A) 5
B) 8
C) 10
D) 20
Correct Answer: A
Solution :
Lens formula is given by \[\frac{1}{f}=\frac{1}{\nu }-\frac{1}{\mu }\] (i) where f is the focal length of lens, \[\nu \] is the image distance and u is the object distance. Given, \[\operatorname{f}\,\,=\,\,10 cm\] (as lens is converging) \[\operatorname{u}= - 8 cm\] (as object is placed on left side of the lens) Substituting these values in Eq. (i), we get \[\frac{1}{10}=\frac{1}{\nu }-\frac{1}{-8}\] \[\Rightarrow \,\,\,\,\frac{1}{\nu }=\frac{1}{10}-\frac{1}{8}\] \[\Rightarrow \,\,\,\,\frac{1}{\nu }=\frac{8-10}{80}\] \[\therefore \,\,\,\,\,\,\,\nu =\frac{80}{-2}=\,\,-\,40\,cm\] Hence, magnification produced by the lens \[m=\frac{\nu }{u}=\frac{-\,40}{-8}=5\]
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