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) $