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A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
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1. Given Data
- Focal length of Objective (fo): 144 cm
- Focal length of Eyepiece (fe): 6.0 cm
- Condition: Normal Adjustment (Image at Infinity)
2. Calculating Magnifying Power
The magnifying power (M) of a telescope in normal adjustment is the ratio of the focal length of the objective to the focal length of the eyepiece.
M = fo / fe
Substituting the values:
M = 144 / 6.0
M = 24
3. Calculating Separation
For a telescope in normal adjustment, the distance between the two lenses (L) is the sum of their focal lengths.
L = fo + fe
Substituting the values:
L = 144 + 6.0
L = 150 cm
Final Answer
- Magnifying Power: 24
- Tube Length (Separation): 150 cm
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