Question

Remote sensing satellites move in an orbit that is at an average height of about $500km$ 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 $50cm$, 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, $\frac{\text{area observed on earth}}{\text{area of screen}}=\frac{A_0}{A}$

$=\frac{\left(\frac{\pi .d_2^2}{4}\right)}{\left(\frac{\pi .d_1^2}{4}\right)}=\frac{d_2^2}{d_1^2}$

$\Rightarrow \frac{A_0}{A}={\left(\frac{h}{f_1}\right)}^2={\left(\frac{500\times 10^{+3}}{50\times 10^{-2}}\right)}^2$

$={\left(10\times 10^3\times 10^3\right)}^2$

$={\left(10^6\right)}^2=10^{12}$