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A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25cm)?
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1. Given Data
- Focal length of Objective (fo): 140 cm
- Focal length of Eyepiece (fe): 5.0 cm
- Least Distance of Distinct Vision (D): 25 cm
2. Case (a): Normal Adjustment (Image at Infinity)
In normal adjustment, the final image is formed at infinity. The magnifying power (M) is given by:
M = fo / fe
Substituting the values:
M = 140 / 5.0
M = 28
3. Case (b): Image at Least Distance (D = 25 cm)
When the final image is formed at the least distance of distinct vision, the magnifying power is calculated as:
M = (fo / fe) × (1 + fe/D)
Substituting the values:
- M = (140 / 5) × (1 + 5/25)
- M = 28 × (1 + 1/5)
- M = 28 × (1 + 0.2)
- M = 28 × 1.2
- M = 33.6
Final Answer
- (a) Normal Adjustment: 28
- (b) Image at Near Point: 33.6
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