Question

A giant telescope in an observatory has an objective of focal length $19m$ and an eye-piece of focal length $1.0cm$. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective ? The diameter of the moon is $3.5\times 10^{6}m$ and the radius of the lunar orbit round the earth is $3.8\times 10^{8}m$.

• A. $10cm$
• B. $12.5cm$
• C. $15cm$
• D. $17.5cm$

Answer 1

Answer: (D)

As $u>>f_0,v=f_0=19m$.
Now $u=-3.8\times 10^{8}m$
Therefore, magnification produced by the objective is
$m_0=\frac{v}{u}$
$=-\frac{19}{3.8\times 10^8}$
$=-0.5\times 10^{-7}$
$\therefore$ Diameter of the image of moon is
$3.5\times 10^6\times 0.5\times 10^{-7}$
$=0.175m$
$=17.5cm$