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Physics · Class 12 · Chapter 10

Wave Optics

Interference, diffraction and polarisation can only be explained if light is a wave. Most numericals come from Young's double-slit experiment and single-slit diffraction, and both reduce to working out a path difference correctly.

In this chapter: wavefronts and Huygens principle, reflection and refraction of plane waves, the Doppler effect for light, coherent and incoherent addition of waves, Young's double-slit experiment and fringe width, single-slit diffraction, the validity of ray optics, polarisation, Malus's law and Brewster's law.

Wavefronts and Huygens principle

A wavefront is a surface of constant phase. Waves from a point source have spherical wavefronts; far from the source a small part of the sphere is effectively a plane wavefront. Rays are the lines perpendicular to the wavefronts and show the direction of energy flow.

Huygens principle: every point on a wavefront acts as a source of secondary wavelets that spread out in the forward direction at the speed of the wave. The common tangent (envelope) to these wavelets after a time t is the new wavefront.

Refraction and reflection of plane waves

Let a plane wavefront meet a surface separating medium 1 (speed v1) from medium 2 (speed v2). While one edge of the wavefront travels a distance v1τ in medium 1, the wavelet from the other edge travels v2τ in medium 2. Constructing the new wavefront by Huygens principle gives

sin isin r = v1v2 = n2n1n = c/v; the frequency is unchanged, and the wavelength in the medium is λ/n

This is Snell's law, and it shows that light travels slower in the denser medium, which the corpuscular theory had got wrong. The same construction for reflection gives i = r. Huygens' construction also explains how a prism, a convex lens and a concave mirror change the shape of a plane wavefront.

Doppler effect for light

When a source moves away from an observer, the observed frequency drops. For speeds small compared with c, Δν/ν = −vradial/c. Light from receding galaxies is shifted towards the red end of the spectrum (red shift); an approaching source shows a blue shift.

Coherent and incoherent addition

Two sources are coherent if the phase difference between them stays constant in time. When waves of intensities I1 and I2 with phase difference φ superpose,

I = I1 + I2 + 2√I1I2 cos φEqual intensities I0: I = 4I0 cos2(φ/2)phase difference φ = (2π/λ) × path difference
  • Constructive interference (bright): path difference nλ, phase difference 2nπ, I = 4I0.
  • Destructive interference (dark): path difference (n + ½)λ, phase difference (2n + 1)π, I = 0.

If the phase difference changes rapidly and randomly, cos φ averages to zero and the intensities simply add: I = I1 + I2. So two separate bulbs never show interference. Coherent sources are obtained by splitting light from one source, as the two slits in Young's experiment do. Energy is not destroyed in interference; it is redistributed from the dark fringes to the bright ones.

Young's double-slit experiment

Light from a narrow source falls on two narrow slits S1 and S2 separated by d. They act as coherent sources, and a pattern of bright and dark fringes forms on a screen at distance D (D ≫ d). For a point P at distance x from the centre O, the path difference is

S2P − S1P ≈ d sin θ ≈ xdDBright fringes: xn = nλDd,   dark fringes: xn = (n + ½)λDdFringe width: β = λDd,   angular width = λ/d
Young's double-slit experimentwww.iitmedicoguide.comSS₁S₂screenOPS₂N = d sin θdDxθβintensityPath difference ≈ xd/D; bright when it is nλ, dark when it is (n + ½)λwww.iitmedicoguide.com
Geometry of Young's double-slit experiment. The path difference at P is d sin θ, which is about xd/D for small angles, and the fringes on the screen are equally spaced with width β.
  • The central fringe (x = 0) is bright, since the path difference is zero.
  • Bright and dark fringes are equally spaced and of equal width.
  • If the whole apparatus is placed in a medium of refractive index n, λ becomes λ/n and β becomes β/n.
  • With white light, the central fringe is white and the fringes on either side are coloured, with blue (shorter λ) closer to the centre than red.
  • If the source slit is too wide, or S is moved too close to the slits, the fringes lose contrast and disappear.
Worked example: In a double-slit experiment, λ = 600 nm, d = 0.50 mm and D = 1.0 m. Find the fringe width, the distance of the third bright fringe from the centre, and the fringe width if the apparatus is immersed in water (n = 4/3).
Solution: β = λD/d = (600 × 10−9 × 1.0)/(0.50 × 10−3) = 1.2 × 10−3 m = 1.2 mm. Third bright fringe: x3 = 3β = 3.6 mm. In water: β′ = β/n = 1.2 × 3/4 = 0.9 mm.

Diffraction at a single slit

When light passes through a single slit of width a comparable to λ, it spreads out and a pattern forms on a screen: a broad, bright central maximum with weaker secondary maxima on either side. Divide the slit into two halves: at the angle where the path difference between the edges is λ, each point in the upper half has a partner in the lower half with path difference λ/2, and they cancel. That is the first minimum.

