Interference, Diffraction, and Polarisation of Light

Wave Optics • Class 12 Physics • NCERT • CBSE

Fringe width β = λD/d. Condition for constructive interference: path difference = nλ. Condition for destructive interference: path difference = (2n−1)λ/2. Resolving power of microscope ∝ 1/λ.

Key Formulas

Frequently Asked Questions

What happens to fringe width if the experiment is immersed in water?
In water, the wavelength of light decreases: λ_water = λ_air/n (where n = 1.33 for water). Since fringe width β = λD/d, immersing in water reduces the wavelength → fringe width decreases by a factor of n. So β_water = β_air/1.33 ≈ 0.75 × β_air.
Why can two separate sodium lamps not produce interference fringes?
Two separate lamps are incoherent sources — their phase difference changes randomly and rapidly (millions of times per second). For sustained, observable interference, sources must be coherent (constant phase difference). In YDSE, both slits are derived from the same source, so they are coherent and produce stable fringes.
What is the difference between interference and diffraction?
Interference: Superposition of waves from two (or more) coherent sources. Produces equally bright and equally spaced fringes. Diffraction: Bending of waves around edges of an obstacle/slit; superposition of wavelets from points within a single wavefront. Central maximum is twice as wide as secondary maxima. In practice, both phenomena occur simultaneously in YDSE.

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