Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 


Can you answer this question?

1 Answer

0 votes
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 $=\large\frac{-1}{f_M}=\frac{-1}{-0.11}$
Further as focal length of lens is 20 cm
ie $0.2\;m$ its power is
$\quad =\frac{1}{20}$$D$
Now as in image formation, light after passing through the lens will be reflected back by curved mirror through lens again .
$\quad= 2 P_L +P_M$
$P= \large\frac{2}{0.2}+\frac{1}{0.11}=\frac{210}{11}$$\;D$
So, focal length of equivalent mirror.
$F= \large\frac{-1}{P} =\frac{11}{210}$$m$
ie silvered lens behaves as a concave mirror of focal length $(\large\frac{110}{21})$$\;cm$
So far object at a distance 10 cm in front of it.
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App