question_answer

The distance between an object and its real image produced by a converging lens is 0.72 m. The magnification is 2. What will be the magnification when the object is moved by 0.04 m towards the lens?

A)  2    

B)  4    

C)  3    

D)  6

Correct Answer: B

Solution :

:                 Converging lens (or convex lens) forms a real image. Let      \[u=-x\] and \[\upsilon =2x\] Then \[3x=0.72m\] \[x=\frac{0.72}{3}m=0.24m=24cm\] \[\therefore \] \[u=-24m\] and \[\upsilon =48cm\] According to thin lens formula \[\frac{1}{\upsilon }-\frac{1}{u}=\frac{1}{f}\] \[\therefore \] \[\frac{1}{48}+\frac{1}{24}=\frac{1}{f}\] \[\frac{3}{48}+\frac{1}{f}\]  or \[f=16cm\] When the object is moved by 0.04 m (= 4 cm) towards the lens, then new object distance is \[u=-20cm\] Let \[\upsilon \] be new image distance. \[\therefore \]   \[\frac{1}{\upsilon }-\frac{1}{-20}=\frac{1}{16}\] or  \[\frac{1}{\upsilon }=\frac{1}{16}-\frac{1}{20}\] \[\frac{1}{\upsilon }=\frac{1}{80}\]  or  \[\upsilon =80cm\] \[+ve\] sign shows that the image formed by the lens is behind the lens. It is real image. New magnification, \[m=\frac{\upsilon }{u}=\frac{80cm}{-20cm}=-4\] \[|m|=4\]