Two convergent lenses A and B (erect) of focal lengths30 cm and 50 cm are separated by d cm apart co- axially. An erect object of height 2 cm is kept in front of lens A at 40 cm from it. The real image is formed at $\frac{100}{3} CM$ cm to the right of lens B (see figure). The value of d is

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20 cm

25 cm

15 cm

10 cm

answer is A.

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Detailed Solution

$μ_{1} = - 40 cm f_{1} = 30 cm$
$∴ v_{1} = 120 cm From v = \frac{uf}{u + f}$
$u_{2} For the II lens = (- d + 120) cm and f_{2} = 50 cm$
$∴ v_{2} = \frac{- 50 (d - 120)}{- (d - 170)} = \frac{100}{3}$
By solving, we get d = 20 cm

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