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 xx 10^(6)m`. and the radius of the lunar orbit round the earth is `3.8 xx 10^(8) m`.

Options

A. `10 cm`
B. `12.5 cm`
C. `15 cm`
D. `17.5 cm`

Answer

D

Explanation

As `u gt gt f_(0), v = f_(0) = 19 m`
Now `u = - 3.8 xx 10^(8) m`. Therefore, magnification porduced by the objective is
`m_(0) = (v)/(u) = - (19)/(3.8 xx 10^(8)) = -0.5 xx 10^(-7)`
`:.` Diameter fo the image of moon is
`3.5 xx 10^(6) xx 0.5 xx 10^(-7) = 0.175 m = 17.5 cm`