First slide

The lens

Question

A point object P (at distance of 15 cm from lens) is approaching towards a thin and silvered equiconvex lens of local length 20 cm and refractive index 1.5 with a speed of $4 c m s^{- 1}$ as shown in figure. Find velocity of image.

Moderate

A

1 $c m s^{- 1}$ away from lens
B

2 $c m s^{- 1}$ towards the lens
C

1 $c m s^{- 1}$ towards the lens
D

4 $c m s^{- 1}$ away from lens

Solution

$\begin{array}{l} \frac{1}{f} = (μ - 1) (\frac{1}{R} + \frac{1}{R}) \\ \Rightarrow \frac{1}{20} = (1.5 - 1) \frac{2}{R} \\ ∴ R = 20 c m \end{array}$
Power of equivalent mirror is $P = 2 P_{1} + P_{m} = \frac{2}{20} - \frac{1}{- 10} = \frac{1}{10} + \frac{1}{10} = \frac{1}{5}$
$\begin{array}{l} ∴ \frac{- 1}{F} = \frac{1}{5} \\ ∴ F = - 5 c m \\ C o m b i n a t i o n b e h a v e s l i k e C o n c a v e M i r r o r \\ ∵ \frac{1}{v} + \frac{1}{u} = \frac{1}{F} \\ o r \frac{1}{v} - \frac{1}{15} = - \frac{1}{5} \\ O r \frac{1}{v} = \frac{1}{15} - \frac{1}{5} = \frac{1 - 3}{15} = \frac{- 2}{15} \\ ∴ v = \frac{- 15}{2} c m \\ ∴ v_{i m a g e} = \frac{- v^{2}}{u^{2}} \frac{d u}{d t} = - \frac{{(- \frac{15}{2})}^{2}}{{(- 15)}^{2}} \times 4 = - 1 c m s^{- 1} \end{array}$

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