The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Nature of lens: Concave lens
Power of lens: -1.25 D
Explanation: A person with myopia has the far point nearer than infinity. In this case, the far point is 80 cm. To correct myopia, a concave lens of suitable power is required. The concave lens will bring the image back onto the retina by diverging the light rays before they enter the eye.
Calculation:
- Object distance (u): Infinity (∞)
- Image distance (v): -80 cm = -0.8 m (negative sign for virtual image formed on the same side)
- Focal length (f): Using lens formula, 1/f = 1/v - 1/u = 1/(-0.8) - 0 = -1/0.8
- f = -0.8 m
- Power (P): P = 1/f = 1/(-0.8) = -1.25 D
The negative power indicates a concave lens is needed to correct the defect.
Explanation
According to the textbook context, myopia (near-sightedness) is corrected using a concave lens of suitable power. The far point of a myopic person is nearer than infinity - here it is 80 cm. For correction, the lens must form a virtual image at the far point (80 cm) for an object at infinity. Using the lens formula and power formula, we calculate the required power as -1.25 D. The negative sign confirms the concave nature of the lens.
Solution Steps
Step 1: Identify that myopia requires a concave lens for correction
Step 2: Note that far point = 80 cm = 0.8 m (this becomes image distance v = -0.8 m)
Step 3: Apply lens formula: 1/f = 1/v - 1/u, where u = ∞ and v = -0.8 m
Step 4: Calculate: 1/f = 1/(-0.8) - 0 = -1.25, so f = -0.8 m
Step 5: Calculate power: P = 1/f = 1/(-0.8) = -1.25 D