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An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
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1. Given Data
Using the Cartesian sign convention for a concave lens:
- Object Size (h1): +3.0 cm
- Object Distance (u): -14 cm
- Focal Length (f): -21 cm (Concave lens)
2. Calculating Image Position
Using the Lens Formula:
1/v - 1/u = 1/f
Rearranging for image distance (v):
1/v = 1/f + 1/u
Substituting the values:
- 1/v = 1/(-21) + 1/(-14)
- 1/v = -(1/21 + 1/14)
- 1/v = -(2/42 + 3/42)
- 1/v = -5 / 42
- v = -42 / 5
- v = -8.4 cm
Image Position (v) = 8.4 cm in front of the lens
3. Nature and Size
Using the Magnification Formula (m):
m = v / u
Substituting values:
- m = (-8.4) / (-14)
- m = +0.6
Now, finding the Image Height (h2):
- h2 = m × h1
- h2 = 0.6 × 3.0
- h2 = +1.8 cm
Final Conclusion
- Location: 8.4 cm in front of the lens (on the same side as the object).
- Nature: Virtual and Erect (indicated by the negative sign of v and positive m).
- Size: 1.8 cm (Diminished).
4. What happens if the object moves further away?
Effect of increasing distance
As the object is moved further away from the lens (towards infinity), the virtual image shifts towards the focus (F) of the lens. The image size will continue to decrease gradually, eventually becoming a point size at the focus, but it will always remain virtual and erect.
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