As astronomical telescope of ten fold angular magnification has a length of 44 cm. The focal length of the objective is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 40 cm
Magnification, $M=\frac{f_{objective}}{f_{eyepiece}}=10$
$⇒f_0=10f_e$
Length of the telescope, $L=f_0+f_e$
$⇒44=f_0+f_e$
$⇒44=10f_e+f_e$
$⇒f_e=4cm$
$∴f_0=f_e×10=4×10$
$=40cm$