The angular magnification of a refracting telescope is 40. When the object and f

Question

The angular magnification of a refracting telescope is 40. When the object and final image are both at infinity, the distance between the eyepiece and the objective is 143.5com. The telescope is used to view a distant radio tower. The real image of the tower, formed by the objective, is 6.0mm in height. The focal point of the eyepiece is positioned at the real image.

A) The focal length of the objective, in cm, is closest to: I know it's 140 cm, but I don't know why
B)The angle subtended by the final image of the tower is closest to: .17rad, but why?

Explanation / Answer

(a)

When the object and final image are both at infinity, the distance between the eyepiece and the objective is 143.5 cm

Thus,

f1+f2=143.5---------------------------(1)

The angular magnification of a refracting telescope is 40

f1/f2=40
f1=40*f2

substituting in (1), we have
40*f2+f2=143.5
41f2=143.5
f2=3.5cm,
f1=40*3.5cm
f1=140cm

focal length of eyepiece = 3.5 cm

focal length of objective = 140 cm

(b)

angle subtended = Hi/fe

=6*10^-3/3.5*10^-2

=0.17 rad

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