An extended object is placed at a point O , 10 cm in front of a convex lens L1 and a concave lens L2 is placed ….

Q: An extended object is placed at a point O , 10 cm in front of a convex lens L1 and a concave lens L2 is placed 10 cm behind it , as shown in the figure . The radii of curvature of all the curved surface in both the lenses are 20 cm .The refractive index of both the lenses is 1.5 . The total magnification of this lens system is

Numerical

(a) 0.4

(b) 0.8

(c) 1.3

(d) 1.6

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Ans: (b)
Sol: Focal length of Convex lens f1

$\displaystyle \frac{1}{f_1} = (\mu -1)[\frac{1}{f_1} – \frac{1}{f_2}]$

$\displaystyle \frac{1}{f_1} = (1.5 -1)[\frac{1}{20} – \frac{1}{-20}]$

f1 = 20 cm

Focal length of Concave lens f2

$\displaystyle \frac{1}{f_2} = (\mu -1)[\frac{1}{f_1} – \frac{1}{f_2}]$

$\displaystyle \frac{1}{f_2} = (1.5 -1)[\frac{1}{-20} – \frac{1}{20}]$

f2 = -20 cm

For Lens 1 :

$\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

v = – 20 cm
$\displaystyle m_1 = \frac{v}{u} = \frac{-20}{-10} = 2$

For Lens 2 :

u = -30 cm , f = -20 cm

$\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

v = -12 cm

$\displaystyle m_2 = \frac{v}{u} = \frac{-12}{-30} = \frac{2}{5}$

Net magnification m = m1 x m2

m = 2 x (2/5) = 4/5 =0.8