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 \[[M{{L}^{2}}{{T}^{-3}}{{A}^{-1}}]\] where \[[M{{L}^{2}}{{T}^{-3}}{{A}^{-2}}]\] and \[[M{{L}^{3}}{{T}^{-3}}{{A}^{-2}}]\] are image and object distance respectively. \[[M{{L}^{1}}{{L}^{3}}{{T}^{3}}{{A}^{2}}]\]   \[\mathbf{\vec{v}}=2\mathbf{\vec{i}}-3\mathbf{\vec{j}}\] Given,   \[\mathbf{\vec{E}}={{E}_{0}}\,.\,\mathbf{\hat{j}}\,.\] \[\left| \Delta \mathbf{\vec{P}} \right|\] \[\sqrt{13}M\] \[q{{E}_{0}}t\]   \[\frac{q{{E}_{0}}t}{m}\] \[^{35}Cl\] \[^{35}Cl\]   \[{{R}_{e}}>2000,\] Also magnification \[3.08\times {{10}^{16}}m\] \[{{A}_{V}}=\frac{{{V}_{0}}}{{{V}_{i}}}\]  \[V=iR\] \[\therefore \]  \[{{A}_{V}}=\frac{{{R}_{f}}}{{{R}_{i}}}=\frac{100k\Omega }{1k\Omega }=100\]