JEE Main & Advanced
JEE Main Paper (Held On 11 April 2015)
question_answer
A thin convex lens of focal length T is put on a plane mirror as shown in the figure. When an object is kept at a distance 'a' from the lens - mirror combination, its image is formed at a distance\[\frac{a}{3}\]in front of the combination. The value of 'a' is :
[JEE Main Online Paper (Held On 11 April 2015)]
A) f
B) 2f
C) 3f
D) \[\frac{3}{2}f\]
Correct Answer:
B
Solution :
This combinations will behave like a mirror of power. \[{{P}_{eq}}=2{{P}_{L}}+{{P}_{M}}\] \[{{P}_{eq}}=2\frac{1}{f}+0\] \[{{F}_{eq}}=-\frac{f}{2}\] so the behaviour will be like a mirror of focal length \[-\frac{f}{2}\] Using mirror equation\[\frac{1}{V}+\frac{1}{U}=\frac{1}{{{f}_{eq}}}\] \[\frac{1}{{{-}^{{}^{9}/{}_{3}}}}+\frac{1}{-a}=\frac{-1}{{}^{f}/{}_{2}}\] \[\frac{1}{a}=\frac{-2}{f}\] \[\]