Question

An astronomical telescope with magnification 50 is to be designed in normal adjustment. Length of the tube is 102 cm. The power of objective and eyepiece are respectively

• A. 2 D, 50 D
• B. 1.5 D, 20 D
• C. 1 D, 40 D
• D. 1 D, 50 D

Answer 1

Answer: (A)

2 D, 50 D

For the astronomical telescope in normal adjustment.

Magnifying power$=m=50,$ length of the tube $=L=102cm$

Let $f_o$ and fe be the focal length of objective and eye piece respectively.

$m=\frac{f_o}{f_e}=50$
$\Rightarrow f_o=50f_e\left(1\right)$

and, $L=f_o+f_e=102cm\left(2\right)$

Putting the value of f0 from equation (1) in (2), we get,

$51f_e=102cm\Rightarrow f_e=2cm=0.02m$

So,$f_o=50f_e{f}_o=50\times 2cm\Rightarrow f_o=100cm=1m$

Power of the objective lens = $\frac{1}{f_o}=1D$

And Power of the eye piece lens = $\frac{1}{f_e}=\frac{1}{0.02}=50D$.