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Light - Reflection and Refraction — Question 14

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Question 14

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer

Given,

Object distance (u) = -20 cm

Object height (h) = 5 cm

Image distance (v) = ?

Image height (h) = ?

Radius of curvature (R) = 30 cm

We know, R = 2f or f = R2\dfrac{R}{2}

Hence, f = 302\dfrac{30}{2} = 15 cm

According to the mirror formula,

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

Substituting the values we get,

1v+1−20=115⇒1v=120+115⇒1v=3+460⇒1v=760⇒v=607=8.57 cm\dfrac{1}{v} + \dfrac{1}{-20} = \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 20} + \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{3+4}{60} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{7}{60} \\[1em] \Rightarrow v = \dfrac{60}{7} = 8.57\text{ cm}

Image is virtual, erect and formed 8.57 cm behind the mirror.

Magnification=−vu=height of imageheight of object−8.57−20=height of image5⇒height of image=−8.57×5−20⇒height of image=2.14 cm\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[0.5em] \dfrac{-8.57}{-20} = \dfrac{\text{height of image}}{5} \\[0.5em] \Rightarrow \text{height of image} = \dfrac{-8.57 \times 5}{-20} \\[0.5em] \Rightarrow \text{height of image} = 2.14 \text{ cm}

Height of image is 2.14 cm

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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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