Question

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

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

Answer 1

Answer: (B)

$28cm$

$\begin{array}{c}{f}_0=140m\\ f_e=5cm\end{array}$

$O=100\times {10}^{2}cm\text{size of object}$

$\begin{array}{c}{v}_e=25cm\\ u_0=-3\times {10}^{5}cm\\ \text{for objective}\\ \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end{array}$

$\begin{array}{c}V=140cm=f_o\text{because}\text{u is very big}\\ \text{magnification}\\ m_0=\frac{v}{u}=\frac{140}{-3\times {10}^5}=-\frac{14}{3}\times {10}^{-4}\end{array}$

$\begin{array}{c}\text{again using lens formula for eye piece}\\ \frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\\ \frac{1}{25}-\frac{1}{u_e}=\frac{1}{5}\\ u_e=\frac{-25}{6}\\ m_e=\frac{v_e}{u_e}=\frac{-25}{\frac{-25}{6}}=6\\ m=m_0\times m_e=-\frac{4}{3}\times {10}^4\times 6\\ m=\frac{I}{O}\\ ⟹I=-28\times {10}^4(100\times {10}^2)=-28cm\end{array}$