A) \[\frac{3}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{3}{2}\]
D) none of the above
Correct Answer: C
Solution :
Here, the object lies along the axis. The two ends of the object should be treated as two point objects and the difference between the corresponding image distances gives the length of the image. When one end of the image touches the rod, this end must be at\[2f\]. In this situation the other end of the rod can be towards the left or right of\[2f\]. Since the image of the rod is elongated, the other end of the rod must lie towards right of\[2f\] (when the object lies between\[f\]and\[2f\], the image is formed more far away behind\[2f\]). So, object distance for closer end of the rod is \[2f-f/3\]and that of the farther end is\[2f\]. The difference between the corresponding image distance is found to be\[\frac{f}{2}\], \[i.e.\]length of the image is\[\frac{f}{2}\]. Now, magnification\[=\frac{length\,\,of\,\,image}{length\,\,of\,\,object}=\frac{f/2}{f/3}=\frac{3}{2}\] Note that the image is elongated and the only option which is greater than 1 is .
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