Wave optics (also known as physical optics) is the branch of optics that studies phenomena that cannot be explained by simple ray optics — especially those that involve interference, diffraction, polarization and wavefronts. In wave optics, light is treated explicitly as a wave, with wavelength, phase, superposition etc., becoming central. These phenomena have many applications: understanding the fine structure of spectra, designing optical instruments (telescopes, microscopes), resolving power, anti-reflection coatings, etc. In this post we will cover all the major topics of wave optics as in CBSE/other equivalent class-12 curricula: definitions, principles, derivations, formulas, and how they are used.
Class 12 Physics Wave Opticsl Notes | Detailed Explanation, Formulas & Questions
Table of Contents
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Wavefronts and Huygens’ Principle
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Superposition, Coherence, Interference
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Young’s Double Slit Experiment
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Diffraction (Single Slit, Multiple Slits & Gratings)
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Polarization
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Mathematical Concepts: Phase, Path Difference, Time Difference
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Important Formulas Summary
1. Wavefronts and Huygens’ Principle
1.1 What is a Wavefront
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A wavefront is the locus (i.e., set) of points in a medium which are in the same phase of oscillation at a given instant.
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Common types: plane wavefronts, spherical wavefronts, cylindrical, etc.
1.2 Huygens’ Principle
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Statement: Every point on a given wavefront acts as a source of secondary wavelets that spread out in all directions with the same speed as the original wave. The new position of the wavefront at any later instant is given by the envelope (tangent surface) to these secondary wavelets.
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Using this principle we can derive/reflection, refraction, propagation in different media etc.
1.3 Reflection & Refraction by Huygens
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By constructing wavelets at a boundary, one obtains Law of Reflection: Angle of incidence = Angle of reflection.
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Also Snell’s Law of Refraction:
n1sini=n2sinrn_1 \sin i = n_2 \sin r
where n1,n2n_1, n_2 are refractive indices of first and second medium, ii the incidence, rr refraction. The speed of wave in medium 1 is v1v_1, in medium 2 is v2v_2; n1=c/v1, n2=c/v2n_1 = c/v_1,\; n_2 = c/v_2.
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Frequency remains constant on refraction; wavelength and speed change.
2. Superposition, Coherence, Interference
2.1 Principle of Superposition
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When two or more waves traverse the same region, the resultant disturbance (displacement, electric field etc.) at any point is the vector sum of individual disturbances.
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This leads to constructive or destructive effects depending on the phase difference.
2.2 Coherent Sources
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Coherent sources are sources that emit waves of same (or nearly same) frequency and maintain a constant phase difference between them. Only coherent sources produce stable interference patterns.
2.3 Phase Difference, Path Difference, Time Difference
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Path Difference (Δ): Difference in distances traveled by two waves from their respective sources to a point of observation.
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Phase Difference (ϕ): Corresponding phase shift between those two waves; often
ϕ=2πλΔ\phi = \frac{2\pi}{\lambda} \Delta
(assuming same wavelength).
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Time Difference: Related to phase difference via period TT:
Time difference=ϕ2π T\text{Time difference} = \frac{\phi}{2\pi} \, T
2.4 Interference
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When two coherent waves overlap, intensity distribution arises due to interference: bright (constructive) and dark (destructive) fringes etc.
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Conditions: same frequency, fixed phase difference, and sufficient coherence.
3. Young’s Double Slit Experiment (YDSE)
3.1 Setup and Observations
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Two narrow slits (separated by distance dd) are illuminated by monochromatic light. The slits act as coherent sources. On a screen placed at distance DD from slits (D≫dD\gg d), a pattern of bright and dark fringes appears.
3.2 Derivation of Fringe Positions
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Constructive Interference (bright fringe at order mm): path difference Δ=mλ\Delta = m\lambda.
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Destructive Interference (dark fringe): Δ=(m+12)λ\Delta = (m + \tfrac{1}{2}) \lambda.
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If yy is the distance of the mm-th bright fringe from central maxima on screen, then approximate (for small angles):
y=mλDdy = \frac{m\lambda D}{d}
3.3 Fringe Width
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Fringe width (distance between two successive bright or dark fringes):
β=λDd\beta = \frac{\lambda D}{d}
4. Diffraction
Diffraction refers to the bending or spreading of waves when they encounter an obstacle or pass through a narrow aperture whose size is comparable to the wavelength.
4.1 Single Slit Diffraction
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When monochromatic light passes through a slit of width aa, the diffraction pattern on a distant screen shows a central bright maximum flanked by weaker maxima and minima.
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Condition for minima (dark fringes):
asinθ=mλ(m=1,2,3,… )a \sin \theta = m\lambda \quad (m = 1,2,3,\dots)
where θ\theta is angle from central axis.
