Defects of Vision and Their Correction for CBSE Class 10 Science

A healthy human eye effortlessly shifts focus between a distant mountain peak and a pocket smartphone screen within milliseconds. However, when the delicate muscular mechanisms of the eye weaken, or when the physical shape of the eyeball deviates even by a millimeter from a perfect sphere, the eye loses its ability to focus light cleanly onto the retina. The resulting optical imperfections are called defects of vision.

In CBSE Class 10 Science, Chapter 10 (The Human Eye and the Colourful World), questions on the three major refractive errors—Myopia, Hypermetropia, and Presbyopia—along with their diagnostic ray diagrams and corrective lens calculations, appear in board exams virtually every year.

What You Will Learn

• The three primary defects of vision: Myopia, Hypermetropia, and Presbyopia
• Symptoms, biological causes, and optical ray diagrams for Myopia
• Symptoms, biological causes, and optical ray diagrams for Hypermetropia
• How to correct vision defects using concave and convex spherical lenses
• The nature and correction of Presbyopia using bifocal lenses
• Step-by-step solved CBSE board examination numerical problems for lens power
• Common algebraic sign traps and diagram presentation rules

1. Myopia (Near-Sightedness / Short-Sightedness)

Definition

Myopia is a common refractive defect of vision in which an individual can see nearby objects clearly, but cannot see distant objects distinctly.

In a myopic eye, the far point is no longer at infinity; it shifts to a finite distance much closer to the eye (e.g., $80\text{ cm}$ or $2\text{ m}$).

    A. Normal Distant Vision (Parallel Rays focus in front of Retina)
         --------------------                              )---* Focus (In FRONT of Retina!)
         --------------------/

    B. Corrected with Concave Lens (Diverges rays so they appear to come from Far Point)
                   ) (
         ---------(   )-------                   ) (         )----* Focus (Sharp on Retina!)
         ---------(   )-------/

The Two Biological Causes of Myopia:

1. Excessive Curvature of the Eye Lens: The lens has become too thick, rounded, and powerful, having too short a focal length.
2. Elongation of the Eyeball: The distance between the eye lens and the retina has physically increased.

Optical Formation:

Parallel light rays coming from a distant object converge too early and form an image in front of the retina, producing a blurred image.

Optical Correction of Myopia:

• Myopia is corrected by placing a Concave (Diverging) Lens of suitable focal length in front of the eye.
• The concave lens diverges the incoming parallel rays slightly, making them appear as if they are originating from the person's defective far point ($F$). The eye lens then focuses these rays cleanly onto the retina.

2. Hypermetropia (Far-Sightedness / Long-Sightedness)

Definition

Hypermetropia is a refractive defect of vision in which an individual can see distant objects clearly, but cannot see nearby objects distinctly.

In a hypermetropic eye, the near point recedes further away than $25\text{ cm}$ (e.g., $50\text{ cm}$ or $1\text{ m}$). The person must hold reading material far away from their eyes to read comfortably.

    A. Defective Near Vision (Rays from normal near point N focus behind Retina)
              N (25 cm) -----                              )----. . . * Focus (BEHIND Retina!)
              N (25 cm) -----/

    B. Corrected with Convex Lens (Converges rays so they appear to come from N')
                   ( )
              N --(   )-------                   ( )         )----* Focus (Sharp on Retina!)
              N --(   )-------/

The Two Biological Causes of Hypermetropia:

1. Focal Length of Eye Lens is Too Long: The lens is too thin, flat, and weak, lacking sufficient converging power.
2. Eyeball Has Become Too Small: The distance between the eye lens and the retina is too short.

Optical Formation:

Diverging light rays from a nearby object at $25\text{ cm}$ are not converged sufficiently and meet at a virtual focal point behind the retina.

Optical Correction of Hypermetropia:

• Hypermetropia is corrected by placing a Convex (Converging) Lens of suitable focal length in front of the eye.
• The convex lens provides additional converging power, bending incoming rays inward so they appear to emanate from the person's receded near point ($N'$).

3. Presbyopia (Old-Age Hypermetropia)

Definition & Causes:

With aging, human vision gradually deteriorates due to two natural physiological changes:

1. Gradual weakening of the ciliary muscles.
2. Diminishing flexibility and elasticity of the crystalline eye lens.

As a result, the eye's power of accommodation drops significantly, and the near point recedes, making it difficult to read comfortably without corrective spectacles.

What are Bi-Focal Lenses?

Sometimes, an elderly person suffers from both myopia and hypermetropia simultaneously:

• They require bi-focal lenses:
  • The upper portion consists of a concave lens to facilitate distant vision.
  • The lower portion consists of a convex lens to facilitate reading nearby text.

4. Solved CBSE Board Examination Numericals

Solved Example 1: Corrective Lens for a Myopic Person (NCERT Classic)

Problem: A person with a myopic eye cannot see objects beyond $1.2\text{ m}$ distinctly. What should be the type and power of the lens used to restore proper vision?

Solution:

1. Analyze the Optics:
   • The myopic person needs to view distant objects at infinity ($u = -\infty$).
   • The corrective concave lens must form a virtual image of the distant object at the person's far point: $v = -1.2\text{ m} = -120\text{ cm}$.
2. Apply the Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{-1.2} - \frac{1}{-\infty} = \frac{1}{f}$ Since $\frac{1}{\infty} = 0$: $-\frac{1}{1.2} - 0 = \frac{1}{f} \implies \mathbf{f = -1.2\text{ m}}$
3. Calculate Power of the Lens ($P$): $P = \frac{1}{f\text{ (in metres)}} = \frac{1}{-1.2} = -\frac{10}{12} = -\frac{5}{6} = \mathbf{-0.83\text{ D}}$
4. Conclusion: <u>The person requires a concave lens of focal length $-1.2\text{ m}$ and power $-0.83\text{ Dioptres}$</u>.

Solved Example 2: Corrective Lens for a Hypermetropic Person

Problem: The near point of a hypermetropic eye is $1\text{ m}$. What is the power of the lens required to read a book held at $25\text{ cm}$?

Solution:

1. Analyze the Optics:
   • The book is placed at the normal near point: $u = -25\text{ cm} = -0.25\text{ m}$.
   • The lens must create a virtual image of the book at the person's defective near point: $v = -1\text{ m} = -100\text{ cm}$.
2. Apply the Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{-100} - \left(\frac{1}{-25}\right) = \frac{1}{f} \implies -\frac{1}{100} + \frac{1}{25} = \frac{1}{f}$ $\frac{-1 + 4}{100} = \frac{1}{f} \implies \frac{3}{100} = \frac{1}{f} \implies f = +\frac{100}{3}\text{ cm} = \mathbf{+\frac{1}{3}\text{ m}}$
3. Calculate Power of the Lens ($P$): $P = \frac{1}{f\text{ (in metres)}} = \frac{1}{+1/3} = \mathbf{+3.0\text{ D}}$
4. Conclusion: <u>The person requires a convex lens of power $+3.0\text{ Dioptres}$</u>.

5. Summary and Examination Tips

Exam Tip: In board exams, drawing the three diagrams for each defect is mandatory:

1. Defective eye ray diagram (showing image in front/behind retina).
2. Far/near point diagram of the defective eye.
3. Corrected eye diagram with the appropriate lens.

Common Mistake: Mixing up object distances in numericals! For Myopia, the object distance is ALWAYS $u = -\infty$. For Hypermetropia, the object distance is ALWAYS $u = -25\text{ cm}$!