In Gallilean telescope, if the powers of an objective and eye lens are respectively $+ 1.25\ D$ and $- 20\ D$, then for relaxed vision, the length and magnification will be

Question

Dr. Abubacker · Accepted Answer

(c) $f_{o} = \frac{1}{1.25} = 0.8\ m$ and $f_{e} = \frac{1}{- 20} = - 0.05\ m$
$	herefore\left| L_{\infty} ight| = \left| f_{o} ight| - \left| f_{e} ight| = 0.8 - 0.05 = 0.75m = 75\ cm$
and $\left| m_{\infty} ight| = \frac{f_{o}}{f_{e}} = \frac{0.8}{0.05} = 16$

Q. In Gallilean telescope, if the powers of an objective and eye lens are respectively $+ 1.25\ D$ and $- 20\ D$, then for relaxed vision, the length and magnification will be

Explanation

Reviewed by Dr. Abubacker

Want more practice?