Question

(i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used. What is the angular magnification of the telescope?
(ii) If this telescope is used to view the moon. What is the diameter of the image of the room formed by the objective lens? The diameter of the moon is $3.48\times {{10}^{8}}m$ and the radius of lunar orbit is $3.8\times {{10}^{8}}m.$

Answer 1

To solve this question, we need to know the basics of optics, including refraction and reflection. We need to know about the angular magnification of a telescope. With the angle subtended by the image, we can estimate the diameter of the image of the moon.

Complete answer:
(I)Before we move on to the solving of the sum, let us define angular magnification.
The angular magnification of the telescope is defined as the negative of the focal length of the objective lens which is divided by the focal length of the eyepiece of the concerned telescope.
Therefore, the angular magnification of a refracting telescope is given by $-\dfrac{{{f}_{0}}}{{{f}_{e}}}$ .
Now putting the values to the above mentioned formula we get,
$-\dfrac{15}{{{10}^{-2}}}=-1500$
So, the angular magnification of the refracting telescope is - 1500.

(II)Let us consider that the diameter of the image id d,
So, the angle subtended by the image is $\dfrac{d}{{{f}_{0}}}=\dfrac{d}{15}$
Now we have to find the angle subtended by the diameter of the moon.
So, the diameter of the moon is $\dfrac{3.48\times {{10}^{6}}}{3.8\times {{10}^{8}}}$.
Now we have to equate the two equations.
So, equating the two equations we get
d = $\dfrac{3.48\times 10^6}{3.8\times 10^8}\times 15m=0.13736m$
Therefore, d = 13.736 cm.
Hence, the diameter of the image of the moon is 13.736 cm.

Note:
We must not confuse ourselves over angle reflection and that of refraction. The basic difference to keep in mind is that angle of refraction is material dependent while angle of reflection is property dependent.