A) 0.95cm
B) 0.85cm
C) 0.75cm
D) None of these
Correct Answer: B
Solution :
Key Idea: Focal length is half of radius of curvature. From the mirror formula \[\frac{1}{f}=\frac{1}{v}=\frac{1}{u}\] where \[f\] is focal length \[\left( =\frac{R}{2} \right)\], v is image distance, and u is object distance. Given, \[f=\frac{R}{2}=\frac{4.8}{2}=2.4\,cm,\,\,\mu =-10\,cm\] \[\therefore \] \[\frac{1}{2.4}=\frac{1}{v}-\frac{1}{10}\] \[\Rightarrow \] \[\frac{1}{v}=\frac{1}{2.4}+\frac{1}{10}=\frac{1}{1.94}\] \[\Rightarrow \,\,\,v=1.94,\,\,cm\] (behind the mirror) In second case, \[\frac{1}{v}=\frac{1}{2.4}+\frac{1}{2}\] \[\Rightarrow \] \[v=1.09\,\,cm\] Hence, shift of image \[=1.94-1.09=0.85\,\,cm\].
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