Answer:
0.095
Explanation:
An astronomical telescope is a telescope used for viewing far distance object. It uses two lenses called the objective lens and the eyepiece lens.
Each lens has its own focal length
Let the focal length of the objective lens be Fo
Focal length of the eyepiece be Fe
Magnification of an astronomical telescope = Fo/Fe
Since the telescope uses a reflecting mirror having radius of curvature of 5.5mm instead of an objective lens, then we will replace Fo as the focal length of the mirror.
Focal length of a mirror Fo = Radius of curvature/2
Fo = R/2
Fo = 5.5/2
Fo = 2.75mm
Converting 2.75mm to cm gives 2.75/10 = 0.275cm
Fo = 0.275cm; Fe = 2.9cm
Magnification of the telescope = 0.275/2.9
Magnification of the astronomical telescope = 0.095
The magnifying power of an astronomical telescope will be:
"0.095".
According to the question,
Radius of curvature, R = 5.5 mm
Focal length of eyepiece, [tex]F_e[/tex] = 2.9 cm
We know that,
→ Focal length of mirror,
F₀ = [tex]\frac{Radius \ of \ curvature}{2}[/tex]
By substituting the values,
= [tex]\frac{5.5}{2}[/tex]
= 2.75 mm or,
= 0.278 cm
hence,
The telescope's magnification be:
= [tex]\frac{F_0}{F_e}[/tex]
= [tex]\frac{0.275}{2.9}[/tex]
= 0.095
Thus the above approach is correct.
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https://brainly.com/question/1477543
