Question

In a compound microscope, the magnified virtual image is formed at a distance of $25cm$ from the eye-piece. The focal length of its objective lens is $1cm$. If the magnification is $100$ and the tube length of the microscope is $20cm$, then the focal length of the eyepiece lens (in cm) is __________.

Answer 1

Step 1: Given data

For the objective lens,

Image distance, $v_o=v_1$

Focal length $f_o=1cm$

Magnification $M=100$

Let object distance = $u_0=x$

For the eyepiece lens,

Object distance,$u_e=-\left(20-v_1\right)$

Image distance,$v_e=-25cm$

Step 2: For the objective lens,

$$\begin{gathered} \frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o} \\ \Rightarrow \frac{1}{v_1}-\frac{1}{x}=\frac{1}{1}\left(v_o=v_1,u_o=x,f_o=1cm\right) \\ \Rightarrow v_1=\frac{x}{x-1}\left(1\right) \end{gathered}$$

Magnification,

$\left|m_o\right|=\left|\frac{v_o}{u_o}\right|=\left|\frac{v_1}{x}\right|=\frac{1}{x-1}$

Substituting the value of $v_1$ from $\left(1\right)$

Step 3: For eye-piece lens,

The image formed by the objective lens is acting as an object,

Here,

$$\begin{gathered} u_e=-\left(20-v_1\right) \\ \Rightarrow u_e=-\left(20-\frac{x}{x-1}\right)\left(2\right) \end{gathered}$$

Magnification,

$\left|m_e\right|=\left|\frac{v_e}{u_e}\right|=\frac{25}{\left|u_e\right|}$

Substituting the value of $u_e$ from $\left(2\right)$

We know,

Total magnification,

$$\begin{gathered} m_o{m}_e=100 \\ \Rightarrow \left(\frac{1}{x-1}\right)\left(\frac{25}{20-{\displaystyle \frac{x}{x-1}}}\right)=100 \\ \Rightarrow \frac{25}{20x-20-x}=100 \\ \Rightarrow 76x-80=1 \\ \Rightarrow x=\frac{81}{76} \\ \Rightarrow u_e=-\left(20-\frac{\left({\displaystyle \frac{81}{76}}\right)}{\left({\displaystyle \frac{81}{76}}-1\right)}\right) \\ \Rightarrow u_e=-\left(20-\frac{\left({\displaystyle \frac{81}{76}}\right)}{\left({\displaystyle \frac{5}{76}}\right)}\right) \\ \Rightarrow u_e=-\left(20-\left(\frac{81}{76}\times \frac{76}{5}\right)\right) \\ \Rightarrow u_e=-\left(20-\frac{81}{5}\right) \\ \Rightarrow u_e=\frac{-19}{5} \end{gathered}$$

By lens formula for eye-piece lens,

$$\begin{gathered} \frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e} \\ \Rightarrow \frac{1}{-25}-\frac{1}{\left({\displaystyle \frac{-19}{5}}\right)}=\frac{1}{f_e} \\ \Rightarrow \frac{1}{-25}+\frac{5}{19}=\frac{1}{f_e} \\ \Rightarrow f_e=\frac{\left(25\times 19\right)}{106} \\ \Rightarrow f_e=4.48cm \end{gathered}$$

Therefore, the focal length of eyepiece lens (in cm) is $4.48cm$.