A) 10 and 7
B) 9 and 6
C) 3 and 4
D) 4 and 3
Correct Answer: A
Solution :
As, linear magnification \[=\frac{v}{u}\] \[=\frac{-36cm}{-36cm}=10\] Now, when object is placed at a distance \[{{u}_{0}}\] from the lens, then angle subtained by the object on the lens is \[\beta =\frac{h}{{{u}_{0}}},\]where h is the height of the object. Also, the maximum angle subtainded on the unaided eye is \[\alpha =\frac{h}{D}\] Thus, angular magnification, \[m=\frac{\beta }{\alpha }=\frac{D}{{{u}_{0}}}=\frac{25cm}{3.6cm}\approx 7\]
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