Optics Question 491
Question: The distance between a point source of light and a screen which is 60 cm is increased to 180 cm. The intensity on the screen as compared with the original intensity will be
[CPMT 1888]
Options:
A) (1 / 9) times
B) (1 / 3) times
C) 3 times
D) 9 times
Show Answer
Answer:
Correct Answer: A
Solution:
$ I\propto \frac{1}{r^{2}} $
$ \Rightarrow \frac{I _{2}}{I _{1}}=\frac{r _{1}^{2}}{r _{2}^{2}} $
$ =\frac{60^{2}}{180^{2}}=\frac{1}{9} $
