The near point of an eye is 50 cm. What focal length lens should be used so that the eye can clearly see an object 25 cm away?

Welcome back ashamuhima626 — missed you this week. I will solve question 16. 16. The near point of an eye is 50 cm. I. What focal length lens should be used so that the eye can clearly see an object 25 cm away? (ANS: f = 50 cm) II. What is the power of this lens? (ANS: 2 diopters) Part I: Determine the focal length of the lens. Step 1: Identify the given values. The near point of the eye is v = -50 cm (image distance, negative because it's a virtual image formed on the same side as the object). The desired object distance is u = -25 cm (negative because it's a real object in front of the lens). Step 2: Use the lens formula to find the focal length f. The lens formula is given by: (1)/(f) = (1)/(v) + (1)/(u) Substitute the values: (1)/(f) = (1)/(-50 cm) + (1)/(-25 cm) (1)/(f) = -(1)/(50 cm) - (1)/(25 cm) Find a common denominator: (1)/(f) = -(1)/(50 cm) - (2)/(50 cm) (1)/(f) = -(3)/(50 cm) Solve for f: f = -(50)/(3) cm This result is for a diverging lens, which is typically used to correct nearsightedness. However, the question implies a converging lens for a farsighted eye (since the near point is further than normal 25 cm). Let's re-evaluate the problem statement. A normal near point is 25 cm. If the near point is 50 cm, the person is farsighted (hyperopic) and needs a converging lens to bring objects closer. For a farsighted person, the lens should form a virtual image of an object placed at the normal near point (25 cm) at the person's actual near point (50 cm). So, the object distance u = -25 cm (where the object is placed). The image distance v = -50 cm (where the eye can see the image clearly). Let's re-calculate using these values: (1)/(f) = (1)/(v) + (1)/(u) (1)/(f) = (1)/(-50 cm) + (1)/(-25 cm) (1)/(f) = -(1)/(50 cm) - (2)/(50 cm) (1)/(f) = -(3)/(50 cm) f = -(50)/(3) cm ≈ -16.67 cm This is still a diverging lens. The answer provided (f = 50 cm) suggests a converging lens. Let's consider the standard convention for correcting farsightedness. For farsightedness, the eye cannot focus on objects closer than its near point. To see an object at 25 cm, the corrective lens must form a virtual image of this object at the person's actual near point (50 cm). So, u = -25 cm (object at normal reading distance). v = -50 cm (virtual image formed at the person's near point). Let's re-check the sign convention for the lens formula. The lens formula is (1)/(f) = (1)/(v) - (1)/(u) for real objects and virtual images, or (1)/(f) = (1)/(u) + (1)/(v) with proper sign conventions for u and v. Using the standard sign convention where u is negative for real objects and v is negative for virtual images on the same side as the object: (1)/(f) = (1)/(v) + (1)/(u) (1)/(f) = (1)/(-50 cm) + (1)/(-25 cm) (1)/(f) = -(1)/(50 cm) - (2)/(50 cm) = -(3)/(50 cm) f = -(50)/(3) cm This is a diverging lens, which is used for myopia (nearsightedness). Let's consider the case where the person is farsighted. The near point is 50 cm. They want to see an object at 25 cm. The lens should create a virtual image of the object at 25 cm, at the 50 cm mark, so the eye can focus on it. So, u = 25 cm (object distance from the lens). v = 50 cm (image distance from the lens, virtual image on the same side as the object, so v = -50 cm). Using the lens formula: (1)/(f) = (1)/(v) + (1)/(u) (1)/(f) = (1)/(-50 cm) + (1)/(25 cm) (1)/(f) = -(1)/(50 cm) + (2)/(50 cm) (1)/(f) = (1)/(50 cm) f = 50 cm This matches the provided answer. This positive focal length indicates a converging lens, which is correct for farsightedness. The focal length of the lens is 50 cm. Part II: Determine the power of the lens. Step 1: Convert the focal length to meters. f = 50 cm = 0.50 m Step 2: Calculate the power of the lens. The power P of a lens is given by the reciprocal of its focal length in meters: P = (1)/(f) P = (1)/(0.50 m) P = 2 diopters The power of the lens is 2 diopters. Send me the next one 📸