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Light - Reflection and Refraction — Question 15

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Question 15

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer

Given,

Object distance (u) = -27 cm [∵ object is placed infront of mirror so object distance will be negative]

Object height (h) = 7 cm

Image distance (v) = ?

Image height (h) = ?

f = -18 cm [∵ focal length of concave mirror is negative by convention]

According to the mirror formula,

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

Substituting the values we get,

1v+1−27=1−18⇒1v=127−118⇒1v=2−354⇒1v=−154⇒v=−54 cm\dfrac{1}{v} + \dfrac{1}{-27} = \dfrac{1}{-18} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 27} - \dfrac{1}{18} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{2-3}{54} \\[1em] \Rightarrow \dfrac{1}{v} = - \dfrac{1}{54} \\[1em] \Rightarrow v = - 54\text{ cm}

∴ The screen should be placed at a distance of 54 cm in front of the mirror on the object side.

Magnification=−vu=height of imageheight of object−−54−27=height of image7⇒height of image=−54×727⇒height of image=−14 cm\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[1em] -\dfrac{-54}{-27} = \dfrac{\text{height of image}}{7} \\[1em] \Rightarrow \text{height of image} = -\dfrac{54 \times 7}{27} \\[1em] \Rightarrow \text{height of image} = -14 \text{ cm}

∴ Height of image is 14 cm and image is real, inverted and magnified.

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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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