Geometrical OpticsEasy
Question
A planet is observed by a astronomical refrecting telescope having an objective of focall length and eyepiece of focal length 2 cm
Options
A.the distance between the objective and the eyepiece is 16.02 m
B.the angular magnification of the planet is - 800
C.the image of the planet is inverted
D.the objective is larger than the eyepiece
Solution
Distance between objective and eyepiece
L = fo + fe = (16 + 0.02) = 16.20 m
Angular magnification
M = - fo / fe = - 16 / 0.02 = - 800
Image is inverted and objective is larger than the eyepiece.
L = fo + fe = (16 + 0.02) = 16.20 m
Angular magnification
M = - fo / fe = - 16 / 0.02 = - 800
Image is inverted and objective is larger than the eyepiece.
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