Question

The length of a compound microscope is 14 cm and its magnifying power when final image is formed at near point is 25. If the focal length of eyepiece is 5 cm, the focal length of objective lens is:

• A. $\frac{59}{31}$ cm
• B. $\frac{59}{25}$ cm
• C. 2 cm
• D. 1.5 cm

Answer 1

The correct option is D 1.5 cm

$\begin{array}{c}L=14cm,m=25F_e=5cm\\ \text{we know that}\\ \begin{array}{cc}L=& v_0+F_e\\ \Rightarrow v_0& =L-F_e\\ & =9cm\end{array}\end{array}$

$\begin{array}{c}\therefore m=\frac{V_0}{U_0}\cdot \frac{D}{F_e}[D=25cm]\\ \Rightarrow 25=\frac{9}{40}\cdot \frac{25}{5}\\ \Rightarrow u_0=\frac{9}{5}cm\end{array}$

and for objective $\mathrm{len}8:-$

$\begin{array}{cc}\frac{1}{f_0}& =\frac{1}{u_0}+\frac{1}{v_0}\\ & =\frac{5}{9}+\frac{1}{9}\end{array}$

$F_0=\frac{9}{6}=1.5cm$