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
(a) 0.4
(b) 0.8
(c) 1.3
(d) 1.6
Click to See Solution :
$\displaystyle \frac{1}{f_1} = (\mu -1)[\frac{1}{f_1} – \frac{1}{f_2}]$
$\displaystyle \frac{1}{f_1} = (1.5 -1)[\frac{1}{20} – \frac{1}{-20}]$
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