The resolving power of an astronomical telesc | vedclass

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Question

(A) The resolving power $(RP)$ of a telescope is given by the formula $RP = \frac{D}{1.22 \lambda}$,where $D$ is the diameter of the objective lens an

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(A) The resolving power $(RP)$ of a telescope is given by the formula $RP = \frac{D}{1.22 \lambda}$,where $D$ is the diameter of the objective lens and $\lambda$ is the wavelength of light.Initially,$RP_1 = \frac{D}{1.22 \lambda} = 0.2 	ext{ seconds}$.When the central half portion of the objective lens is covered,the effective diameter of the lens remains the same because the light passes through the annular region. However,the intensity of the image decreases,but the resolving power depends on the aperture diameter $D$.If the question implies the aperture diameter is reduced to half $(D' = D/2)$,then the new resolving power $RP_2 = \frac{D/2}{1.22 \lambda} = \frac{1}{2} RP_1 = 0.1 	ext{ seconds}$.Thus,the correct option is $A$.

The resolving power of an astronomical telescope is $0.2$ seconds. If the central half portion of the objective lens is covered,the resolving power will be......$sec$

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