First slide

The lens

Question

The figure shows the initial position of a point source of light S (100 cm away from lens), a detector D and a lens L. Now at t = 0, all the three start moving towards right with different velocities as shown. The time(s) at which the detector receives the maximum light is (velocity of lens is 10cm $s^{- 1}$ )

Moderate

A

0.56 s and 8.94 s
B

3.8 s
C

8.94s and 19.62s
D

0.56s

Solution

The detector will receive the maximum light when the image of point source of light coincides with the position of detector.
Velocity of object and detector are 10 and 20 with respect to lens .
Let at any time t, image of S and detector coincides then, for $u = - (100 - 10 t) c m$
$v = 20 t c m$
$f = 10 c m$
The detector receives maximum light.
From, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{20 t} - \frac{1}{- (100 - 10 t)} = \frac{1}{10}$
$\Rightarrow 2 t^{2} - 19 t + 10 = 0$
$\Rightarrow t = 0.56 s a n d 8.94 s$

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