(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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(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.

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