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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 ............... $A$
$2 \times 10^3$
$10^6$
$10^{12}$
$4 \times 10^{12}$
Solution
(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, $\frac{area \,observed \,on \,earth}{area \,of \,screen}=\frac{A_0}{A}$
$=\frac{\left(\frac{\pi \cdot d_2^2}{4}\right)}{\left(\frac{\pi \cdot d_1^2}{4}\right)}=\frac{d_2^2}{d_1^2}$
$\Rightarrow \quad \frac{A_0}{A}=\left(\frac{h}{f_1}\right)^2=\left(\frac{500 \times 10^{+3}}{50 \times 10^{-2}}\right)^2$
$=\left(10 \times 10^3 \times 10^2\right)^2=\left(10^6\right)^2=10^{12}$
