MCQ
Many exoplanets have been discovered by the transit method, where in one monitors, a dip in the intensity of the parent star as the exoplanet moves in front of it. The exoplanet has a radius $R$ and the parent star has radius $100 \,R$. If $I_0$ is the intensity observed on earth due to the parent star, then as the exoplanet transits
  • A
    the minimum observed intensity of the parent star is $0.9 \,I_0$
  • B
    the minimum observed intensity of the parent star is $0.99 \,I_0$
  • C
    the minimum observed intensity of the parent star is $0.999 \,I_0$
  • the minimum observed intensity of the parent star is $0.9999 \,I_0$

Answer

Correct option: D.
the minimum observed intensity of the parent star is $0.9999 \,I_0$
d
(d)

Intensity of radiation (mainly visible light) emitted from surface of a star is proportional to its area.

So, $\quad I \propto A$ or $I=k A$

where, $k=$ constant.

Now, if $I_0=$ intensity of parent star.

Then, $I_0=k \pi(100 R)^2=k \pi R^2 \times 10000$

When exoplanet is in front of star,

observed intensity will be minimum. Let intensity minimum is $I_{\min }$, then

$I_{\min }=k\left[\pi(100 R)^2-\pi R^2\right]$

$\Rightarrow I_{\min } =k \pi R^2(10000-1)$

$=k \pi R^2 \times 9999$

$\text { So, } \frac{I_{min}}{I_0} =\frac{k \pi R^2 \times 9999}{k \pi R^2 \times 10000}$

$\Rightarrow I_{\min }=I_0 \times 0.9999$

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