A) \[-\,4.8\text{ }cm,\,-3.3\text{ }D\]
B) \[-5.8\text{ }cm,\,\,-4.3\text{ }D\]
C) \[-7.5\text{ }cm,\,\,-6.3\text{ }D\]
D) \[-15.8\text{ }cm,\,\,-6.33\text{ }D\]
Correct Answer: D
Solution :
The student should use a lens which forms image at distance of 15 cm of the object placed at 3m i.e., object distance\[u=-\,3m=-\,300\,cm,\] image distance \[v=-15\,cm\] From lens formula \[\frac{1}{\upsilon }-\frac{1}{u}=\frac{1}{f}\] We get \[\frac{1}{-15}-\frac{1}{-\,300}=\frac{1}{f}\] or \[\frac{1}{f}=\frac{1}{300}-\frac{1}{15}=-\frac{19}{300}\] or \[f=\frac{-\,300}{19}\] \[=-15.8\,cm\] (concave lens) Now power of the lens is \[P=\frac{100}{f(cm)}=\frac{100}{-\,015.8}\] \[=-\,6.33\,D\]
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