First slide

The lens

Question

When an object is at distance 10 cm and 30 cm from a thin converging lens, virtual and real image of same size is formed. The power of lens is

Moderate

A

2D
B

4D
C

5D
D

10D

Solution

Here, $u = - 10 c m, f = + F, m = + m_{0}$
$∵ \frac{v}{u} = + m_{0} o r v = - 10 m_{0}$
Or    $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} o r - \frac{1}{10 m_{0}} + \frac{1}{10} = \frac{1}{F}$
Or    $\frac{- 1 + m_{0}}{10 m_{0}} = \frac{1}{F} ................................. (i)$
If $u' = - 30 c m, f' = F$
$∴ m' = - m_{0} = \frac{v'}{u'}$
$∴ v' = - m_{0} u'$
$= - m_{0} (- 30) = 30 m_{0} ∴ \frac{1}{30 m_{0}} + \frac{1}{30} = \frac{1}{F} ............................... (i i)$
From Eqs. (i) and (ii) we get, $\frac{- 1 + m_{0}}{10 m_{0}} = \frac{1}{30 m_{0}} + \frac{1}{30}$
Or    $\frac{- 1 + m_{0}}{10 m_{0}} = \frac{1 + m_{0}}{30 m_{0}}$
Or $- 30 m_{0} + 30 m_{0}^{2} = 10 m_{0} + 10 m_{0}^{2}$
Or $20 m_{0}^{2} = 40 m_{0}$
$∴ m_{0} = 2$
$∴ \frac{1}{F} = \frac{1}{60} + \frac{1}{30} = \frac{1 + 2}{60} = \frac{1}{20}$
P=1/F = 100/20=5D

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