Question

The optical length of an astronomical telescope with magnifying power of $10$. For normal vision is $44cm$, what is focal length of the objective?

• A. 4 cm
• B. 40 cm
• C. 44 cm
• D. 440 cm

Answer 1

Answer: (B)

To see with relaxed eye final image should be formed at infinity.
The distance between the objective and eyepiece is adjusted so that image $A^{\prime }{B}^{\prime }$ formed by objective is at focus $F_e^{\prime }$ of the eyepiece. This adjustment of telescope is called normal adjustment. In this position the length of the telescope is
$f_o+f_e(f_o$ is focal length of objective, $f_e$ of eyepiece)

$\therefore 44=f_0+f_e$ ... (i)
Also, magnifying power $m=-\frac{f_o}{f_e}$
$-10=-\frac{f_e}{f_e}$
$\Rightarrow f_o=10f_e$ ... (ii)
Using Eqs. (i) and (ii), we get
$10f_e+f_e=44$
$f_e=4cm$
Also, focal length of objective is
$f_0=10f_e$
$=10\times 4=40cm$
Therefore, focal length of objective is $40cm$.