Question

In a compound microscope, the focal length of two lenses are $1.5\text{cm}\text{and}6.25\text{cm}$. An object is placed at $2\text{cm}$ from objective and the final image is formed at $25\text{cm}$ from the eyepiece. The distance between the two lens is (in $\text{cm}$ )

• A. $11$
• B. $9.25$
• C. $7.75$
• D. $6$

Answer 1

The correct option is A $11$

Given that
Focal length of objective $f_o=1.5\text{cm}$
focal length of eyepiece $fe=6.25\text{cm}$
$u_0=-2\text{cm}{v}_e=25\text{cm}$

$\left(i\right)$ Lens formula at objective piece
$\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}\Rightarrow \frac{1}{1.5}=\frac{1}{v_0}+\frac{1}{2}\Rightarrow \frac{1}{1.5}-\frac{1}{2}=\frac{1}{v_0}$
$\Rightarrow \frac{2-1.5}{3}=\frac{1}{v_0}\Rightarrow v_0=\frac{3}{0.5}=6\text{cm}$

$\left(ii\right)$ lens formula at eye piece
$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}\Rightarrow \frac{1}{6.25}=-\frac{1}{25}-\frac{1}{u_e}$
$\Rightarrow \frac{1}{u_e}=-\frac{1}{25}-\frac{1}{6.25}=-\left[\frac{1+4}{25}\right]=-\frac{5}{25}$
$\Rightarrow u_e=-5\text{cm}$
Length of the microscope $L=v_0+u_e$
$L=6+5$
$L=11\text{cm}$