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  4. The aperture diameter of a telescope is 5 m. The separation between the moon and the ear this 4 × 105 km. With light of wavelength of 5500 mathringA, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :

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Q. 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 $\mathring{A}$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :

JEE MainJEE Main 2020Wave Optics

Solution:

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 $60m.$