Question

The transverse magnification produced by the objective lens and the angular magnification of the eye lens of a compound microscope are 25 and 6, respectively. The magnifying power of this microscope is
A. 19
B. 31
C. 150
D. $\sqrt{150}$

Answer 1

Hint: A microscope consists of an objective lens and an eye lens. The total magnification of a compound microscope is given as the product of the magnification of each of the lenses. Put the values given in the question in the obtained expression to find the required answer.

Complete step-by-step answer:
Magnification can be defined as the measure of the increase in size of the image compared to the size of the object. For compound lenses the magnification is the ratio of the height of the image formed to the height of the object placed.
A compound microscope consists of two lenses. One is the objective lens and the other is the eye lens. Both lenses have their own magnifying power. The magnifying power of a compound microscope can be found as a total magnification by the system of the two lenses.
Let the magnification of the objective lens is ${{M}_{o}}$ and the magnification of the eye lens is ${{M}_{e}}$ .
So, the total magnification produced by the compound microscope can be mathematically expressed in terms of the magnification of the objective lens and the eye lens as,
$M={{M}_{o}}\times {{M}_{e}}$
Now, the magnification of the objective lens is ${{M}_{o}}=25$
The magnification of the eye lens is ${{M}_{e}}=6$
So, the magnification of the compound microscope will be,
$\begin{align}
  & M={{M}_{o}}\times {{M}_{e}} \\
 & M=25\times 6 \\
 & M=150 \\
\end{align}$

So, the correct answer is “Option C”.

Note: We can also define the magnification of the compound microscope in terms of the focal length of the objective lens and the focal length of the eye lens.
$M={{M}_{o}}{{M}_{e}}=\left( \dfrac{L}{{{f}_{o}}} \right)\left( \dfrac{D}{{{f}_{e}}} \right)$