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A pin is placed $10 cm$ in front of a convex lens of focal length $20 cm$ made of a material having refractive index $1.5$. The surface of lens is further away from P in is silvered and has a radius of curvature $22 \;cm$ Then position of final image will be :

(a) 10 cm in front of silvered lens

(b) 10 cm backward of silvered lens

(c) 11 cm in front of silvered lens

(d) 11 cm backward of silvered lens 

 

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The curved silvered surface will behave as a concave lens of focal length $f_m =\large\frac{R}{2}=\frac{-22}{2}$
$\qquad= -11 \;cm$
$\qquad= -0.11\;m$
And hence
$P_M $=the power of mirror $=\large\frac{-1}{f_M}=\frac{-1}{-0.11}$
$\qquad=\large\frac{1}{0.11}$$D$
Further as focal length of lens is 20 cm
ie $0.2\;m$ its power is
$P_2=\large\frac{1}{f_2}$
$\quad =\frac{1}{20}$$D$
Now as in image formation, light after passing through the lens will be reflected back by curved mirror through lens again .
$P=P_L+P_M+P_L$
$\quad= 2 P_L +P_M$
$P= \large\frac{2}{0.2}+\frac{1}{0.11}=\frac{210}{11}$$\;D$
So, focal length of equivalent mirror.
$F= \large\frac{-1}{P} =\frac{11}{210}$$m$
$\qquad=\large\frac{-110}{21}\;cm$
ie silvered lens behaves as a concave mirror of focal length $(\large\frac{110}{21})$$\;cm$
So far object at a distance 10 cm in front of it.
$\large\frac{1}{V}+\frac{1}{-10}=\frac{21}{110}$
i,e $V= -11 \;cm$
The image will be 11 cm in front of silvered lens and will be real.
Hence c is the correct answer.
answered Feb 13, 2014 by meena.p
 

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