An astronomical telescope has an eyepiece of focal length 5 cm. If the angular magnification in normal adjustment is 10, when final image is at least distance of distinct vision (25cm) from eyepiece, then angular magnification will be :
Answers
Answer:
Angular magnification will be 12.
Explanation:
=> It is given that,
An astronomical telescope has an eyepiece of focal length, fe = 5 cm.
If the angular magnification in normal adjustment, m = 10 cm
final image is at least distance of distinct vision from eyepiece = 25 cm
=> Magnification for normal adjustment:
m = -f₀/fe = |f₀/fe| = 10 cm
f₀ = 10 * fe
= 10 * 5
= 50 cm.
=> When final image is at least distance of distinct vision from eyepiece, magnification m' is:
m' = -f₀/fe (1 + fe / D)
m' = f₀/fe (1 + fe / D)
= 10 (1 + 5/25 )
= 10 (25 + 5 / 25)
= 10 (30 / 25)
= 10 * 6 / 5
= 60 / 5
= 12
Thus, the angular magnification will be 12.
Learn more:
Q:1 A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The least distant of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the object. Calculate:
(a) the separation between the objective and eyepiece
(b) the magnification produced?
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Q:2 The focal length of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively find the position of the object related to the objective in order to obtain an angular magnification of 30 in normal adjustment.
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