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Class 12 Science (PCM)

Physics Part 2

Wave Optics

Wave Optics

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Conservation of Energy in Interference and Diffraction
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Frequency, Wavelength and Speed in Refraction
Accuracy: 33%
Total Attempted9
Total Correct3
Total Incorrect6
Difficulty Level:
High
Polarised and Unpolarised Light
Accuracy: 40%
Total Attempted5
Total Correct2
Total Incorrect3
Difficulty Level:
High
Reflection by Huygens’ Principle
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Snell's Law and Refraction
Accuracy: 50%
Total Attempted16
Total Correct8
Total Incorrect8
Difficulty Level:
Medium
Polarisation of Light
Accuracy: 53%
Total Attempted15
Total Correct8
Total Incorrect7
Difficulty Level:
Medium
Wave Theory of Light and Limits of Ray Optics
Accuracy: 55%
Total Attempted11
Total Correct6
Total Incorrect5
Difficulty Level:
Medium
Need for Coherent Sources in Young’s Experiment
Accuracy: 62%
Total Attempted13
Total Correct8
Total Incorrect5
Difficulty Level:
Medium
Huygens' Principle and Wavefront Construction
Accuracy: 64%
Total Attempted22
Total Correct14
Total Incorrect8
Difficulty Level:
Medium
Polaroids and Intensity Control
Accuracy: 67%
Total Attempted6
Total Correct4
Total Incorrect2
Difficulty Level:
Medium
Superposition Principle and Interference Intensity
Accuracy: 67%
Total Attempted12
Total Correct8
Total Incorrect4
Difficulty Level:
Medium
Interference of Light Waves
Accuracy: 68%
Total Attempted19
Total Correct13
Total Incorrect6
Difficulty Level:
Medium
Coherent and Incoherent Sources
Accuracy: 69%
Total Attempted16
Total Correct11
Total Incorrect5
Difficulty Level:
Medium
Conditions for Constructive and Destructive Interference
Accuracy: 69%
Total Attempted16
Total Correct11
Total Incorrect5
Difficulty Level:
Medium
Fringe Width in Young’s Double-Slit Experiment
Accuracy: 73%
Total Attempted11
Total Correct8
Total Incorrect3
Difficulty Level:
Medium
Malus' Law
Accuracy: 75%
Total Attempted8
Total Correct6
Total Incorrect2
Difficulty Level:
Low
Young's Double-Slit Experiment
Accuracy: 76%
Total Attempted21
Total Correct16
Total Incorrect5
Difficulty Level:
Low
Diffraction and Single Slit Diffraction Pattern
Accuracy: 78%
Total Attempted18
Total Correct14
Total Incorrect4
Difficulty Level:
Low
Wavefronts and Rays
Accuracy: 87%
Total Attempted23
Total Correct20
Total Incorrect3
Difficulty Level:
Low
Refraction at Denser and Rarer Media
Accuracy: 88%
Total Attempted8
Total Correct7
Total Incorrect1
Difficulty Level:
Low
Single-Slit Diffraction Minima and Maxima
Accuracy: 88%
Total Attempted16
Total Correct14
Total Incorrect2
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the historical development of light theories, including the corpuscular and wave models.
  • Explain Huygens' principle and its application in wavefront analysis.
  • Analyze the behavior of light during reflection and refraction using wave optics.
  • Describe the phenomenon of interference and its significance in wave optics.
  • Calculate the wavelength, frequency, and speed of light in different media.
  • Investigate the conditions for constructive and destructive interference in Young's double-slit experiment.
  • Explore the effects of diffraction and polarization on light waves.

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Revision Notes & Summary

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Chapter Ten: Wave Optics

10.1 Introduction

  • Corpuscular Model of Light: Proposed by Descartes in 1637, explaining reflection and refraction.
  • Wave Theory of Light: Introduced by Huygens in 1678, explaining light behavior through wave models.
  • Contradiction: Wave model predicts light slows down in denser media, confirmed by Foucault's experiment in 1850.

