First slide

The lens

Question

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.)

Difficult

A

0.56 s and 4.45s
B

0.56s & 8.94s
C

8.94s and 19.62s
D

0.56s

Solution

The detector receives maximum light when image of ‘S’ falls on ‘D’.
u=-[100-(20-10)t] =-(100-10t)
v=(30-10)t=20t
$\begin{array}{l} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \Rightarrow \frac{1}{10} = \frac{1}{20 t} - \frac{1}{- (100 - 10 t)} \\ \Rightarrow \frac{1}{10} = \frac{1}{20 t} + \frac{1}{(100 - 10 t)} \\ \Rightarrow \frac{1}{10} = \frac{(100 - 10 t) + 20 t}{20 t (100 - 10 t)} \\ \Rightarrow 200 t - 20 t^{2} = 10 t + 100 \\ \Rightarrow 2 t^{2} - 19 t + 10 = 0 \end{array}$
$\begin{array}{l} c o m p a r i n g w i t h q u a d r a t i c e q u a t i o n a t^{2} + b t + c = 0 \\ a = 2, b = - 19, c = 10 \\ \Rightarrow t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ \Rightarrow t = \frac{- (- 19) \pm \sqrt{{(- 19)}^{2} - 4 (2) (10)}}{2 (2)} \end{array}$
$\begin{array}{l} \Rightarrow t = \frac{19 \pm \sqrt{361 - 80}}{4} = \frac{19 \pm 16.7}{4} \\ \Rightarrow t = \frac{19 + 16.7}{4} = 8.94 s \\ \Rightarrow t = \frac{19 - 16.7}{4} = 0.56 s \end{array}$

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