The most commonly used formula in amateur astronomy is used to calculate the magnification of a telescope: magnification = focal length of telescope / focal length of eyepiece. Example: using a 10mm eyepiece in a telescope with a focal length of 1000mm results in a magnification of 100x (1000 / 10 = 100)
42) A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? The diameter of the moon is 3.5??106?m and the radius of the lunar orbit round the earth is 3.8??18?m
Answer is:
17.5 cm
Explanation:
Related Optics MCQ with Answers
Answer is:
both objective and eye-piece have short focal lengths
Explanation:
Magnifying power of a compound microscope is high if both objective and eye-piece have short focal lengths
Answer is:
the objective has a long focal length and the eye-piece has a short focal length
Explanation:
The magnifying power of telescope is high if the objective has a long focal length and the eye-piece has a short focal length
Answer is:
a converging lens of focal length 100 cm
Explanation:
A lens of power + 2.0 D is placed in contact with another lens of power ? 1.0 D. The combination will behave like a converging lens of focal length 100 cm
Answer is:
10 s
Explanation:
Suppose W is the power of the lamp. Then amount of energy needed to print photo is E=2.5*W/4??602 Now suppose t is the required time when the lamp is kept at 1.5 m away. So we have E=W/4??1202t Equating energy for both we get t=10 s.