Question

A compound microscope consists of an objective lens of focal length $1cm$ and an eyepiece of focal length $5cm$ with a separation of $10cm$. The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{n}{40}cm$. The value of $n$ is

Answer 1

Step 1. Given data:

$f_o=1cm$, Focal length of objective lens

$f_e=5cm$, Focal length of eyepiece

$L=10cm$, Distance between eyepiece and objective lens

$v_o=?$ ,Distance between image and objective lens

$u_o=\frac{n}{40}cm$, Distance between object and objective lens.

Step 2. Calculating value of $n$

As, $L=v_o+f_e$

$$\begin{gathered} \Rightarrow 10=v_o+5 \\ \therefore v_o=5cm \end{gathered}$$

Now, Using Lens formula

$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$

Putting values in the above equation,

$\begin{array}{rcl}\frac{1}{5}-\frac{1}{u_o}& =& \frac{1}{1}\\ \frac{1}{u_o}& =& \frac{1}{5}-1\\ \frac{1}{u_o}& =& \left(-\frac{4}{5}\right)\\ u_o& =& \left(-\frac{5}{4}\right)\\ \left|u_o\right|& =& \frac{5}{4}\end{array}$

$\begin{array}{rcl}\frac{n}{40}& =& \frac{5}{4}\\ n& =& 50\end{array}$ $\left[u_o=\frac{n}{40}cm\right]$

Hence, the value of $\begin{array}{rcl}\mathit{n}\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{50}\end{array}$.