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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

10D

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed 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