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A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?
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1. Given Data
- Object Distance (u): +12 cm (Virtual Object)
- (Since the beam was converging at P, P acts as a virtual object behind the lens.)
2. Case (a): Convex Lens
- Focal Length (f): +20 cm
Using the Lens Formula:
1/v - 1/u = 1/f
Rearranging for image distance (v):
1/v = 1/f + 1/u
Substituting values:
- 1/v = 1/20 + 1/12
- 1/v = (3 + 5) / 60
- 1/v = 8 / 60
- v = 60 / 8
- v = +7.5 cm
(a) Image converges at 7.5 cm behind the lens
3. Case (b): Concave Lens
- Focal Length (f): -16 cm
Using the same Lens Formula:
1/v = 1/f + 1/u
Substituting values:
- 1/v = 1/(-16) + 1/12
- 1/v = -1/16 + 1/12
- 1/v = (-3 + 4) / 48
- 1/v = 1 / 48
- v = +48 cm
(b) Image converges at 48 cm behind the lens
Final Conclusion
- Convex Lens: The beam converges closer to the lens at 7.5 cm.
- Concave Lens: The beam converges further away at 48 cm.
Why is 'u' positive here?
Normally, object distance 'u' is negative. However, in this problem, the incident rays are converging towards a point behind the lens. This point P acts as a Virtual Object. Since it is on the right side (direction of light travel), we take u = +12 cm.
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