The diameter of the moon is $3.5 \times 10^{3} k m$ and its distance from earth is $3.8 \times 10^{5} k m .$ When seen by a telescope having focal lengths of the objective and the eye piece as 40 mm and 10 mm respectively, the angle subtended by the image of the moon at the eyes for normal vision will be approximately

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$2^{0}$

$10^{0}$

$20^{0}$

$4^{0}$

answer is A.

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Detailed Solution

(1)

Angular magnification = $\frac{β}{α} = \frac{f_{0}}{f_{e}}$

$\begin{array}{l} N o w α = \frac{3.5 \times 10^{3}}{3.8 \times 10^{5}} r a d = 9.210 \times 10^{- 3} r a d . \\ ∴ β = 9.210 \times 10^{- 3} \times \frac{40}{10} r a d ≃ 2^{0} \end{array}$

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