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Width of central maximum: from first minimum on one side to first minimum on other side, approx 2λD/a2\lambda D / a on screen (for small angle approx).
4.2 Multiple Slits and Diffraction Grating
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A diffraction grating has many slits regularly spaced. It produces very sharp maxima (due to many-slit interference).
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Grating equation:
dsinθ=mλd \sin \theta = m \lambda
where dd is separation between adjacent slits (grating spacing), mm is order of maxima.
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Angular separation between orders, resolving power etc. can be studied using grating.
4.3 Fresnel vs Fraunhofer Diffraction
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Fraunhofer diffraction: far field diffraction; the source and screen are effectively at infinite distances, or used with collimating lenses. Simplifies computation.
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Fresnel diffraction: near field diffraction; distances are finite, wavefront curvature matters.
5. Polarization
5.1 Nature of Polarization
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Light waves are transverse electromagnetic waves; polarization refers to the direction of oscillation of the electric field. Natural light is unpolarized (oscillations in all transverse directions), whereas polarized light has a preferential direction.
5.2 Types of Polarization
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Linear polarization: electric field oscillates in single plane.
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Circular and elliptical polarization (often beyond basic class 12, but linear is the core).
5.3 Polarizer-Analyzer and Malus’ Law
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If polarized light of intensity I0I_0 is incident on an analyzer whose axis makes an angle θ\theta with the polarization direction, then transmitted intensity is:
I=I0cos2θI = I_0 \cos^2 \theta
5.4 Applications of Polarization
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Reduction of glare (polarizing sunglasses), stress analysis (photoelasticity), optical filters, etc.
6. Mathematical Concepts & Derivations
6.1 Path Difference, Phase Difference
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Path difference Δ\Delta between two waves leads to a phase difference:
ϕ=2πλΔ\phi = \frac{2\pi}{\lambda} \Delta
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If Δ=nλ\Delta = n\lambda → ϕ=2πn\phi = 2\pi n → constructive;
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If Δ=(n+12)λ\Delta = (n + \tfrac{1}{2})\lambda → ϕ=(2n+1)π\phi = (2n +1)\pi → destructive.
6.2 Resultant Amplitude & Intensity for Two Waves
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Let two waves of amplitudes a1a_1 and a2a_2, same frequency, meeting with phase difference ϕ\phi. The resultant amplitude AA is given by:
A=a12+a22+2a1a2cosϕA = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos\phi}
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Since intensity II is proportional to square of amplitude:
I=I1+I2+2I1I2 cosϕI = I_1 + I_2 + 2 \sqrt{I_1 I_2}\, \cos \phi
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For identical amplitudes a1=a2=aa_1 = a_2 = a (or identical intensities I1=I2=I0I_1 = I_2 = I_0), these simplify.
7. Important Formulas Summary
Here is a consolidated list of the most important formulas with heading styling.
## Snell’s Law (Refraction)
n1sini=n2sinrn_1 \sin i = n_2 \sin r
## Relation between phase and path difference
ϕ=2πλ Δ\phi = \frac{2\pi}{\lambda} \,\Delta
## Time difference related to phase difference
Time difference=ϕ2πT\text{Time difference} = \frac{\phi}{2\pi} T
where TT = period = 1/f1/f.
## Resultant Amplitude of Two Waves
A=a12+a22+2 a1a2cosϕA = \sqrt{a_1^2 + a_2^2 + 2\,a_1 a_2 \cos\phi}
## Resultant Intensity for Two Waves
I=I1+I2+2I1I2 cosϕI = I_1 + I_2 + 2 \sqrt{I_1 I_2}\, \cos\phi
For identical waves (I1=I2=I0I_1 = I_2 = I_0):
I=2I0(1+cosϕ)=4I0cos2(ϕ2)I = 2I_0(1 + \cos\phi) = 4 I_0 \cos^2\left(\frac{\phi}{2}\right)
## Young’s Double Slit – Fringe Position
ym=mλDdy_m = \frac{m \lambda D}{d}
where
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mm = order of bright fringe,
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DD = distance from slits to screen,
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dd = separation between slits.
## Fringe Width
β=λDd\beta = \frac{\lambda D}{d}
## Single Slit Diffraction Minima
asinθ=mλ(m=1,2,3,…)a \sin \theta = m \lambda \quad (m =1,2,3,\ldots)
## Diffraction Grating Equation
dsinθ=mλd \sin \theta = m \lambda
## Malus’ Law (Polarization)
I=I0cos2θI = I_0 \cos^2\theta
8. Practical Tips & Remarks
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Always check whether small angle approximation (sinθ≈θ\sin \theta \approx \theta, tanθ≈θ\tan\theta \approx \theta in radians) is acceptable. This simplifies many formulae in YDSE and diffraction.
