question_answer

A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If this telescope is used to see a 50 m tall building at a distance of 2 km, what is the height of the image of the building formed by the objective lens?                               

A) 5 cm                

B) 10 cm    

C) 1 cm                

D) 2 cm    

Correct Answer: A

Solution :

A telescope is an optical instrument used to see distant objects. Since convex lens is used, from lens formula we have \[\frac{1}{{{f}_{o}}}=\frac{1}{{{\nu }_{o}}}-\frac{1}{{{U}_{o}}}\] where \[{{v}_{o}}\,and\,{{u}_{o}}\] are image and object distance respectively. \[\therefore \,\,\,\,\,\,\,\,\frac{1}{{{\nu }_{o}}}=\frac{1}{{{f}_{o}}}+\frac{1}{{{u}_{o}}}\] Given, \[{{f}_{o}}= 200 cm\] \[{{u}_{o}}=-\,2 km=-\,2 \times 1{{0}^{5}}\,cm\] \[O=50\,\,m=5\times 1{{0}^{3}}\,\,cm\] \[\therefore \,\,\,\,\,\,\,\frac{1}{{{\nu }_{o}}}=\frac{1}{200}+\frac{1}{-200\times {{10}^{3}}}\] \[=\,\,\,\,\frac{{{10}^{3}}-1}{200\times {{10}^{3}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,{{\nu }_{o}}=\frac{200\times {{10}^{3}}}{999}\,\,cm\] Also magnification \[m=\left| \frac{{{\nu }_{o}}}{{{u}_{o}}} \right|=\frac{I}{O}\] \[\therefore \,\,\,\,\,\,\,\frac{200\times {{10}^{3}}}{999\times 200\times {{10}^{3}}}=\frac{I}{5\times {{10}^{3}}}\] \[\Rightarrow \,\,\,\,\,\,I=\frac{5\times {{10}^{3}}}{999}\,\,=\,\,5\,\,cm\]