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

Step-by-step solution

Intensity of radiation (mainly visible light) emitted from surface of a star is proportional to its area. So , I ∝ A or I = k A where, k = constant. Now, if I 0 = intensity of parent star. Then, I 0 = k π ( 100 R ) 2 = k π R 2 × 10000 When exoplanet is in front of star, observed intensity will be minimum. Let intensity minimum is I min , then I min = k π ( 100 R ) 2 - π R 2 ⇒ I min = k π R 2 ( 10000 - 1 ) = k π R 2 × 9999 So, I min I 0 = k π R 2 × 9999 k π R 2 × 10000 ⇒ I min = I 0 × 0 . 9999