Question:
The aperture diameter of a telescope is 5 m. The separation between the moon and the ear this $4 \times 10^5$ km. With light of wavelength of 5500 $?$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :
The aperture diameter of a telescope is 5 m. The separation between the moon and the ear this $4 \times 10^5$ km. With light of wavelength of 5500 $?$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :
Updated On: Aug 21, 2024
- 600 m
- 20 m
- 60 m
- 200 m
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The Correct Option is C
Solution and Explanation
Let distance is $x$ then
$d\theta=\frac{1.22\lambda}{D}$ ($D =$ diameter)
$\frac{x}{d}=\frac{1.22\lambda}{D}$ ($d =$ distance between earth & moon)
$x=\frac{1.22\times\left(5500\times10^{-10}\right)\times\left(4\times10^{8}\right)}{5}=53.68\,m$
most appropriate is $60m.$
$d\theta=\frac{1.22\lambda}{D}$ ($D =$ diameter)
$\frac{x}{d}=\frac{1.22\lambda}{D}$ ($d =$ distance between earth & moon)
$x=\frac{1.22\times\left(5500\times10^{-10}\right)\times\left(4\times10^{8}\right)}{5}=53.68\,m$
most appropriate is $60m.$
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