Minima: a sin θ = nλ   (n = ±1, ±2, ±3, …)Secondary maxima: approximately θ = (n + ½)λ/aAngular width of central maximum = 2λ/a; linear width = 2λD/a
Intensity pattern of single-slit diffractionwww.iitmedicoguide.comsin θI−3λ/a−2λ/a−λ/aλ/a2λ/a3λ/a0angular width 2λ/asecondary maximaminima at a sin θ = nλ (n = ±1, ±2, …); secondary maxima much weaker, near (n + ½)λ/awww.iitmedicoguide.com
Intensity in single-slit diffraction. The central maximum is twice as wide as the others and far brighter; the secondary maxima fall off quickly in intensity.

Narrowing the slit makes the central maximum wider. Compared with double-slit interference, the diffraction fringes are of unequal width and intensity, whereas interference fringes (for narrow slits) are equally spaced and of nearly equal intensity. In fact, the double-slit pattern is an interference pattern modulated by the diffraction pattern of each slit.

Worked example: Light of wavelength 500 nm falls on a slit 0.2 mm wide, and the pattern is seen on a screen 1 m away. Find the distance of the first minimum from the centre and the width of the central maximum.
Solution: y1 = λD/a = (500 × 10−9 × 1)/(0.2 × 10−3) = 2.5 × 10−3 m = 2.5 mm. Width of the central maximum = 2 × 2.5 = 5.0 mm.

Validity of ray optics

A beam of width a spreads by diffraction to a width of about λz/a after travelling a distance z. The spreading becomes comparable to a at the Fresnel distance, zF ≈ a2/λ. For distances much less than zF, diffraction can be ignored and ray optics works. For a 3 mm aperture and visible light (λ ≈ 500 nm), zF is about 18 m.

Polarisation

Light is a transverse wave; in ordinary (unpolarised) light the electric field vibrates in all directions perpendicular to the direction of travel. A polaroid transmits only the component of E along its pass axis, giving plane polarised light. Only transverse waves can be polarised, so polarisation proves that light is transverse.

Unpolarised light through a polaroid: I = I0/2Malus's law: I = I0 cos2θθ is the angle between the direction of polarisation of the incident light and the pass axis

Light scattered at 90° by molecules of the atmosphere is polarised, which is why the blue sky seen through a rotating polaroid changes in brightness. Light reflected from a transparent surface is completely polarised at one angle of incidence, the Brewster angle, for which the reflected and refracted rays are perpendicular:

tan iB = n   (Brewster's law)
Worked example: Unpolarised light of intensity I0 passes through two polaroids whose pass axes are at 60° to each other. Find the transmitted intensity.
Solution: After the first, I0/2. After the second, (I0/2) cos260° = (I0/2)(1/4) = I0/8.
Common mistakes: (1) Using the bright-fringe condition for a single slit; a sin θ = nλ gives minima in diffraction but maxima in double-slit interference. (2) Forgetting that λ, not the frequency, changes in a medium. (3) Adding intensities of coherent waves directly; add amplitudes and include the phase. (4) Taking the central maximum width as λD/a; it is 2λD/a. (5) Applying Malus's law to unpolarised light; the first polaroid simply halves the intensity.

JEE and NEET focus

  • Huygens construction for refraction and the result that light is slower in a denser medium.
  • Resultant intensity for a given phase or path difference, and the ratio of maximum to minimum intensity.
  • Fringe width in YDSE and how it changes with d, D, λ and the medium.
  • Positions of minima and width of the central maximum in single-slit diffraction.
  • Malus's law with two or three polaroids, and Brewster's angle.

Practice questions

In YDSE, the slit separation is doubled and the screen distance is halved. The fringe width becomes:

  1. β
  2. β/2
  3. β/4
  4. 4β
Show answer
C. β = λD/d; D/2 and 2d give β/4.

A YDSE set-up is immersed in water of refractive index 4/3. The fringe width becomes:

  1. 4/3 times
  2. 3/4 times
  3. unchanged
  4. 9/16 times
Show answer
B. λ in water is λ/n, so β becomes β/n.

Two coherent waves of equal intensity I0 meet with a path difference of λ/4. The resultant intensity is:

  1. 4I0
  2. 2I0
  3. I0
  4. zero
Show answer
B. φ = π/2, so I = 4I0 cos2(π/4) = 2I0.

In single-slit diffraction, the slit width is halved. The angular width of the central maximum:

  1. halves
  2. doubles
  3. stays the same
  4. becomes four times
Show answer
B. Angular width = 2λ/a.

The Brewster angle for glass of refractive index √3 is:

  1. 30°
  2. 45°
  3. 60°
  4. 90°
Show answer
C. tan iB = √3.

Light from two independent sodium lamps does not produce a steady interference pattern because the lamps are:

  1. of different intensity
  2. not coherent
  3. monochromatic
  4. too far apart
Show answer
B. Their phase difference changes randomly and rapidly.

Unpolarised light of intensity 32 W m−2 passes through a polaroid. The transmitted intensity is:

  1. 32 W m−2
  2. 16 W m−2
  3. 8 W m−2
  4. zero
Show answer
B. A polaroid transmits half of unpolarised light.

In YDSE, the central fringe is:

  1. always dark
  2. bright, since the path difference is zero
  3. bright only for white light
  4. absent
Show answer
B. Waves from both slits arrive in phase at O.
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