The objective and the eyepiece of a refracting astronomical telescope have focal lengths of 320 cm and 4.0 cm,respectively.The telescope is used to view Neptune and the final image is set at infinity.The diameter of Neptune is 4.96 $\times$ 10 7 m and the distance from Earth at the time of observation is 4.4 $\times$ 10 12 m.The angle subtended by the final telescopic image of Neptune,in m rad,is closest to:
A)0.9
B)1.1
C)1.3
D)1.5
E)1.7

Question

Quizplus · Accepted Answer

The angular magnification of a telescope is given by the ratio of the focal length of the objective to the focal length of the eyepiece:

m = fo/fe = 320/4.0 = 80

The apparent angular size of Neptune as seen from Earth is given by:

θ = D/d

where D is the diameter of Neptune and d is the distance from Earth.

θ = (4.96 x 10^7)/(4.4 x 10^12) = 1.13 x 10^-5 rad

The final angular size of the image formed by the telescope is:

θ' = θ/m = (1.13 x 10^-5)/80 = 1.41 x 10^-7 rad

Note that we have used radians as the unit of measurement for angle.

To convert radians to milliradians (mrad), we multiply by 1000:

θ' = 1.41 x 10^-7 x 1000 = 0.141 mrad

Therefore, the answer is closest to 0.9 mrad, which is choice A.

The Objective and the Eyepiece of a Refracting Astronomical Telescope ×\times×

The Objective and the Eyepiece of a Refracting Astronomical Telescope $\times$