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Q21 (NDA-I/2008) Science & Technology › Basic Science (Physics, Chemistry, Biology) › Geometrical optics Answer Verified

A far sighted person has a near point at 100 cm. What must be the power of the correcting lens?

Result
Your answer: —  ·  Correct: D
Explanation

A far-sighted person (hypermetropia) has a near point further than the normal 25 cm. To correct this, a convex (converging) lens is used to form a virtual image of an object placed at the normal near point (u = -25 cm) at the person's actual near point (v = -100 cm) [t2, t5]. Using the lens formula 1/f = 1/v - 1/u, we substitute the values: 1/f = 1/(-100) - 1/(-25) [t2]. This simplifies to 1/f = -0.01 + 0.04 = 0.03 cm⁻¹ [t2]. Converting to meters for power calculation, P = 1/f(m). Thus, P = 1/(-1.0 m) - 1/(-0.25 m) = -1 + 4 = +3.0 D [t5, t7]. The positive sign confirms a convex lens is required to increase the eye's converging power [t3, t8]. Therefore, the power of the correcting lens must be +3.0 D.

Sources

  1. [1] Science , class X (NCERT 2025 ed.) > Chapter 10: The Human Eye and the Colourful World > 10.1.1 Power of Accommodation > p. 162
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