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 $4m$ and $10cm$ respectively. The angle subtended by the diameter of the image of the moon will be approximately

• A. $2^o$
• B. ${20}^o$
• C. ${40}^o$
• D. ${50}^o$

Answer 1

The correct option is B ${20}^o$

Diameter of the moon = $3.5\times {10}^3\text{km}$

Distance from earth = $3.8\times {10}^5\text{km}$

The angle subtended by the diameter of the moon on the objective will be

${\theta _0 } = \frac{{{\text{Diameter of the moon}}}}{{{\text{Distance from earth}}}}$

${\theta }_0=\frac{3.5\times {10}^3}{3.8\times {10}^5}=9.2\times {10}^{-3}\text{rad = 0}{\text{.5}}^{\circ }$

Magnification (m) = $\frac{{\text{f}}_0}{{\text{f}}_{\text{e}}}$

$\Rightarrow \text{m =}\frac{\text{400}}{\text{10}}\text{= 40}$

Also, we know that

$\text{m}=\frac{\theta }{{\theta }_0}$

Substituting the values,

$40=\frac{\theta }{0.5}$

$\Rightarrow \theta ={20}^{\circ }$

The angle subtended by the diameter of the image of the moon will be ${20}^{\circ }$