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 × 10⁷ m,and its distance from Earth at the time of observation is 4.4 × 10¹² m.What is the angle (in milliradians)subtended by the final telescopic image of Neptune? A)0.90 mrad B)1.1 mrad C)1.3 mrad D)1.5 mrad E)1.7 mrad

Question

Quizplus · Accepted Answer

The angular magnification of a telescope is given by the ratio of the focal lengths of the objective and eyepiece lenses. In this case, the angular magnification is 320 cm / 4.0 cm = 80. The angle subtended by the final telescopic image of Neptune is therefore 80 times the angle subtended by Neptune at the Earth. The angle subtended by Neptune at the Earth is given by the ratio of the diameter of Neptune to its distance from Earth. In this case, the angle subtended by Neptune at the Earth is 4.96 × 10^7 m / 4.4 × 10^12 m = 1.13 × 10^-5 radians. Therefore, the angle subtended by the final telescopic image of Neptune is 80 × 1.13 × 10^-5 radians = 9.04 × 10^-4 radians = 0.90 mrad.

The Objective and the Eyepiece of a Refracting Astronomical Telescope