Physics - Geometrical Optics Question with Solution | TestHub

According to question, the situation is shown in the figure below$E_1=\frac{I}{(LO{)}^2}=\frac{I}{h^2}...(i)$ The illumination at the edge $A$ is given by $E_2=\frac{I\cos \theta }{(LA{)}^2}=\frac{I\cos \theta }{\left(h^2+r^2\right)}...(ii)$ From figure, $\cos \theta =\frac{h}{\sqrt{\left(h^2+r^2\right)}}$ $\therefore E_2=\frac{Ih}{{\left(h^2+r^2\right)}^{3/2}}$ Dividing Eq. (i) by Eq. (ii), we get $\begin{array}{rl}\frac{E_1}{E_2}& =\frac{I/h^2}{Ih/{\left(h^2+r^2\right)}^{3/2}}=\frac{{\left(h^2+r^2\right)}^{3/2}}{h^3}\\ & ={\left(\frac{h^2+r^2}{h^2}\right)}^{3/2}={\left(1+\frac{r^2}{h^2}\right)}^{3/2}\end{array}$