MCQ
The average distance between the earth and moon is $38.6 \times 10^4 \mathrm{~km}$. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of $5 \mathrm{~m}$ with $\lambda=6000 \mathcal{A}$ is
  • A
    $5.65 \mathrm{~m}$
  • B
    $28.25 \mathrm{~m}$
  • C
    $11.30 \mathrm{~m}$
  • $56.51 \mathrm{~m}$

Answer

Correct option: D.
$56.51 \mathrm{~m}$
Resolving power $=\frac{1.22 \lambda}{a}=\frac{1.22 \times 6000 \times 10^{-10}}{5}$
Also resolving power $=\frac{d}{D}=\frac{d}{38.6 \times 10^7}$
$\therefore \frac{1.22 \times 6 \times 10^{-7}}{5}=\frac{d}{38.6 \times 10^7} $
$\Rightarrow d=\frac{1.22 \times 6 \times 10^{-7} \times 38.6 \times 10^7}{5} m =56.51 m$

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