• # question_answer A photosensitive material is at 9 m to the left of the origin and the source of light is at 7 m to the right of the origin along $x$-axis. The photosensitive material and the source of light start from rest and moves with 8i m/s and 4i m/s respectively. The ratio of intensity at $t=0$ to$t=3$s as received by the photosensitive material is A)  16: 1                           B)  1 : 16            C)  2 : 7                            D)  7 : 2

The   separation   between   source   and photosensitive material at ${{\phi }_{1}}=$ is 16m, therefore, intensity received by photosensitive material at $=BA$ is ${{\phi }_{2}}=$, where P is the power of source of light. At $||$s, the source is at (15, 0) and detector is at (19, 0), so the separation between them is 4 m. $=0$             So,       $\therefore \,\,|\xi |\,=\,\left| \frac{-\Delta \phi }{\Delta t} \right|=\,\left| -\left( \frac{0-BA}{\frac{T}{4}-0} \right) \right|=\frac{4BA}{T}$