A) 120
B) 115
C) 110
D) 105
Correct Answer: C
Solution :
The final image is formed at infinity, ie, image formed by eye-piece is at infinity. It mean that objective has formed the image at focus of eye lens. So, \[{{v}_{0}}=20.5-2.5\] \[=18\,cm,\] \[{{f}_{0}}=1.5\,cm\] From \[\frac{1}{{{f}_{o}}}=\frac{1}{{{v}_{o}}}-\frac{1}{{{u}_{o}}}\] \[\frac{1}{1.5}=\frac{1}{18}-\frac{1}{{{u}_{o}}}\] \[\frac{1}{{{u}_{o}}}=\frac{1}{18}-\frac{1}{1.5}\] \[=\frac{5-60}{90}\] \[=-\frac{55}{90}=-\frac{11}{18}\] \[{{u}_{o}}=-\frac{18}{11}cm\] Magnification \[M=-\frac{{{v}_{o}}}{{{u}_{o}}}\cdot \frac{D}{{{f}_{e}}}\] \[=-\frac{18}{18/11}\times \frac{25}{2.5}\] \[=110\]
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