
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.
Answer
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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.
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.
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