A) \[25\]
B) \[6\]
C) \[6\]
D) \[150\]
Correct Answer: D
Solution :
: Here, Focal length of objective, \[{{f}_{0}}=1cm\] Focal length of eyepiece, \[{{f}_{e}}=6\,cm\] Least distance of distinct vision, \[D=25\text{ }cm\] Tube length, \[L=30\text{ }cm\] When the image is formed at the least distance of distinct vision, the magnification is \[m=\frac{L}{{{f}_{0}}}\left( 1+\frac{D}{{{f}_{e}}} \right)=\frac{30}{1}\left( 1+\frac{25}{6} \right)\] \[=155\approx 150\]
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