Question

A compound microscope has a magnifying power of $100$ when the image is formed at infinity. The objective has a focal length of $0.5\text{cm}$ and the tube length is $6.5\text{cm},$ then the focal length of the eye-piece is -

• A. $0.2\text{cm}$
• B. $2.5\text{cm}$
• C. $2\text{cm}$
• D. $4.0\text{cm}$

Answer 1

The correct option is C $2\text{cm}$

Given,

$M_{\infty }=100;f_0=0.5\text{cm};L=6.5\text{cm}$

For the give compound microscope,
Since, the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece.



Thus, $L=v_0+f_e=6.5\text{cm}.......\left(1\right)$

For normal adjustment,

$M_{\infty }=-\frac{v_0}{u_0}\times \frac{D}{f_e}$

$\Rightarrow M=-[1-\frac{v_0}{f_0}]\frac{D}{f_e}[\because \frac{v_0}{u_0}=1-\frac{v_0}{f_0}]$

$\Rightarrow 100=-[1-\frac{v_0}{0.5}]\times \frac{25}{f_e}\text{[Taking D= 25~ text{cm}]}$

$\Rightarrow 2v_0-4f_e=1........\left(2\right)$

$\text{Solving equation (1)and (2) we can get,}$

$v_0=4.5\text{cm}\text{and}{f}_e=2\text{cm}$

So, the focal length of the eye piece is $2\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.