A telescope has an objective of focal length 50 cm and an eyepiece of focal length 5 cm . The least

Question

A telescope has an objective of focal length 50 cm and an eyepiece of focal length 5 cm . The least distance of distinct vision is 25 cm . The telescope is focussed for distinct vision on a scale 200 cm away from the objective. The separation between the objective and the eyepiece is $\_\_\_\_$ cm (Round off to nearest integer)

Professor PC Thomas Classes · Answer

Given $u=-200 \mathrm{~cm}, f_{o}=50 \mathrm{~cm}$. The distance of the image $I$ formed by the objective is given by\ $\frac{1}{v}=\frac{1}{f_{o}}+\frac{1}{u}=\frac{1}{50}-\frac{1}{200}$\ which gives $v=\frac{200}{3} \mathrm{~cm}$.\ Therefore, magnification by objective is\ $m_{o}=\frac{|v|}{|u|}=\frac{200 / 3}{200}=\frac{1}{3}$\ The image $I$ serves as the object for the eyepiece. Thus $v^{\prime}=-25 \mathrm{~cm}, f_{e}=5 \mathrm{~cm}$. The object distance $u^{\prime}$ is given by\ $\frac{1}{u^{\prime}}=\frac{1}{v^{\prime}}-\frac{1}{f_{e}}=-\frac{1}{25}-\frac{1}{5} \Rightarrow u^{\prime}=-\frac{25}{6} \mathrm{~cm}$.\ Magnification by eye-piece is\ $m_{e}=\frac{\left|v^{\prime}ight|}{\left|u^{\prime}ight|}=\frac{25}{25 / 6}=6$\ Net magnification, $m=m_{o} 	imes m_{e}=\frac{1}{3} 	imes 6=2$\ Separation between objective and eye-piece\ $=|v|+\left|u^{\prime}ight|=\frac{200}{3}+\frac{25}{6}=70.8 \mathrm{~cm}$\