Question

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 $0.999\, I_{0}$
• D. the minimum observed intensity of the parent star is $0.9999 \, I_{0}$

Answer 1

Answer: (D)

Intensity of radiation (mainly visible light) emitted from surface of a star is proportional to its area
So, $I \propto A$ or $I= KA$
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
$I_{\min}=K[\pi (100\,R)^{2}-\pi\,R^{2}]$

$\Rightarrow I_\min =k \pi R^{2}(10000-1)$
$=k \pi R^{2} \times 9999$
So, $\frac{I_{\min}}{I_{0}}$
$=\frac{k\pi\,R^{2}\times 9999}{k\pi\,R^{2}\times 10000}$
$\Rightarrow I_\min=I_{0}\times 0.9999$