A hypermetropic (far-sighted) person has a near point located at a distance of 100\text{ c… — JAMB UTME

A hypermetropic (far-sighted) person has a near point located at a distance of $100\text{ cm}$ from the eye. What power of spectacle lens, in dioptres, is required to enable this person to read print comfortably held at the standard near point of $25\text{ cm}$ ?

$+3.0\text{ D}$ Answer
B
$-3.0\text{ D}$
C
$+5.0\text{ D}$
D
$+0.75\text{ D}$

Answer

+3.0\text{ D}

To correct hypermetropia, a converging (convex) lens is required to bend incoming rays so that an object placed at the standard near point of

25\text{ cm}

(

+0.25\text{ m}

) forms a virtual image at the defective eye's near point of

100\text{ cm}

(

-1.0\text{ m}

). Substituting

u = +0.25\text{ m}

and

v = -1.0\text{ m}

into the power formula

P = \frac{1}{u} + \frac{1}{v}

yields

P = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

Step-by-Step Solution

Identify object distance (

u

) and required virtual image distance (

v

)

u = +25\text{ cm} = +0.25\text{ m}

and

v = -100\text{ cm} = -1.0\text{ m}

The lens must create a virtual image (on the same side as the object) at the person's actual near point when an object is placed at the standard reading distance.

Apply the thin lens formula to determine lens power

P

P = \frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Power in dioptres is the reciprocal of the focal length in metres.

Calculate the numerical value of lens power

P = \frac{1}{0.25\text{ m}} + \frac{1}{-1.0\text{ m}} = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

Adding the reciprocal quantities yields a positive focal power of

+3.0\text{ D}

Key Concept

Correction of Hypermetropia using Converging Lenses

Estimated Time:1m 30s

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