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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

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.