First slide

Resolving power of optical instruments

Question

The aperture diameter of a telescope is 5 m. The separation between the moon and the earth is $4 \times 10^{5} k m$ . With light of wavelength of $5500 A^{0}$ , the minimum separation between objects on the surface of moon, so that they are just resolved, is close to:

Moderate

A

20 m
B

200 m
C

60 m
D

600 m

Solution

$\begin{array}{l} Let say the minimum separation between objects on the surface of moon = x \\ Separation between the moon and the earth = d = 4 \times 10^{8} m \\ Wavelength = λ = 5500 \times 10^{- 10} m \\ Aperture Diameter = 5 m \end{array}$
$\begin{array}{l} ∴ \frac{x}{d} = \frac{1.22 λ}{a} \\ \Rightarrow x = \frac{1.22 λ}{a} d \\ \Rightarrow x = \frac{1.22 \times 5500 \times 10^{- 10} \times 4 \times 10^{8}}{5} = 53.68 ≃ 60 m \end{array}$

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