Question

The focal lengths of the objective and of the eye-piece of a compound microscope are $f_o$ and $f_e$ respectively. If $L$ is the tube length and $D$, the least distance of distinct vision, then its angular magnification, when the image is formed at infinity, is

• A. $\left(1-\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)$
• B. $\left(1+\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)$
• C. $\frac{L}{f_o}\left(1-\frac{D}{f_e}\right)$
• D. $\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)$
• E. $\frac{L}{f_o}\left(\frac{D}{f_e}\right)$

Answer 1

Answer: (E)

In compound microscope, the final image is formed at infinite. Magnification
$M_{\infty }=-\frac{v_o}{u_o}\left(\frac{D}{f_e}\right)$ Or $m_{\infty }=\frac{L}{f_o}\left(\frac{D}{f_e}\right)$