1.
Based on the graph above, determine the telescope magnification when the viewing eye is relaxed.
Known :
The lens closest to the object is called the objective lens (focal length fo) and forms a real image. The second lens called the eyepiece (focal length fe) acts as a magnifier.
The focal length of the objective lens (fo) = 100 cm
The focal length of the eyepiece lens (fe) = 8 cm
Wanted : Telescope magnification when eye is relaxed
Solution :
M = -fo / fe
M = -100 cm / 8 cm
M = -12.5 X
Minus sign indicates that the image is inverted.
[irp]
2.
Based on the figure above, what is the telescope magnification when the eye is relaxed?
Known :
The focal length of the objective lens (fo) = 200 cm
The focal length of the eyepiece lens (fe) = 5 cm
Wanted : Total magnification
Solution :
When eye is relaxed, the telescope magnification :
M = -fo / fe
M = -200 cm / 5 cm
M = -40 X
Minus sign indicates that the image is inverted.
[irp]
3.
Based on above figure, if the distance between the eyepiece lens and the objective lens (d) = 11 times the focal length of the eyepiece lens, then what is the total magnification when the eye is relaxed.
Known :
The distance between the eyepiece lens and the objective lens = the length of telescope = 11 fe
The focal length of the eyepiece lens (fe) = fe
The focal length of the objective lens (fo) = the length of a telescope – the focal length of the eyepiece lens = 11 fe – fe = 10 fe
Wanted: The telescope magnification
Solution :
M = -fo / fe
M = -10 fe / fe
M = -10 X
Minus sign indicates that the image is inverted.
[irp]
4. What is the total magnification of the astronomical telescope?
Known :
The distance between objective and eyepiece lens (l) = 126 cm
The focal length of the eyepiece lens (fe) = 6 cm
The focal length of the objective lens (fo) = 120 cm
Wanted: The total magnification
Solution :
Based on the above figure, the final image at infinity, so the eye is relaxed.
The total magnification of the astronomical telescope :