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Remember distinction: interference requires coherent sources; diffraction can happen with a single source and obstacle/aperture.
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Understand difference between Fraunhofer (far field) and Fresnel (near field) diffraction. Usually class 12 deals mostly with Fraunhofer.
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For polarization: know that natural light has all polarization directions; polarizer selects one; analyzer then gives intensity variation via Malus’ law.
OBJECTIVE QUESTION -Wave Optics
Q1. The phenomenon of bending of light waves around obstacles is known as
(a) Interference
(b) Diffraction
(c) Polarization
(d) Reflection
Correct Answer: (b) Diffraction
Q2. In Young’s Double Slit Experiment (YDSE), the fringe width (β) is directly proportional to
(a) Wavelength of light
(b) Distance between screen and slits
(c) Distance between the slits
(d) Both (a) and (b)
Correct Answer: (d) Both (a) and (b)
Q3. The condition for constructive interference is
(a) Path difference = (2n+1)λ2(2n+1)\frac{\lambda}{2}
(b) Path difference = nλn\lambda
(c) Phase difference = π\pi
(d) Path difference = λ/4\lambda/4
Correct Answer: (b) Path difference = nλn\lambda
Q4. The condition for destructive interference is
(a) Phase difference = 00
(b) Phase difference = 2πn2\pi n
(c) Phase difference = (2n+1)π(2n+1)\pi
(d) Phase difference = 4πn4\pi n
Correct Answer: (c) Phase difference = (2n+1)π(2n+1)\pi
Q5. Malus’ law is related to
(a) Interference
(b) Polarization
(c) Diffraction
(d) Reflection
Correct Answer: (b) Polarization
Q6. Polarization proves that light waves are
(a) Longitudinal
(b) Transverse
(c) Mechanical
(d) Scalar
Correct Answer: (b) Transverse
Q7. In single-slit diffraction, the angular width of the central maximum is given by
(a) λa\frac{\lambda}{a}
(b) 2λa\frac{2\lambda}{a}
(c) aλ\frac{a}{\lambda}
(d) a2λ\frac{a}{2\lambda}
Correct Answer: (b) 2λa\frac{2\lambda}{a}
Q8. The width of the central bright fringe in YDSE is
(a) λDd\frac{\lambda D}{d}
(b) dλD\frac{d}{\lambda D}
(c) 2λDd\frac{2\lambda D}{d}
(d) λdD\frac{\lambda d}{D}
Correct Answer: (a) λDd\frac{\lambda D}{d}
Q9. Huygens’ principle explains
(a) Snell’s law
(b) Newton’s law
(c) Kirchhoff’s law
(d) Gauss’ law
Correct Answer: (a) Snell’s law
Q10. The intensity at any point in interference is given by
(a) I=I1+I2+2I1I2cosϕI = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos \phi
(b) I=I1+I2I = I_1 + I_2
(c) I=I1−I2I = I_1 – I_2
(d) I=I12+I22I = \sqrt{I_1^2 + I_2^2}
Correct Answer: (a) I=I1+I2+2I1I2cosϕI = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos \phi
Q11. The central bright fringe in YDSE is called
(a) Principal maximum
(b) Secondary maximum
(c) Central minimum
(d) Secondary minimum
Correct Answer: (a) Principal maximum
Q12. The coherent sources in YDSE can be obtained by
(a) Two independent sources
(b) Splitting a single source into two
(c) Two different colored sources
(d) Two sources with different frequencies
Correct Answer: (b) Splitting a single source into two
Q13. The angular width of principal maxima in diffraction grating
(a) Decreases with increasing number of slits
(b) Increases with wavelength
(c) Increases with number of slits
(d) Remains constant
Correct Answer: (a) Decreases with increasing number of slits
Q14. The phase difference between two points on a wave is given by
(a) 2πλ×path difference\frac{2\pi}{\lambda} \times \text{path difference}
(b) λ2π×path difference\frac{\lambda}{2\pi} \times \text{path difference}
(c) πλ×path difference\frac{\pi}{\lambda} \times \text{path difference}
(d) None of these
Correct Answer: (a) 2πλ×path difference\frac{2\pi}{\lambda} \times \text{path difference}
Q15. The concept of secondary wavelets was given by
(a) Huygens
(b) Newton
(c) Fresnel
(d) Maxwell
Correct Answer: (a) Huygens
Q16. The width of the central maximum in a single slit diffraction pattern is
(a) Directly proportional to λ\lambda
(b) Inversely proportional to λ\lambda
(c) Inversely proportional to slit width
(d) Both (a) and (c)
Correct Answer: (d) Both (a) and (c)
Q17. In YDSE, if the slit separation is doubled, the fringe width becomes
(a) Double
(b) Half
(c) Unchanged
(d) Four times
Correct Answer: (b) Half
Q18. The shape of wavefront for a point source is
(a) Plane
(b) Spherical
(c) Cylindrical
(d) Elliptical
Correct Answer: (b) Spherical
Q19. The unit of fringe width is
(a) Meter
(b) Radian
(c) No unit
(d) Second
Correct Answer: (a) Meter
Q20. The first minimum in single slit diffraction occurs when
(a) asinθ=λa \sin \theta = \lambda
(b) asinθ=0a \sin \theta = 0
(c) dsinθ=mλd \sin \theta = m\lambda
(d) θ=0\theta = 0
Correct Answer: (a) asinθ=λa \sin \theta = \lambda
Short Answer Questions-Wave Optics
Q1. What is the principle of Young’s Double Slit Experiment (YDSE)?
