Question

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s08_qp_1.pdf Q26

Answer 1

The the intensity is inversely proportional to the square of the distance\[I\propto \frac1{x^2}\]
Let the proportionality constant be alpha (\(\alpha\))\[I(x)=\frac{\alpha}{x^2}\]
At point P, \(x=r\), and the intensity is given
\[I_P=I(r)=\frac\alpha{r^2}=8.0[\text{ μm}]\]
Use this to find an expression for alpha
\[\alpha=r^28.0[\text{ μm}]\]
At point Q, \(x=2r\)
\[I_Q=I(2r)=\frac\alpha{(2r)^2}=\frac\alpha{4r^2}\]
substituting in alpha\[=\frac{r^28.0[\text{ μm}]}{4r^2}\]
simplify
\[I_Q=\]