A) 5.5 cm
B) 7.5 cm
C) 12.0 cm
D) 20.0 cm
Correct Answer: B
Solution :
As shown in figure, the point on the right side of the lens at which rays converge will behave as virtual object of the lens. u = + 12 cm, / = 20 cm From the relation \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] \[\therefore \] \[\frac{1}{20}=\frac{1}{v}-\frac{1}{12}\] \[\Rightarrow \] \[\frac{1}{v}=\frac{1}{20}+\frac{1}{12}\] \[=\frac{3+5}{60}=\frac{8}{60}\] \[\therefore \] \[v=\frac{60}{8}=7.5\,\,cm\] So, image will be formed on same side of the virtual object at a distance of 7.5 cm from the lens.
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