NDA (II) 2023 General Ability Test · Question 85 of 146

Correct — B, −0.5 D. Being unable to see distinctly beyond 2 m means the far point has moved in from infinity to 2 m, which is myopia or near-sightedness. The correcting lens must take an object at infinity and place its image at the eye's own far point, so the object distance is infinite and the image distance is −2 m, which makes the required focal length f = −2 m. Power is the reciprocal of the focal length in metres, so P = 1 ÷ (−2) = −0.5 dioptre. The negative sign says the lens is concave, and a concave lens of suitable power is exactly what is prescribed for myopia.

• (a)+ 0·5 D — The magnitude is right but the sign is not. A positive power means a convex, converging lens, which is the correction for far-sightedness; on a myopic eye it would push the image even further in front of the retina and make matters worse.
• (c)+ 0·2 D — Wrong on both counts. The positive sign makes it a convex lens, and +0.2 D corresponds to a focal length of 5 m, which matches nothing in the question.
• (d)– 0·2 D — The right sign but the wrong size. A power of −0.2 D is a concave lens of focal length −5 m, which would suit a person whose far point is 5 m, not 2 m. This option catches anyone who mishandles the reciprocal of 2.

A normal eye can see clearly from a near point of about 25 cm out to a far point at infinity. In myopia the far point comes closer, either because the eye lens is too strongly curved or because the eyeball is elongated, so the image of a distant object forms in front of the retina instead of on it. A diverging lens placed before the eye spreads the incoming parallel rays just enough that the eye's own optics then land them on the retina.

The whole item reduces to one line of arithmetic, and the two places candidates lose it are the sign and the reciprocal. Carry the rule that for myopia the focal length of the correcting lens equals the far-point distance, taken as negative, so a far point of 2 m gives f = −2 m and P = −0.5 D. Check the arithmetic by going the other way as well — 1 ÷ 0.5 is 2, whereas 1 ÷ 0.2 is 5, which is how you eliminate option (d). And keep the near point out of it; 25 cm belongs to far-sightedness problems, not to this one.

• The far point of a normal eye is infinity; in myopia the far point is nearer than infinity, so distant objects are focused in front of the retina.
• Myopia is corrected by a concave lens of suitable power, which brings the image back onto the retina.
• The power of a lens in dioptre is the reciprocal of its focal length in metres, and a concave lens has negative power.
• For a far point of 2 m the correcting lens has f = −2 m and P = −0.5 D; the parallel textbook exercise with a far point of 80 cm gives −1.25 D.

• Dropping the negative sign and choosing the positive option of the same size.
• Bringing the 25 cm near point into a far-point problem.

As a one-step power calculation like this one, as a nature-of-lens question, or as a defect-identification item from a described symptom.

• practice — not a real PYQ

  The far point of a myopic person is 50 cm. The power of the lens required to correct the defect is

  • (a)− 0.5 D
  • (b)− 2 D
  • (c)+ 2 D
  • (d)− 5 D

  Answer(b) − 2 D — the focal length must be −0.5 m, and the reciprocal of −0.5 is −2.

• practice — not a real PYQ

  Hypermetropia, or far-sightedness, is corrected by using

  • (a)a concave lens
  • (b)a convex lens
  • (c)a cylindrical lens
  • (d)a plane glass

  Answer(b) a convex lens — the converging lens supplies the extra focusing power needed to bring near objects onto the retina.