Question

The objective lens of a certain refracting telescope has a diameter of 63.5 cm. The telescope is mounted in a satellite that orbits the Earth at an altitude of 265 km to view objects on the Earth's surface. Assuming an average wavelength of 500 nm, find the minimum distance between two objects on the ground if their images are to be resolved by this lens.

Answer 1

Answer:

0.255 m

Explanation:

We are given that

Diameter=d=63.5 cm=$63.5\times 10^{-2} m$

$1 cm=10^{-2} m$

L=265 km =$265\times 1000=265000 m$

Wavelength,$\lambda=500nm=500\times 10^{-9} m$

$1nm=10^{-9} m$

We have to find the minimum distance between two objects on the ground if their images are to be resolved by this lens.

$sin\theta=1.22\frac{\lambda}{d}$

$sin\theta=\frac{1.22\times 500\times 10^{-9}}{63.5\times 10^{-2}}$

$sin\theta=\approx \theta=9.606\times 10^{-7} rad$

$\frac{y}{L}=tan\theta\approx \theta$

$y=L\theta=265000\times 9.606\times 10^{-7}=0.255 m$