10.2 Key Principles

  1. Huygens' Principle: Each point on a wavefront acts as a source of secondary waves.
  2. Superposition of Waves: Intensity at a point is the sum of individual intensities plus an interference term.
  3. Young's Double Slit Experiment: Produces equally spaced interference fringes.
  4. Single Slit Diffraction: Results in a central maximum with weaker secondary maxima.
  5. Polarization: Natural light is unpolarized; polaroids transmit only one component, resulting in linearly polarized light.

10.3 Points to Ponder

  • Waves spread from point sources; light travels in rays.
  • Interference can be constructive or destructive.
  • Diffraction limits optical resolution.
  • Polarization is unique to transverse waves like light.

10.4 Exercises

  1. Monochromatic Light Incident on Water: Calculate wavelength, frequency, and speed of reflected and refracted light.
  2. Wavefront Shapes: Determine shapes for light from point sources, lenses, and distant stars.
  3. Speed of Light in Glass: Calculate speed and discuss color dependence.
  4. Young's Double-Slit Experiment: Calculate wavelength based on fringe measurements.
  5. Intensity Calculations: Analyze intensity at different path differences in interference.
  6. Interference of Two Wavelengths: Find distances of bright fringes and coincidences.

10.5 Important Diagrams

  • Reflection and Refraction: Illustrates wavefronts reflecting and refracting at surfaces.
  • Young's Double-Slit Experiment: Setup showing light source, slits, and interference pattern.
  • Single Slit Diffraction: Shows intensity distribution with central maximum and secondary maxima.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

Interference

B.

Diffraction

C.

Polarisation

D.

Reflection
Correct Answer: B

Solution:

Diffraction sets the limit of ray optics because it causes light to deviate from the straight-line path, which is a fundamental assumption in geometrical optics.

Chapter Concept:

Wave Theory of Light and Limits of Ray Optics

A.

It doubles.

B.

It remains the same.

C.

It halves.

D.

It quadruples.
Correct Answer: C

Solution:

The fringe width β\beta is given by β=λDd\beta = \frac{\lambda D}{d}, where λ\lambda is the wavelength, DD is the distance to the screen, and dd is the slit separation. Doubling dd will halve β\beta.

Chapter Concept:

Fringe Width in Young’s Double-Slit Experiment

A.

Energy is lost in dark regions and gained in bright regions.

B.

Energy is redistributed with no net gain or loss.

C.

Energy is only present in bright regions.

D.

Energy is absorbed by the medium in dark regions.
Correct Answer: B

Solution:

In interference and diffraction, light energy is redistributed such that dark regions correspond to reduced intensity while bright regions gain intensity, with no net gain or loss of energy.

Chapter Concept:

Conservation of Energy in Interference and Diffraction

A.

The experiment demonstrates the wave nature of light.

B.

Dark fringes occur at x=(n+1/2)λDdx = (n + 1/2) \frac{\lambda D}{d}.

C.

The experiment can be performed with incoherent light sources.

D.

The fringe pattern is a result of interference.
Correct Answer: C

Solution:

Young's double-slit experiment requires coherent light sources to produce a stable interference pattern.

Chapter Concept:

Young's Double-Slit Experiment

A.

By assuming light travels faster in denser media.

B.

By assuming light travels slower in denser media.

C.

By considering the change in frequency of light.

D.

By considering the change in wavelength of light.
Correct Answer: B

Solution:

Huygens' principle explains refraction by considering that light travels slower in denser media, causing the wavefronts to bend towards the normal.

Chapter Concept:

Huygens' Principle and Wavefront Construction

A.

The angle of incidence is equal to the angle of reflection because the secondary wavelets form a backward wavefront.

B.

The angle of incidence is equal to the angle of reflection because the secondary wavelets form a forward wavefront.

C.

The angle of incidence is equal to the angle of reflection because the speed of light changes upon reflection.

D.

The angle of incidence is equal to the angle of reflection because the wavefronts become spherical after reflection.
Correct Answer: B

Solution:

According to Huygens' principle, each point on the wavefront acts as a source of secondary wavelets. The new wavefront is the forward envelope of these wavelets. This construction leads to the law of reflection, where the angle of incidence is equal to the angle of reflection.

