = \dfrac{1}{{0.95}} \\
  \Rightarrow \dfrac{6}{{95}} - \dfrac{1}{{{v_o}}} = \dfrac{{100}}{{95}} \\
  \Rightarrow - \dfrac{1}{{{v_o}}} = \dfrac{{94}}{{95}} \\
  \Rightarrow {v_o} = - \dfrac{{95}}{{94}} \\
 $
Thus, the distance of the image from the objective lens is $ - \dfrac{{95}}{{94}}$.
The expression for calculating the magnification of the lens is given as,
$M = \dfrac{{{u_o}}}{{{v_o}}}\left( {1 + \dfrac{D}{{{f_e}}}} \right)$
Where, ${u_o}$ is the distance of object from objective lens, ${v_o}$ is the distance of image from the objective lens, $D$ is the least distance of distinct vision and ${f_e}$ is the focal length of eyepiece lens.
Substituting the values in the above expression gives,
$
  M = \dfrac{{\dfrac{{95}}{6}}}{{ - \dfrac{{95}}{{94}}}}\left( {1 + \dfrac{{25}}{5}} \right) \\
   = 95 \\
$
Thus, the total magnification is $95$.

The answer is option A.

Note: We have to note that the magnification is the ratio of object distance to the image distance. In a compound microscope the total magnification depends on the focal length of the eyepiece lens also.