The aperture diameter of a telescope is 5; m. | vedclass

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Question

(C) The angular resolution of a telescope is given by the formula $\Delta 	heta = \frac{1.22 \lambda}{D}$, where $\lambda$ is the wavelength of light

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The angular resolution of a telescope is given by the formula $\Delta 	heta = \frac{1.22 \lambda}{D}$, where $\lambda$ is the wavelength of light and $D$ is the diameter of the aperture.Given: $\lambda = 5500 \; \mathring{A} = 5500 	imes 10^{-10} \; m$, $D = 5 \; m$, and distance $d = 4 	imes 10^{5} \; km = 4 	imes 10^{8} \; m$.The linear separation $x$ between two objects on the moon that can be just resolved is given by $x = d \cdot \Delta 	heta$.Substituting the values:$x = d 	imes \frac{1.22 \lambda}{D} = \frac{4 	imes 10^{8} 	imes 1.22 	imes 5500 	imes 10^{-10}}{5}$.$x = \frac{4 	imes 1.22 	imes 5.5 	imes 10^{-2}}{5} 	imes 10^{8} = 0.8 	imes 1.22 	imes 5.5 	imes 10^{-2} 	imes 10^{8} = 53.68 \; m$.Rounding this value to the nearest provided option, we get $60 \; m$.

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

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