Question

The large space telescope that has been placed into an Earth orbit has an aperture diameter of 2.40 meters. Determine
(a) the angular resolution that this telescope will archieve for visible light of wavelength 4.80*10^-7 m, and (b) the maximum distance (in kilometers) for which this telescope can distinguish two objects 150 kilometers apart.

NOTE:AS FAST AS POSSIBLE

Answer 1

(a) The angular resolution is

$$\theta=\text{arcsin}\bigg(1.22\frac{\lambda}{D}\bigg)=\text{arcsin}\bigg(1.22\frac{4.80\cdot10^{-7} }{2.4}\bigg)=1.4\cdot10^{-5}°.$$

(b) The maximum distance (in km) for which this telescope can distinguish two objects 150 km apart can be determined from the following figure:

So, we see that

$$\text{ sin}\frac{\theta}{2}=\frac{150/2}{d},\\\space\\ d=\frac{150/2}{\text{sin}(\theta/2)}=614\cdot10^6\text{ km}.$$