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Question

Question: A planet is observed by an astronomical refracting telescope having an objective of focal length \(1...

A planet is observed by an astronomical refracting telescope having an objective of focal length 16m16m and an eyepiece of focal length2cm2cm. Then,
This question has multiple correct options.
A. The distance between the object and the eyepiece is16.02m16.02m.
B. The angular magnification of the planet is 800-800 .
C. The image of the planet is inverted.
D. The object is larger than the eyepiece.

Explanation

Solution

In this question we are going to find, among the options, which options are correct. Use formula (f0+fe)\left( {{f}_{0}}+{{f}_{e}} \right) (distance between object of focal length and eyepiece of focal length) and (f0fe)\left( -\dfrac{{{f}_{0}}}{{{f}_{e}}} \right) (angular of magnification), to solve this problem.

Complete step by step answer:
Let’s discuss the above option and find the answer.
The distance between the object and the eyepiece is16.02m16.02m : Here object of focal length f0=16m{{f}_{0}}=16m and eyepiece of focal length 2cm=2100m=0.02m2cm=\dfrac{2}{100}m=0.02m . Now the distance between the object and the eyepiece is f0+fe=16+0.02=16.02m\Rightarrow {{f}_{0}}+{{f}_{e}}=16+0.02=16.02m . Hence option A is correct.
The angular magnification of the planet is800-800 : Here object of focal lengthf0=16m{{f}_{0}}=16m and eyepiece of focal length2cm=2100m=0.02m2cm=\dfrac{2}{100}m=0.02m . now the angular of magnification of the planet is f0fe=160.02=162100=16002=800\Rightarrow -\dfrac{{{f}_{0}}}{{{f}_{e}}}=-\dfrac{16}{0.02}=-\dfrac{16}{\dfrac{2}{100}}=-\dfrac{1600}{2}=-800 . Hence option B is correct.
The image of the planet is inverted: from option B we know that the angular magnification of the planet is negative so the final image of the planet is inverted. Hence option C is correct.
The object is larger than the eyepiece: From the above question focal length of object f0=16m{{f}_{0}}=16m and focal length of eyepiece is 2cm=2100m=0.02m2cm=\dfrac{2}{100}m=0.02m i.e, object is larger than the eyepiece. Hence option D is correct.

Hence option A, B, C and D are correct.

Note:
Alternative method: first we need to know the distance between them and the value of angular magnification. If the magnification value is negative then the planet is inverted.