JAMB 2025 Physics · Question 88

A short-sighted person's far point is 95cm. The defect can be corrected using

Options

  1. A.Diverging mirror of focal length 95cm
  2. B.Plano convex lens of focal length 90cm
  3. C.Diverging lens of focal length 95cm
  4. D.Converging lens of focal length 95cm

Answer: C. Diverging lens of focal length 95cm

Explanation written with AI and checked against the answer key

Analysis of the Question

The question describes a person with short-sightedness (myopia). A person with this defect can see near objects clearly but cannot see distant objects beyond a certain point (the far point). To correct this, we need to use a lens that will take an object at infinity (where a normal eye should see) and form a virtual image at the person's actual far point.

Concept Explanation
  1. The Defect: Myopia occurs when the eyeball is too long or the lens is too powerful, causing light from distant objects to focus in front of the retina.
  2. The Correction: We use a diverging lens (concave lens) to spread out the light rays before they enter the eye, effectively moving the image back onto the retina.
  3. The Formula: We use the thin lens formula: $1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ • uu (Object distance): Since we want the person to see objects at infinity, u=∞u = \infty. • vv (Image distance): The lens must form a virtual image at the person's far point. Since it is a virtual image, v=−95 cmv = -95\text{ cm}. • ff (Focal length): This is what we need to find.
Step-by-Step Solution
  1. Substitute the values into the lens formula: $1f=1∞+1−95\frac{1}{f} = \frac{1}{\infty} + \frac{1}{-95}$
  2. Recall that 1∞=0\frac{1}{\infty} = 0: $1f=0−195\frac{1}{f} = 0 - \frac{1}{95}$ $1f=−195\frac{1}{f} = -\frac{1}{95}$
  3. Solving for ff: $f=−95 cmf = -95\text{ cm}$
  4. The negative sign indicates that the lens is diverging (concave). Therefore, a diverging lens with a focal length of 95 cm is required.
Conclusion

Comparing our result to the given options, the correct choice is a diverging lens with a focal length of 95 cm.

Exam Tip

Always remember: Myopia is corrected with a Concave (Diverging) lens, and Hypermetropia (long-sightedness) is corrected with a Convex (Converging) lens. A quick mnemonic is "Myopia" has 6 letters, and "Concave" has 7; "Hypermetropia" is a long word, and "Convex" is a short word—or simply remember the acronym "CON-CAVE" for the "near" defect!

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