Chapter Concept:

Reflection by Huygens’ Principle

A.

Because they have different wavelengths.

B.

Due to rapid random phase changes.

C.

Because they emit light of different colors.

D.

Due to the difference in their intensities.
Correct Answer: B

Solution:

Two independent light sources do not give stable fringes due to rapid random phase changes.

Chapter Concept:

Need for Coherent Sources in Young’s Experiment

A.

Spherical

B.

Plane

C.

Cylindrical

D.

Elliptical
Correct Answer: A

Solution:

Huygens' principle states that light diverging from a point source forms a spherical wavefront.

Chapter Concept:

Huygens' Principle and Wavefront Construction

A.

Constructive interference

B.

Destructive interference

C.

No interference

D.

Amplified interference
Correct Answer: B

Solution:

Destructive interference occurs when the path difference is an odd multiple of half-wavelengths, causing the waves to cancel each other out.

Chapter Concept:

Polarised and Unpolarised Light

A.

2.0×108 m/s2.0 \times 10^8 \text{ m/s}

B.

2.5×108 m/s2.5 \times 10^8 \text{ m/s}

C.

1.5×108 m/s1.5 \times 10^8 \text{ m/s}

D.

3.0×108 m/s3.0 \times 10^8 \text{ m/s}
Correct Answer: A

Solution:

The speed of light in a medium is given by v=cnv = \frac{c}{n} where cc is the speed of light in vacuum and nn is the refractive index. Thus, v=3.0×1081.5=2.0×108 m/sv = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8 \text{ m/s}.

Chapter Concept:

Frequency, Wavelength and Speed in Refraction

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True or False

Correct Answer: True

Solution:

The intensity formula for two coherent waves with equal amplitude and a phase difference ϕ\phi is indeed I=4I0cos⁡2(ϕ/2)I = 4I_0 \cos^2(\phi/2), where I0I_0 is the intensity of each wave.

Chapter Concept :

Superposition Principle and Interference Intensity

Correct Answer: True

Solution:

Malus' Law describes how the intensity of light changes as it passes through a polarising filter, depending on the angle between the light's initial polarisation direction and the filter's axis.

Chapter Concept :

Malus' Law

Correct Answer: True

Solution:

The frequency of light remains constant when transitioning between media, while the speed and wavelength adjust according to the medium's properties. In a denser medium, the speed and wavelength of light decrease as described by the equation v=νλv = \nu \lambda.

Chapter Concept :

Frequency, Wavelength and Speed in Refraction

Correct Answer: True

Solution:

Diffraction limits the applicability of ray optics as it accounts for the wave nature of light, which becomes significant when the wavelength is comparable to the size of the optical elements.

Chapter Concept :

Wave Theory of Light and Limits of Ray Optics

Correct Answer: True

Solution:

Huygens' principle states that each point on a wavefront is a source of secondary waves, which add up to give the wavefront at a later time.

Chapter Concept :

Huygens' Principle and Wavefront Construction

Correct Answer: False

Solution:

In linearly polarised light, the electric field oscillates in one fixed plane, not multiple planes.

Chapter Concept :

Polarisation of Light

Correct Answer: False

Solution:

In single-slit diffraction, the central maximum is actually the broadest and brightest region, with the intensity decreasing for subsequent maxima.

Chapter Concept :

Diffraction and Single Slit Diffraction Pattern

Correct Answer: True

Solution:

Bright fringes in Young's double-slit experiment are located at positions where the path difference is an integer multiple of the wavelength, given by x=nλD/dx = n\lambda D/d.

Chapter Concept :

Young's Double-Slit Experiment

Correct Answer: True

Solution:

Constructive interference occurs for path difference nλn\lambda, where nn is an integer, resulting in bright regions.

Chapter Concept :

Interference of Light Waves

Correct Answer: True

Solution:

Constructive interference results in bright regions when the path difference is nλn\lambda, and destructive interference results in dark regions when the path difference is (n+1/2)λ(n + 1/2)\lambda.

Chapter Concept :

Interference of Light Waves

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