The diameter of the objective of a telescope | vedclass

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(D) The resolving power of a telescope is given by the formula:$RP = \frac{D}{1.22 \lambda}$Where $D$ is the diameter of the objective and $\lambda$ i

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$3.28 	imes 10^{6}$

Vedclass · Answer

(D) The resolving power of a telescope is given by the formula:$RP = \frac{D}{1.22 \lambda}$Where $D$ is the diameter of the objective and $\lambda$ is the wavelength of light.Given:$D = 200 	ext{ cm} = 2 	ext{ m}$$\lambda = 5000 	ext{ \AA} = 5000 	imes 10^{-10} 	ext{ m} = 5 	imes 10^{-7} 	ext{ m}$Substituting the values:$RP = \frac{2}{1.22 	imes 5 	imes 10^{-7}}$$RP = \frac{2}{6.1 	imes 10^{-7}}$$RP = \frac{2}{6.1} 	imes 10^{7}$$RP \approx 0.32786 	imes 10^{7}$$RP \approx 3.28 	imes 10^{6}$Therefore, the resolving power of the telescope is $3.28 	imes 10^{6}$.

The diameter of the objective of a telescope is $200 \text{ cm}$. What is the resolving power of the telescope? Take the wavelength of light $\lambda = 5000 \text{ \AA}$.

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