A lens of large focal length and large apertu | vedclass

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(D) The magnifying power of an astronomical telescope is given by $MP = \frac{f_o}{f_e}$, where $f_o$ is the focal length of the objective and $f_e$ i

Vedclass · Accepted Answer

all of the above.

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(D) The magnifying power of an astronomical telescope is given by $MP = \frac{f_o}{f_e}$, where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.The resolving power $(R.P.)$ of a telescope is given by $R.P. = \frac{a}{1.22 \lambda}$, where $a$ is the aperture of the objective lens and $\lambda$ is the wavelength of light.$1$. $A$ large aperture $(a)$ increases the resolving power, allowing the telescope to distinguish between two closely spaced objects.$2$. $A$ large aperture increases the light-gathering power of the objective, which is proportional to the area of the lens $(\pi r^2)$. This ensures that faint objects are visible and the image quality is improved.Since all these factors contribute to the performance of an astronomical telescope, the correct answer is $D$.

$A$ lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since :

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