Question

The diameter of moon is $3.5 \times 10^{3}\, km$ and its distance from the earth is $3.8 \times 10^{5} \,km$. The focal length of the objective and eyepiece are $4 \,m$ and $10 \,cm$ respectively. The limit of resolution of diameter of the image of the moon will be approximately

• A. $2^{\circ}$
• B. $21^{\circ}$
• C. $40^{\circ}$
• D. $50^{\circ}$

Answer 1

Answer: (B)

Here, $\alpha=\frac{3.5 \times 10^{3}}{3.8 \times 10^{5}}$,rad
$=\frac{3.5}{3.8 \times 100} \times \frac{180^{\circ}}{\pi}$
$=\frac{3.5 \times 180 \times 7^{\circ}}{38 \times 100 \times 22}$
Also, $M=\frac{f_{o}}{f_{ e }}=\frac{400}{10}=40$
$\therefore $ Limit of resolution, $ \beta =\frac{40 \times 35 \times 180 \times 7^{\circ}}{35 \times 100 \times 22} $
$=21 . 1^{\circ} \approx 21^{\circ}$