Human Eye and Colourful World Solved Examples (Class 10 Physics)
Understand the lens formula and power of lenses as applied to the human eye. These problems cover accommodation, defects of vision, and corrective lens cal
TL;DR: Understand the lens formula and power of lenses as applied to the human eye. These problems cover accommodation, defects of vision, and corrective len…
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Understand the lens formula and power of lenses as applied to the human eye. These problems cover accommodation, defects of vision, and corrective lens cal
Human Eye and Colourful World — Solved Numerical Examples (Step by Step)
Example 1: A person has a near point distance of 60 cm. Calculate the power of the lens required to correct this defect.
Solution: Near point should be at 25 cm for normal vision
Object distance u = -60 cm (where the person can see clearly)
Required image distance v = -25 cm (normal near point)
Using lens formula: 1/f = 1/v + 1/u
1/f = 1/(-25) + 1/(-60)
1/f = -1/25 - 1/60 = (-12 - 5) / 300 = -17/300
f = -300/17 = -17.6 cm = -0.176 m
Power P = 1/f = -5.68 D
Example 2: A person can see clearly up to 2 m away but cannot see distant objects. What is the power of the corrective lens needed?
Solution: This is myopia (short-sightedness)
Far point = 2 m, but should be at infinity
For clear vision of distant objects, the image should form at the far point
u = -∞ (object at infinity), v = -2 m
1/f = 1/v - 1/u = 1/(-2) - 1/(-∞) = -1/2 + 0 = -1/2
f = -2 m
Power P = 1/f = -0.5 D
Example 3: A convex lens of power 2 diopters forms a real image at a distance of 40 cm. If the object is placed at 30 cm from the lens, find the focal length and verify with the lens formula.
Solution: Power P = 2 D
Focal length f = 1/P = 1/2 = 0.5 m = 50 cm
Object distance u = -30 cm
Image distance v = 40 cm
Using lens formula: 1/f = 1/v + 1/u
1/50 = 1/40 + 1/(-30)
1/50 = 1/40 - 1/30 = (3 - 4) / 120 = -1/120
This gives f = -120 cm, which contradicts. Let me recalculate:
If v = 40 cm (real image), u must satisfy: 1/f = 1/40 + 1/u
1/0.5 = 1/40 + 1/u
2 = 0.025 + 1/u
1/u = 1.975
u = 0.506 m ≈ 50.6 cm (approximately)
Example 4: A concave lens of power -1.5 diopters is used by a person with myopia. What is its focal length?
Solution: Power P = -1.5 D
Focal length f = 1/P = 1/(-1.5) = -0.667 m = -66.7 cm
Example 5: A person has a far point at 1.5 m. Calculate the power of the lens to correct myopia.
Solution: For clear distant vision, the corrective lens should form an image at the far point when object is at infinity
u = -∞, v = -1.5 m
1/f = 1/v - 1/u = 1/(-1.5) - 0 = -1/1.5 = -2/3
f = -1.5 m = -150 cm
Power P = 1/f = -1/1.5 = -0.667 D
Example 6: What is the focal length of a lens with power 4 diopters?
Solution: Power P = 4 D
Focal length f = 1/P = 1/4 = 0.25 m = 25 cm
Tips
- Power of a lens P = 1/f (in meters), measured in diopters (D); positive for convex, negative for concave.
- Myopia (short-sightedness): far point closer than infinity; corrected with concave lens.
- Hypermetropia (long-sightedness): near point farther than 25 cm; corrected with convex lens.
Frequently Asked Questions
How does a bifocal lens help presbyopia?
A bifocal lens has two focal lengths: the lower part (convex) for near vision and the upper part for distant vision, allowing the wearer to see clearly at both distances.
Why does the eye's lens need to be flexible?
The ciliary muscles change the lens shape to adjust focal length, allowing the eye to focus on objects at varying distances—this is called accommodation.
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