The figure shows the initial position of a point source of light S, a detector D and Lens L. Now at t=0 all three start moving towards right with different velocities as shown in figure. The times at which detector receives the maximum light are (initially detector D is on lens as shown in figure.)

 The detector receives maximum light when image of ‘S’ falls on ‘D’. 

u=-[100-(20-10)t] =-(100-10t)

v=(30-10)t=20t

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\text{}\\ \Rightarrow \frac{1}{10}=\frac{1}{20t}-\frac{1}{-\left(100-10t\right)}\text{}\\ \Rightarrow \frac{1}{10}=\frac{1}{20t}+\frac{1}{\left(100-10t\right)}\text{}\\ \Rightarrow \frac{1}{10}=\frac{\left(100-10t\right)+20t}{20t\left(100-10t\right)}\\ \Rightarrow \text{}200t-20t^2=10t+100\text{}\\ \Rightarrow 2t^2-19t+10=0\text{}\end{array}$

$\begin{array}{l}comparing with quadratic equation a{t}^2+bt+c=0\\ \text{}a=2, b=-19, c=10\text{}\\ \Rightarrow t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\text{}\\ \Rightarrow t=\frac{-\left(-19\right)\pm \sqrt{{\left(-19\right)}^2-4\left(2\right)\left(10\right)}}{2\left(2\right)}\text{}\end{array}$

$\begin{array}{l}\Rightarrow t=\frac{19\pm \sqrt{361-80}}{4}=\frac{19\pm 16.7}{4}\\ \Rightarrow \text{}t=\frac{19+16.7}{4}=8.94s\\ \Rightarrow \text{}t=\frac{19-16.7}{4}=0.56s\end{array}$