Chapter 10: Waves Optics

Physics Part 2 • Class 12

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Chapter Analysis

Advanced20 pages • English

Quick Summary

Chapter 10 of the Class 12 NCERT Physics Part 2 textbook focuses on wave optics, exploring core concepts such as interference, diffraction, and polarization of light. It delves into the wave theory of light, building on foundational principles like Huygens' theory, and explains how wave phenomena manifest. The chapter further examines the importance of coherence in light waves for creating stable interference patterns, and the chapter closes with a discussion on the applications and implications of these wave phenomena in real-world optics.

Key Topics

  • Huygens' Principle
  • Interference of Light
  • Diffraction
  • Polarization
  • Young’s Double Slit Experiment
  • Coherent Sources
  • Superposition Principle
  • Electromagnetic Waves

Learning Objectives

  • Understand the wave nature of light and its phenomena.
  • Explain Huygens' Principle and its applications.
  • Analyze the effects of interference and diffraction.
  • Describe polarization and its applications.
  • Conduct experiments related to wave optics.
  • Apply the principles of wave optics to solve physics problems.

Questions in Chapter

Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

Page 273

What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Page 273

The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 m s–1) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

Page 273

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Page 273

In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?

Page 273

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

Page 273

Additional Practice Questions

Explain how diffraction affects the resolution of optical instruments.

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Answer: Diffraction limits the resolution of optical instruments. When light passes through an aperture or around an edge, it spreads out, creating a diffraction pattern. This can limit the ability of telescopes and microscopes to distinguish between closely spaced objects, as the diffraction pattern can cause images of nearby objects to overlap.

Describe an experiment to demonstrate the interference of sound waves.

easy

Answer: Set up two loudspeakers connected to the same audio source, placed a few meters apart. As they emit coherent sound waves, move around the space in front of the speakers. You will observe regions of maximum and minimum sound intensity due to constructive and destructive interference, respectively.

What applications utilize the polarization of light, and how does it work?

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Answer: Applications of polarized light include sunglasses, photographic filters, and liquid-crystal displays (LCDs). Polarization limits the vibrations of light waves to a single plane, reducing glare and improving contrast in visual displays. Polaroid lenses, for example, block horizontally polarized light, cutting down on glare from reflective surfaces.

Calculate the critical angle for total internal reflection for light passing from water (n=1.33) to air (n=1.00).

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Answer: The critical angle θc is given by sinθc = n2/n1, where n1 > n2. For water to air, sinθc = 1.00/1.33. Thus, θc = arcsin(0.75) ≈ 48.75 degrees.

What is the principle of superposition, and how does it relate to wave interference?

easy

Answer: The principle of superposition states that when two or more waves overlap, the resultant displacement is the vector sum of the individual displacements. This principle underlies wave interference, where overlapping waves can amplify (constructive interference) or diminish (destructive interference) each other's effects.