Question

A) A giant refracting telescope at an observatory has an objective lens of focal length $15m$. If an eyepiece of focal length $1.0\text{cm}$ is used, what is the angular magnification of the telescope?

B) A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length $1.0\text{cm}$. If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48\times {10}^{6}m$, and the radius of lunar orbit is $3.8\times {10}^{8}m$.

Answer 1

A) Step 1: Draw the diagram.

Step 2: Find the angular magnification.
Given,
Focal length of the objective lens,
$f_o=15m=1500\text{cm}$
Focal length of the eyepiece, $f_e=1.0\text{cm}$

Angular magnification,
$m=\frac{\tan \beta }{\tan \alpha }=\frac{f_o}{f_e}=\frac{1500}{1}=1500$
Final Answer: $m=1500$.

B) Step 1: Draw the diagram.

Step 2:Find the diameter of the image of the moon.
Given:
Diameter of the moon,
$l_1=3.48\times {10}^{6}m$
Radius of the lunar orbit, $r_1=3.8\times {10}^{8}m$
If 𝑑 is the diameter of the image, then the angle subtended by diameter of moon,
${\theta }_1=\frac{l_1}{r_1}$
${\theta }_1=\frac{3.48\times {10}^6}{3.8\times {10}^8}$
and the angle subtended by image,
${\theta }_2=\frac{l_2}{r_2}$

If focal length of objective lens,
$f_o=15m$
${\theta }_2=\frac{d}{f_0}=\frac{d}{15}$

Step 3: Find the diameter of the image of the moon.
Then,
Angle subtended by the diameter of the moon $\left({\theta }_1\right)$ is equal to the angle subtended by the image $\left({\theta }_2\right)$,
${\theta }_1={\theta }_2$
Therefore,
$\frac{d}{15}=\frac{3.48\times {10}^6}{3.8\times {10}^8}=\frac{3.48\times 15\times {10}^{-2}}{3.8}$$=13.73\times {10}^{-2}m$
$d=13.73\text{cm}$
Final Answer : $d=13.73\text{cm}$.