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The Human Eye and the Colourful World — Question 7

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Question 7

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

Hypermetropia can be corrected by using a convex lens as shown in the diagram below:

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. NCERT Class 10 Science CBSE Solutions.

An object at 25 cm forms an image at the near point of hypermetropic eye.

Given,

Near point of hypermetropic eye = 1 m =100 cm

Object distance, u = -25 cm

Image distance, v = -100 cm

According to the formula,

⇒1v−1u=1f⇒1−100−1−25=1f⇒−1+4100=1f⇒3100=1f⇒f=1003 cm⇒f=13 mPower=1f=113∴Power=3 D\phantom{\Rightarrow} \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{1}{-100} - \dfrac{1}{-25} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{-1 + 4}{100} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{3}{100} = \dfrac{1}{f} \\[1 em] \Rightarrow f = \dfrac{100}{3} \text{ cm} \\[1 em] \Rightarrow f = \dfrac{1}{3} \text{ m} \\[1 em] \text{Power} = \dfrac{1}{f} = \dfrac{1}{\dfrac{1}{3}} \\[1 em] \therefore \text{Power} = 3 \text{ D}

Hence, the power of the lens required to correct this defect is 3 dioptre.

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Science | Chapter 10: The Human Eye and the Colourful WorldWeb Content

Chapter 10: The Human Eye and the Colourful World — Quick Revision Guide

Introduction

The human eye is a natural optical instrument. This chapter explains how the eye works, defects of vision and their correction, and beautiful phenomena like dispersion, atmospheric refraction, and scattering of light.

Key Points at a Glance

  1. Eye: cornea (refraction) → iris/pupil (light control) → lens (fine focus) → retina (image) → optic nerve → brain
  2. Accommodation: eye lens changes focal length using ciliary muscles; near point = 25 cm, far point = ∞
  3. Myopia: can't see far; image before retina; corrected by concave lens
  4. Hypermetropia: can't see near; image behind retina; corrected by convex lens
  5. Presbyopia: age-related; corrected by bifocal lens; Cataract: opaque lens; corrected by surgery
  6. Dispersion: white light splits into VIBGYOR through prism; violet deviates most, red least
  7. Atmospheric refraction: twinkling of stars, advanced sunrise/delayed sunset (~2 min each)
  8. Scattering: intensity ∝ 1/λ4; blue sky (short λ scattered more); red sunrise/sunset (blue scattered away); white clouds (all λ scattered equally by large droplets)
  9. Tyndall effect: scattering by colloidal particles; visible beam in dusty room, fog

Real-World Connections

Eye donations restore sight; LASIK surgery reshapes cornea; blue colour of sky and red sunsets explained by scattering; danger signals are red because red light travels farthest.

Quick Self-Test (5 Questions)

  1. What is the most important concept you learned from this chapter?
  2. Can you write three key equations/formulae from this chapter from memory?
  3. Draw a labelled diagram relevant to this chapter without looking at your notes.
  4. Explain one real-world application of a concept from this chapter.
  5. What is one common mistake students make in this chapter, and how can you avoid it?

Further Study

  • NCERT Textbook Chapter 10
  • NCERT Exemplar Problems
  • Bright Tutorials Detailed Notes: ch10-human-eye.html
  • Bright Tutorials Practice Questions: ch10-human-eye.html
  • Previous Year CBSE Board Papers

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