Question

If we need a magnification of $375$ from a compound microscope of tube length $150$ mm and an objective of focal length $5$ mm, the focal length of the eye-piece should be close to:

• A. $22$ mm
• B. $12$ mm
• C. $2$ mm
• D. $33$ mm

Answer 1

The correct option is A

$22$ mm

Explanation for the correct option.

Step 1. Given Data:

$M=375,L=150mm,f_o=5mm,D=250mm$ and find the value of $f_e$.

Step 2. Concept used

The magnification for the least distance of distinct vision for a compound microscope is given as: $M=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)$.

In the formula, $L$ is the tube length, $f_o$ is the focal length of the objective, $f_e$ is the focal length of the eyepiece, and $D$ is the least distance of distinct vision which is equal to $25\text{cm}=250\text{mm}$.

Step 3. Find the focal length of eyepiece.

It is given that tube length is $150$ mm and the focal length of the objective is $5$ mm also a magnification of $375$ is needed.

So in the formula $M=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)$,

substitute $M=375,L=150mm,f_o=5mm,D=250mm$ and find the value of $f_e$.

$$\begin{gathered} 375=\frac{150mm}{5mm}\left(1+\frac{250mm}{f_e}\right) \\ \Rightarrow 375=30\left(1+\frac{250}{f_e}\right) \\ \Rightarrow 12.5=1+\frac{250}{f_e} \\ \Rightarrow 11.5=\frac{250}{f_e} \\ \Rightarrow f_e=\frac{250}{11.5} \\ \Rightarrow f_e\approx 22mm \end{gathered}$$

So, the focal length of the eyepiece is approximately $22$ mm.

Hence, the correct option is A.