A) 3 and 1
B) 3 and 3
C) 3 and 2
D) 1 and 3
Correct Answer: A
Solution :
The focal length of each portion of lens corresponding to a certain material will be given by \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{1}}} \right)\] \[\therefore \] \[\frac{1}{f}\propto (\mu -1)\] \[\left[ \because \,\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\,\text{is}\,\text{constant}\,\text{for}\,\text{all}\,\text{lenses} \right]\] Thus, for lens P, three different portions passes three different focal lengths and hence corresponding to the same object distance there will be three images for each. For lens Q, the combination of the three lenses would effectively act as a single lens and hence only one image is formed.
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