Question

Remote sensing satellites move in an orbit that is at an average height of about $500\, km$ from the surface of the earth. The camera onboard one such satellite has a screen of area A on which the images captured by it are formed. If the focal length of the camera lens is $50 \,cm$, then the terrestrial area that can be observed from the satellite is close to

• A. $2\times 10^{3}\,A$
• B. $10^{6}\, A$
• C. $10^{12}\,A$
• D. $4 \times 10^{12}\,A$

Answer 1

Answer: (C)

Consider the given diagram,

Assuming area observed and screen both circular, we have
$\theta_{1}=\theta_{2} \Rightarrow \frac{d_{1}}f=\frac{d_{2}}{h}$
$\Rightarrow \frac{d_{2}}{d_{1}}=\frac{h}{f}$
where, $d_{1}$ = diameter of camera screen
and $d_{2}$= diameter of area on earth
Now, $\frac{\text{area observed on earth}}{\text{area of screen}}=\frac{A_{0}}{A}$
$=\frac{\left(\frac{\pi.d_{2}^{2}}{4}\right)}{\left(\frac{\pi. d_{1}^{2}}{4}\right)}=\frac{d_{2}^{2}}{d_{1}^{2}}$
$\Rightarrow \frac{A_{0}}{A}=\left(\frac{h}{f_{1}}\right)^{2}=\left(\frac{500\times10^{+3}}{50\times10^{-2}}\right)^{2}$
$=\left(10\times10^{3}\times10^{3}\right)^{2}$
$=\left(10^{6}\right)^{2}=10^{12}$