Question

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

• A. $5\text{cm}$
• B. $10\text{cm}$
• C. $1\text{cm}$
• D. $2\text{cm}$

Answer 1

The correct option is A $5\text{cm}$

Given,

$u=-2000\text{m};f_0=200\text{cm}=2\text{m}{f}_e=2\text{cm}$

Where, $f_0=\text{Focal length of the objective}{f}_e=\text{Focal length of the eyepiece}$

Now, using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_0}$

$\Rightarrow \frac{1}{v}+\frac{1}{2000}=\frac{1}{2}$

$\Rightarrow \frac{1}{v}=\frac{1}{2}-\frac{1}{2000}=\frac{1000-1}{2000}$

$\Rightarrow \frac{1}{v}=\frac{999}{2000}$

$\Rightarrow v=\frac{2000}{900}\text{m}$

Since, magnification $m=-\frac{v}{u}=\frac{h_i}{h_0}$

$\Rightarrow m=\frac{2000}{999\times 2000}=\frac{h_i}{50}$

$\therefore$ Size of the image is,

$h_i=\frac{1}{999}\times 50\times 100\approx 5\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question ? Concept and Application: It involves use of focal length formula and its application in real-time scenario.