Question

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Answer 1

Hypermetropia, or farsightedness, is a vision condition where distant objects are seen clearly, but near objects appear blurry. This occurs because the eye's lens focuses incoming light rays, which diverge from near objects, behind the retina. A convex lens corrects this defect by converging light rays so that the image forms on the retina, as illustrated in the accompanying figure.

The convex lens effectively creates a virtual image (N') of a near object at the person's natural near point of vision (N). If the near point of a normal eye is 25 cm, and the hypermetropic individual's near point is 1 m, they can see an object at 25 cm clearly if its image is formed at 1 m.

Object distance, u = −25 cm
Image distance, v = −1 m = −100 cm 
Focal length, f
Applying the lens formula:
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\(\frac{1}{-100}-\frac{1}{-25}=\frac{1}{f}\)

\(⇒\frac{1}{f}=\frac{1}{25}-\frac{1}{100}\)

\(⇒\frac{1}{f}=\frac{4-1}{100}\)

\(⇒f=\frac{100}{3} \approx 33.3\) cm
\(\approx 0.33\) m

∴\text{Power, P}=\frac{1}{f}\text{(in metres)}\)
\(=\frac{1}{0.33} \approx +3.0\) D
A convex lens with a power of +3.0 D is required to correct this vision impairment.