Question

Mars $(6800kmdiameter)$is viewed through a telescope on a night when it is $1.1\times {10}^{8}km$from the earth. Its angular size as seen through the eyepiece is $0.50°$, the same size as the full moon seen by the naked eye. If the eyepiece focal length is $25mm$, how long is the telescope?

Short answer

The telescope is $3529.15mm$long.

Step-by-step solution

Given Information.

We need to find out the telescope focal length.

Calculation

Consider,

$$\begin{gathered} D=6800km \\ r=1.1\times {10}^{8}km \\ \theta =0.50° \\ f=25mm \end{gathered}$$

Here,$D$is diameter, $r$is distance between Earth and Mars, $\theta$is is the angular size and $f$is the focal length of the eyepeice.

The mars angular size is:

$\frac{6800km}{1.1\times {10}^{8}km}=6.1818\times {10}^{-5}$.

The angular size:

$$\begin{gathered} arc\tan (6.1818\times {10}^{-5}) \\ =0.00354192° \end{gathered}$$

The mag required to give $0.5°$in eye piece:

$\frac{0.5°}{0.00354192}=141.166$

So, Let us find the telescope focal length:

localid="1648791238247" $$\begin{gathered} mag=\frac{L}{f} \\ 141.161=\frac{L}{25} \\ L=141.161\times 25 \\ L=3529.15mm \end{gathered}$$