A person is unable to see objects distinctly closer than 50 cm. (a) Identify the defect of vision the person is suffering from. (b) State two possible causes of this defect. (c) Calculate the focal length and power of the corrective lens required to enable him to read a book placed at 25 cm clearly. Take the normal near point to be 25 cm.

Q: A person is unable to see objects distinctly closer than 50 cm. (a) Identify the defect of vision the person is suffering from. (b) State two possible causes of this defect. (c) Calculate the focal length and power of the corrective lens required to enable him to read a book placed at 25 cm clearly. Take the normal near point to be 25 cm.

(a) Identification of Defect:\nThe person is suffering from Hypermetropia, also known as far-sightedness. In this defect, a person can see distant objects clearly, but cannot see nearby objects distinctly. (b) Causes of Hypermetropia: Excessive shortening of the eyeball: The distance between the eye lens and the retina decreases, causing the image to form behind the retina. Focal length of the eye lens is too large: The converging power of the eye lens decreases, or the focal length increases so that light rays from a nearby point focus behind the retina instead of on it. (c) Numerical Calculation: Given: Object distance, $u = -25$ cm (Normal near point where the book is placed) Image distance, $v = -50$ cm (The near point of the defective eye where the image must be formed) Using the Lens Formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$| Substituting the values: $$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25}$$ $$\frac{1}{f} = -\frac{1}{50} + \frac{1}{25}$| $$\frac{1}{f} = \frac{-1 + 2}{50} = \frac{1}{50}$| Therefore, focal length $f = +50$ cm $= +0.5$ m. Calculating Power of the Lens ($P$): $$P = \frac{1}{f \text{ (in meters)}}$$ $$P = \frac{1}{0.5} = +2.0 \text{ D}$$ Conclusion:\nA convex lens of focal length $+50$ cm and power $+2.0$ D is required to correct this defect.

Question

Accepted Answer

(a) Identification of Defect:
The person is suffering from Hypermetropia, also known as far-sightedness. In this defect, a person can see distant objects clearly, but cannot see nearby objects distinctly. (b) Causes of Hypermetropia: Excessive shortening of the eyeball: The distance between the eye lens and the retina decreases, causing the image to form behind the retina. Focal length of the eye lens is too large: The converging power of the eye lens decreases, or the focal length increases so that light rays from a nearby point focus behind the retina instead of on it. (c) Numerical Calculation: Given: Object distance, $u = -25$ cm (Normal near point where the book is placed) Image distance, $v = -50$ cm (The near point of the defective eye where the image must be formed) Using the Lens Formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$| Substituting the values: $$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25}$$ $$\frac{1}{f} = -\frac{1}{50} + \frac{1}{25}$| $$\frac{1}{f} = \frac{-1 + 2}{50} = \frac{1}{50}$| Therefore, focal length $f = +50$ cm $= +0.5$ m. Calculating Power of the Lens ($P$): $$P = \frac{1}{f 	ext{ (in meters)}}$$ $$P = \frac{1}{0.5} = +2.0 	ext{ D}$$ Conclusion:
A convex lens of focal length $+50$ cm and power $+2.0$ D is required to correct this defect.