Class 12 Physics Notes
Complete, exam-ready notes on wave optics: Huygens' principle, reflection and refraction of plane waves, interference and Young's double-slit experiment, diffraction and polarisation — written for CBSE boards, JEE and NEET revision.
Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali
Wave optics treats light as a wave, explaining interference, diffraction and polarisation — phenomena that the ray (geometrical) model cannot account for.
Every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront at a later time is the envelope of all these secondary (Huygens') wavelets travelling at the wave speed.
Reflection and refraction
Huygens' construction shows the angle of incidence equals the angle of reflection, and gives Snell's law with refractive index .
Sources with a constant phase difference emit coherent waves that produce a stable interference pattern of bright and dark fringes. Two independent light sources are incoherent — they need a single source split in two.
The spacing between successive bright (or dark) fringes. Narrower slits and longer wavelengths widen the fringes; moving the screen farther increases .
Monochrome needed
A clear, high-contrast fringe pattern needs coherent monochromatic light. White light gives overlapping coloured fringes with a white central maximum.
The bending and spreading of light as it passes an edge or through a narrow slit. Single-slit diffraction produces a central maximum flanked by weaker secondary maxima.
Interference vs diffraction
Interference arises from a few coherent sources (bright fringes roughly equal width and intensity). Diffraction arises from many sources on one slit (central maximum broad and brightest, secondary maxima taper off).
Confining the vibrations of a light wave to a single plane. This is possible for transverse (light) waves, not longitudinal waves — polarisation proves light is transverse.
Example: In Young's experiment, slits are 0.2 mm apart and the screen is 1 m away with green light of wavelength 500 nm. Find the fringe width.
Solution: .
Revision
Memorise these before attempting numericals — most exam questions hinge on one of them.
Fringe width
Bright fringe
Single-slit minima
Malus's law
Brewster's law
Refractive index
Exam tips
Where this topic appears in CBSE, JEE Main and NEET papers.
FAQ
Interference results from a few coherent sources and gives equally spaced fringes; diffraction results from many sources on a single slit and gives a broad central maximum with decaying secondary maxima.
Each wavelength interferes at slightly different path differences, so the bright fringes for different colours land at slightly different positions, producing coloured bands around a white centre.
Only transverse waves can be polarised, so the ability to polarise light confirms it is a transverse wave and not longitudinal.
Increase the wavelength or the screen distance, or decrease the slit separation — fringe width β = λD/d.
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