A telescope of aperture 3 times 10^{-2} , m d | vedclass

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Question

(A) The limit of resolution of the telescope is given by $\Delta 	heta = \frac{1.22 \lambda}{d}$.Substituting the values: $\Delta 	heta = \frac{1.22

Vedclass · Accepted Answer

Yes,it is possible with the same aperture size.

Vedclass · Answer

(A) The limit of resolution of the telescope is given by $\Delta 	heta = \frac{1.22 \lambda}{d}$.Substituting the values: $\Delta 	heta = \frac{1.22 	imes 5.5 	imes 10^{-7}}{3 	imes 10^{-2}} = 2.236 	imes 10^{-5} \, 	ext{rad}$.The minimum separation $x$ that the telescope can resolve at a distance $D = 80 \, m$ is $x = \Delta 	heta 	imes D$.$x = 2.236 	imes 10^{-5} 	imes 80 = 1.788 	imes 10^{-3} \, m$.The spacing of the wire mesh is $2 	imes 10^{-3} \, m$.Since the spacing $(2 	imes 10^{-3} \, m)$ is greater than the limit of resolution $(1.788 	imes 10^{-3} \, m)$,the mesh can be resolved.If the aperture is halved,the new limit of resolution $\Delta 	heta' = 2 	imes \Delta 	heta = 4.472 	imes 10^{-5} \, 	ext{rad}$.The new minimum separation $x' = 4.472 	imes 10^{-5} 	imes 80 = 3.577 	imes 10^{-3} \, m$.Since $3.577 	imes 10^{-3} \, m > 2 	imes 10^{-3} \, m$,the mesh cannot be resolved if the aperture is halved.Therefore,it is possible with the current aperture,but not with half the diameter.

$A$ telescope of aperture $3 \times 10^{-2} \, m$ diameter is focused on a window at $80 \, m$ distance fitted with a wire mesh of spacing $2 \times 10^{-3} \, m$. Given: $\lambda = 5.5 \times 10^{-7} \, m$,which of the following is true for observing the mesh through the telescope?

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