Question

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

Answer 1

Answer: (A)

5 cm

Answer 2

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