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Difficulty: MediumThin Lenses, Optical Instruments, and Defects of Vision

A short-sighted person cannot see objects clearly beyond a distance of 80 cm80\text{ cm}. What type of lens, of what focal length and power, is required to correct this vision defect so that the person can view distant objects clearly?

  1. Diverging lens of focal length 80 cm80\text{ cm} and power −1.25 D-1.25\text{ D}Answer
  2. B
    Converging lens of focal length 80 cm80\text{ cm} and power +1.25 D+1.25\text{ D}
  3. C
    Converging lens of focal length 36.4 cm36.4\text{ cm} and power +2.75 D+2.75\text{ D}
  4. D
    Diverging lens of focal length 80 cm80\text{ cm} and power −0.0125 D-0.0125\text{ D}

Answer

Diverging lens of focal length 80 cm80\text{ cm} and power −1.25 D-1.25\text{ D}
For a myopic person with a far point at 80 cm80\text{ cm}, parallel rays from a distant object (u=∞u = \infty) must be diverged so they appear to come from the far point (v=−80 cm=−0.8 mv = -80\text{ cm} = -0.8\text{ m}). Using 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}, we obtain f=−0.8 m=−80 cmf = -0.8\text{ m} = -80\text{ cm}. The negative focal length corresponds to a diverging lens, and its power is P=1−0.8 m=−1.25 DP = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}.

Step-by-Step Solution

1
Identify the optical requirements for correcting myopia (short-sightedness).
For a distant object (u=∞u = \infty), the corrective lens must form a virtual image at the eye's far point (v=−80 cm=−0.8 mv = -80\text{ cm} = -0.8\text{ m}).
Myopic eyes focus rays from infinity in front of the retina; placing a virtual image at the far point allows the eye lens to focus it correctly onto the retina.
2
Apply the thin lens formula to calculate the required focal length.
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{\infty} + \frac{1}{-0.8\text{ m}} = 0 - 1.25\text{ m}^{-1} \implies f = -0.8\text{ m} = -80\text{ cm}\).
The negative sign indicates that a concave (diverging) lens is required.
3
Calculate the power of the corrective lens in dioptres.
\(P = \frac{1}{f\text{ (in metres)}} = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}\).
Lens power in dioptres (D) is the reciprocal of the focal length expressed in meters.

Key Concept

Correction of Myopia (Short-Sightedness) using Diverging Lenses
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Topics

Thin Lenses, Optical Instruments, and Defects of Vision

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