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} $



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