The aperture of the objective of a telescope | vedclass

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(C) The formula for the resolving power $(R.P.)$ of a telescope is given by $R.P. = \frac{D}{1.22 \lambda}$,where $D$ is the aperture of the objective

Vedclass · Accepted Answer

$8.2 	imes 10^{5}$

Vedclass · Answer

(C) The formula for the resolving power $(R.P.)$ of a telescope is given by $R.P. = \frac{D}{1.22 \lambda}$,where $D$ is the aperture of the objective and $\lambda$ is the wavelength of light.Given: $D = 24.4 \, cm = 24.4 	imes 10^{-2} \, m$ and $\lambda = 2440 \, \mathring{A} = 2440 	imes 10^{-10} \, m$.Substituting the values:$R.P. = \frac{24.4 	imes 10^{-2}}{1.22 	imes 2440 	imes 10^{-10}}$$R.P. = \frac{24.4 	imes 10^{-2}}{2976.8 	imes 10^{-10}}$$R.P. = \frac{24.4}{2976.8} 	imes 10^{8}$$R.P. \approx 0.0082 	imes 10^{8} = 8.2 	imes 10^{5}$.

The aperture of the objective of a telescope is $24.4 \, cm$. What is the resolving power of this telescope if light of wavelength $2440 \, \mathring{A}$ is used to view the object?

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