Answer:
The principle of YDSE is based on the phenomenon of interference of light. When two coherent sources emit light waves of the same frequency and wavelength, they superimpose to form an interference pattern of bright and dark fringes on a screen.
Q2. Define coherence in the context of wave optics.
Answer:
Coherence refers to a fixed phase relationship between two waves of the same frequency. Two sources are said to be coherent if the phase difference between them remains constant over time.
Q3. What is the condition for constructive interference?
Answer:
Constructive interference occurs when the path difference between two waves is an integral multiple of the wavelength, i.e., Δx=nλ\Delta x = n\lambda, where n=0,1,2,3,…n = 0, 1, 2, 3, \dots.
Q4. State Malus’ Law for polarization of light.
Answer:
Malus’ Law states that when polarized light is incident on an analyzer, the intensity of light transmitted varies as the square of the cosine of the angle between the transmission axis of the analyzer and the direction of polarization of light:
I=I0cos2θI = I_0 \cos^2 \theta
Q5. What is the shape of the wavefront produced by a point source?
Answer:
The wavefront produced by a point source is spherical in shape because the light waves spread out equally in all directions from the source.
Long Answer Questions-Wave Optics
Q1. Derive the expression for fringe width in Young’s Double Slit Experiment.
Answer:
Consider two coherent sources S1S_1 and S2S_2 separated by distance dd, and a screen placed at distance DD. The point PP on the screen has a path difference Δx=d⋅xD\Delta x = \frac{d \cdot x}{D}.
For constructive interference, Δx=nλ\Delta x = n\lambda.
Fringe width β\beta is the distance between two consecutive bright fringes:
β=λDd\beta = \frac{\lambda D}{d}
This shows that fringe width is directly proportional to wavelength and distance between the screen and slits, but inversely proportional to the distance between the slits.
Q2. Explain Huygens’ Principle and use it to derive the law of refraction.
Answer:
Huygens’ Principle states that every point on a wavefront acts as a source of secondary wavelets that spread out in all directions with the speed of the wave. The new wavefront is the tangent to these secondary wavelets.
Using this principle, Snell’s Law sinisinr=v1v2=n2n1\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1} can be derived, which relates the angle of incidence and refraction with the speed of light in two media.
Q3. What is diffraction of light? Explain the diffraction pattern due to a single slit.
Answer:
Diffraction is the bending of light around the corners of an obstacle or aperture. In single-slit diffraction, a central bright fringe (principal maximum) is formed at θ=0\theta = 0, and secondary maxima and minima occur on both sides. The first minimum occurs at asinθ=±λa \sin \theta = \pm \lambda.
Q4. Write a note on the polarization of light and its applications.
Answer:
Polarization is the phenomenon in which vibrations of light waves are restricted to a single plane perpendicular to the direction of propagation. It proves that light waves are transverse in nature. Applications include polaroid sunglasses, optical instruments, 3D movies, and stress analysis in engineering materials.
Q5. Differentiate between interference and diffraction of light.
Answer:
Interference occurs due to the superposition of two coherent light waves, producing a regular pattern of bright and dark fringes. Diffraction occurs when light bends around edges, producing a broad central maximum with diminishing secondary fringes. Interference fringes are equally spaced, while diffraction fringes are not equally spaced.
Conclusion
Wave optics brings a deeper understanding of light than ray optics. By studying wavefronts, interference, diffraction, and polarization, we see many phenomena that explain why light behaves the way it does in real settings (thin films, resolving images, glare, etc.). Mastery of the formulas, when and how to use approximations, and coherent/incoherent superposition is essential.
Class 12 Physics Ray Optics and Optical Instruments Notes | Complete Guide