A planet is observed by an astronomical refracting telescope having an objective of focal length $16m$ and an eyepiece of focal length$2cm$. Then,
This question has multiple correct options.
A. The distance between the object and the eyepiece is$16.02m$.
B. The angular magnification of the planet is $-800$ .
C. The image of the planet is inverted.
D. The object is larger than the eyepiece.

Hint: In this question we are going to find, among the options, which options are correct. Use formula $\left( {{f}_{0}}+{{f}_{e}} \right)$ (distance between object of focal length and eyepiece of focal length) and $\left( -\dfrac{{{f}_{0}}}{{{f}_{e}}} \right)$ (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 is$16.02m$ : Here object of focal length ${{f}_{0}}=16m$ and eyepiece of focal length $2cm=\dfrac{2}{100}m=0.02m$ . Now the distance between the object and the eyepiece is $\Rightarrow {{f}_{0}}+{{f}_{e}}=16+0.02=16.02m$ . Hence option A is correct.
The angular magnification of the planet is$-800$ : Here object of focal length${{f}_{0}}=16m$ and eyepiece of focal length$2cm=\dfrac{2}{100}m=0.02m$ . now the angular of magnification of the planet is $\Rightarrow -\dfrac{{{f}_{0}}}{{{f}_{e}}}=-\dfrac{16}{0.02}=-\dfrac{16}{\dfrac{2}{100}}=-\dfrac{1600}{2}=-800$ . 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 ${{f}_{0}}=16m$ and focal length of eyepiece is $2cm=\dfrac{2}{100}m=0.02m$ 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.