A person with a myopic eye cannot see objects beyond 1.2 m distinctly. (a) What is the nature of the lens used to correct this defect? (b) Calculate the focal length and power of the corrective lens required to restore normal vision.

Q: A person with a myopic eye cannot see objects beyond 1.2 m distinctly. (a) What is the nature of the lens used to correct this defect? (b) Calculate the focal length and power of the corrective lens required to restore normal vision.

Solution: (a) Nature of the Lens:\nTo correct myopia (near-sightedness), a concave lens is used because it diverges the incoming light rays so that the image is formed back on the retina. (b) Calculation of Focal Length and Power: Given: Far point of the myopic person ($v$) = $-1.2 \text{ m}$ (The image must be formed at the person's far point so that they can see it). Object distance ($u$) = $-\infty$ (Infinity, because a normal eye sees objects up to infinity). Using the Lens Formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ Substituting the values: $$\frac{1}{f} = \frac{1}{-1.2} - \frac{1}{-\infty}$$ Since $\frac{1}{-\infty} = 0$, we have: $$\frac{1}{f} = -\frac{1}{1.2}$$ $$f = -1.2 \text{ m}$$ Calculation of Power: The power ($P$) of a lens is given by the formula: $$P = \frac{1}{f \text{ (in meters)}}$$ $$P = \frac{1}{-1.2 \text{ m}}$$ $$P = -\frac{10}{12} = -0.833 \text{ D}$$ Conclusion: The focal length of the corrective concave lens is $-1.2 \text{ m}$ (or $-120 \text{ cm}$), and its power is $-0.833 \text{ D}$.

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Accepted Answer

Solution: (a) Nature of the Lens:
To correct myopia (near-sightedness), a concave lens is used because it diverges the incoming light rays so that the image is formed back on the retina. (b) Calculation of Focal Length and Power: Given: Far point of the myopic person ($v$) = $-1.2 	ext{ m}$ (The image must be formed at the person's far point so that they can see it). Object distance ($u$) = $-\infty$ (Infinity, because a normal eye sees objects up to infinity). Using the Lens Formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ Substituting the values: $$\frac{1}{f} = \frac{1}{-1.2} - \frac{1}{-\infty}$$ Since $\frac{1}{-\infty} = 0$, we have: $$\frac{1}{f} = -\frac{1}{1.2}$$ $$f = -1.2 	ext{ m}$$ Calculation of Power: The power ($P$) of a lens is given by the formula: $$P = \frac{1}{f 	ext{ (in meters)}}$$ $$P = \frac{1}{-1.2 	ext{ m}}$$ $$P = -\frac{10}{12} = -0.833 	ext{ D}$$ Conclusion: The focal length of the corrective concave lens is $-1.2 	ext{ m}$ (or $-120 	ext{ cm}$), and its power is $-0.833 	ext{ D}$.