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=20t1f=1v−1u​⇒110=120t−1−100−10t​⇒110=120t+1100−10t​⇒110=100−10t+20t20t100−10t⇒​200t−20t2=10t+100​⇒2t2−19t+10=0​​comparing with quadratic equation at2+bt+c=0​a=2, b=−19, c=10​⇒t=−b±b2−4ac2a​⇒t=−−19±−192−421022​⇒t=19±361−804=19±16.74⇒​t=19+16.74=8.94s⇒​t=19−16.74=0.56s