Question

A tower $100m$ tall at a distance of $3$km is seen through a telescope having objective of focal length $140$cm and eyepiece of focal length $5cm$. Then the size of final image if it is at $25$cm from the eye?

• A. 14 cm
• B. 28 cm
• C. 42 cm
• D. 56 cm

Answer 1

The correct option is B 28 cm

For objective lens

$\frac{1}{v}-\frac{1}{3\times {10}^5}=\frac{1}{140}\left\{\begin{array}{c}v=140cm=f0\\ approx\end{array}\right\}$

$v=140$

and as final image is at least distance of dishnct vision, so for eye lens,

we have

$\begin{array}{cc}\frac{1}{-25}& -\frac{1}{4e}=\frac{1}{6},(i.e{.4}_e=-\frac{25}{6})\\ U_e& =-\frac{25}{6}\\ m_0=\frac{V_0}{4}& =\frac{140}{-3\times {10}^5}\\ & =\frac{-14}{3}\times {10}^{-4}\\ m_e& =\frac{Ve}{u_e}=\frac{-25}{-25/6}=6\\ m& =m_0\times m_e\\ & =-\frac{14}{3}\times {10}^{-4}\times 6=-28\times {10}^{-4}\end{array}$

We know that

$\begin{array}{cc}M& =\frac{T}{0}\\ \Rightarrow T& =M\times 0\\ & =-28\times {10}^{-4}\times 100\times {10}^2\text{(Object}h=100m)\end{array}$

$=-28cm$ (-ve sign indicates invert image)

Hence (B) is correct ophon