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(A) The magnifying power $(m)$ of a telescope is given by the ratio of the focal length of the objective $(f_o)$ to the focal length of the eyepiece $

Vedclass · Accepted Answer

$6.1 	imes 10^{-6} \, 	ext{rad}$ and $12$

Vedclass · Answer

(A) The magnifying power $(m)$ of a telescope is given by the ratio of the focal length of the objective $(f_o)$ to the focal length of the eyepiece $(f_e)$: $m = \frac{f_o}{f_e} = \frac{30 \, 	ext{cm}}{2.5 \, 	ext{cm}} = 12$. The resolving limit $(\Delta 	heta)$ 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 objective aperture. Given $\lambda = 5000 \, \mathring{A} = 5000 	imes 10^{-10} \, 	ext{m} = 5 	imes 10^{-7} \, 	ext{m}$ and $D = 10 \, 	ext{cm} = 0.1 \, 	ext{m}$. Substituting the values: $\Delta 	heta = \frac{1.22 	imes 5 	imes 10^{-7}}{0.1} = 1.22 	imes 5 	imes 10^{-6} = 6.1 	imes 10^{-6} \, 	ext{rad}$. Thus,the resolving limit is $6.1 	imes 10^{-6} \, 	ext{rad}$ and the magnifying power is $12$.

$A$ telescope uses light having a wavelength of $5000 \, \mathring{A}$ and lenses with focal lengths of $2.5 \, \text{cm}$ and $30 \, \text{cm}$. If the diameter of the aperture of the objective is $10 \, \text{cm}$,what are the resolving limit and the magnifying power of the telescope,respectively?

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