Question 11 of 16intermediate🔧 ApplyDiagram5 marks

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.

Correct Answer

Hypermetropia (far-sightedness) is a defect where a person can see distant objects clearly but cannot see nearby objects distinctly. In this condition, the light rays from a close object are focused at a point behind the retina instead of on it. This defect arises because the focal length of the eye lens is too long, or the eyeball has become too small.

To correct hypermetropia, a convex lens of appropriate power is used. The convex lens is a converging lens that provides the additional focusing power required for forming the image on the retina. In the diagram, the convex lens placed before the eye converges the light rays from a nearby object, and then the eye lens focuses them properly on the retina.

Calculation of Power of Lens:

Given: Near point of hypermetropic eye = 1 m = 100 cm; Near point of normal eye = 25 cm

Object distance (u) = -25 cm (object at normal near point)

Image distance (v) = -100 cm (virtual image at defective near point)

Using lens formula: 1/f = 1/v - 1/u

1/f = 1/(-100) - 1/(-25)

1/f = -1/100 + 1/25 = -1/100 + 4/100 = 3/100

Focal length (f) = 100/3 cm = 1/3 m

Power (P) = 1/f = 1/(1/3) = +3 D

Therefore, a convex lens of power +3 dioptre is required to correct this defect.

Exercise: EXERCISES | Q: 7 | (Page Page 10)
For More Understanding

Explanation

This question tests understanding of hypermetropia correction and lens power calculation. The textbook explains that hypermetropia is corrected using a convex lens which provides additional converging power. For calculation, the corrective lens must form a virtual image at the defective near point (1 m) when the object is at the normal near point (25 cm). Using the lens formula with proper sign conventions gives focal length, and power = 1/f (in metres). The positive power indicates a convex lens, consistent with the textbook's statement that converging lenses correct hypermetropia.

Solution Steps

  1. Step 1: Identify that hypermetropia requires a convex lens for correction

  2. Step 2: Set object distance u = -25 cm (normal near point)

  3. Step 3: Set image distance v = -100 cm (defective near point, virtual image)

  4. Step 4: Apply lens formula: 1/f = 1/v - 1/u

  5. Step 5: Calculate: 1/f = -1/100 + 1/25 = 3/100

  6. Step 6: Find focal length: f = 100/3 cm = 1/3 m

  7. Step 7: Calculate power: P = 1/f = 1/(1/3) = +3 D

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