Question

A telescope has an objective of focal length $100cm$ and eye piece of focal length $6cm$ and the least distance of distinct vision is $25cm$. The telescope is focused for distinct vision of an object at a distance $100m$ from the objective. What is the distance of separation between the objective and the eye-piece?

• A. $2.5m$
• B. $1.5m$
• C. $1.06m$
• D. $0.98m$

Answer 1

Answer: (C)

$1.06m$

Telescope least distance of distinct vision $=25cm$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\Rightarrow$lens formula

$\frac{1}{100\times {10}^{-2}}=\frac{1}{v_1}-\frac{1}{100}$

$\frac{1}{v_1}=\frac{1}{100}+1$

$v_1=\frac{100}{101}m=0.99$

$\frac{1}{0.06}=\frac{1}{25}-\frac{1}{-u_2}$

$\frac{100}{6}-\frac{1}{25}=\frac{1}{u_2}{u}_2=\frac{6\times 25}{2500-6}m=0.067$

Distance between objective and eye piece

$\begin{array}{cc}=& v_1+u_2\\ =& 0.99+0.067\\ & =1.06m\\ \text{Hence option}c\text{)}1.06m\text{is the}& \\ \text{correct option.}& \end{array}$