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Light - Reflection and Refraction — Question 10

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Question 10

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed.

Answer

Given,

Object height (ho) = 5 cm

Object distance (u) = -25 cm

focal length (f) = 10 cm

Ray diagram is shown below:

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed. NCERT Class 10 Science CBSE Solutions.

Image distance (v) = ?

Image height (hi) = ?

According to the Lens formula,

1v\dfrac{1}{v} - 1u\dfrac{1}{u} = 1f\dfrac{1}{f}

Substituting the values we get,

1v−1−25=1101v+125=110⇒1v=110−125⇒1v=5−250⇒1v=350⇒v=16.66 cm\dfrac{1}{v} - \dfrac{1}{-25} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{25} = \dfrac{1}{10} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} - \dfrac{1}{25} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{5-2}{50} \\[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{3}{50} \\[0.5em] \Rightarrow v = 16.66 \text{ cm}

Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.

Magnification (m) = vu\dfrac{\text{v}}{\text{u}} = height of imageheight of object\dfrac{\text{height of image}}{\text{height of object}}

Substituting the values we get,

16.66−25\dfrac{16.66}{-25} = height of image5\dfrac{\text{height of image}}{5}

height of image = 16.66×5−25\dfrac{16.66 \times 5}{-25} = -3.3

Negative sign of the height of image means that an inverted image is formed.

The image is reduced, real and inverted.

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Science | Chapter 9: Light: Reflection and RefractionWeb Content

Chapter 9: Light: Reflection and Refraction — Quick Revision Guide

Introduction

Light behaves predictably when it reflects off mirrors and refracts through lenses. This chapter covers the laws of reflection and refraction, image formation by spherical mirrors and lenses, and numerical problem-solving using mirror and lens formulae.

Key Points at a Glance

  1. Laws of reflection: angle of incidence = angle of reflection; incident ray, reflected ray, normal in same plane
  2. Concave mirror: converging; image depends on object position (6 cases from ∞ to between P and F)
  3. Convex mirror: always virtual, erect, diminished image; used as rear-view mirror (wider field of view)
  4. Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h'/h
  5. Sign convention: object on left → u negative; concave f negative; convex f positive
  6. Refraction: bending of light at interface; Snell's law: sin i / sin r = n21; denser medium bends towards normal
  7. Convex lens: converging; 6 image positions; concave lens: always virtual, erect, diminished
  8. Lens formula: 1/v − 1/u = 1/f; m = v/u; Power P = 1/f(m); unit: dioptre (D)
  9. Convex lens: +f, +P; Concave lens: −f, −P; combination: P = P1 + P2

Real-World Connections

Concave mirrors in solar cookers and headlights; convex mirrors for vehicle rear-view; magnifying glass is a convex lens; spectacles correct vision defects; cameras and projectors use lenses.

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 9
  • NCERT Exemplar Problems
  • Bright Tutorials Detailed Notes: ch09-light-reflection-refraction.html
  • Bright Tutorials Practice Questions: ch09-light-reflection-refraction.html
  • Previous Year CBSE Board Papers

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