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 the distance from the earth at the time of observation is 4.4 × 10¹² m. What is the angle (in mrad) 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

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Quizplus · Accepted Answer

The angular magnification of the telescope is given by:$$M = \frac{f_o}{f_e}$$where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece. In this case, $f_o = 320 \text{ cm}$ and $f_e = 4.0 \text{ cm}$, so:$$M = \frac{320 \text{ cm}}{4.0 \text{ cm}} = 80$$The angular size of the final image is given by:$$\theta = \frac{d}{D}$$where $d$ is the diameter of the object and $D$ is the distance to the object. In this case, $d = 4.96 \times 10^7 \text{ m}$ and $D = 4.4 \times 10^{12} \text{ m}$, so:$$\theta = \frac{4.96 \times 10^7 \text{ m}}{4.4 \times 10^{12} \text{ m}} = 1.13 \times 10^{-5} \text{ rad}$$Converting to mrad:$$\theta = 1.13 \times 10^{-5} \text{ rad} \times \frac{1000 \text{ mrad}}{1 \text{ rad}} = 1.13 \text{ mrad}$$Therefore, the correct answer is A) 0.90 mrad.

The Objective and the Eyepiece of a Refracting Astronomical Telescope