Question

An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eye piece is 36 cms and the final images is formed at infinity. The focal length $f_o$ of the objective and $f_e$ of the eye piece are

• A. $f_0=45cm$ and $f_e=-9cm$
• B. $f_0=50cm$ and $f_e=-10cm$
• C. $f_0=7.2cm$ and $f_e=-5cm$
• D. $f_0=30cm$ and $f_e=-6cm$

Answer 1

The correct option is D $f_0=30cm$ and $f_e=-6cm$

$m=\frac{f_0}{f_e}$
$5=\frac{f_0}{f_e}$
$f_0=5f_e$
$L=f_0+f_e$
$36=f_0+f_e$
$36=5f_e+f_e$
$f_e=6cm$
$f_0=5f_e$
$=6\times 5$
$=30cm$