Question

A giant telescope in an observatory has an objective of focal length $19 \,m$ and an eye-piece of focal length $1.0 \,cm$. 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. $10\, cm$
• B. $12.5\, cm$
• C. $15\, cm$
• D. $17.5\, cm$

Answer 1

Answer: (D)

As $u > > f_0, v = f_0 = 19\,m$.
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} $
$ = 0.175 \,m$
$ = 17.5\,cm$