Interference vs diffraction — conditions, pattern differences, applications

Question

Compare interference and diffraction of light. How do their fringe patterns differ? Why are interference fringes equally spaced while diffraction fringes are not?

(CBSE 12 & JEE Main — comparison question)

Solution — Step by Step

Interference: Superposition of waves from two or more coherent sources (e.g., two slits). Produces alternating bright and dark bands.

Diffraction: Bending and spreading of waves from a single slit or obstacle. The slit itself acts as many tiny sources, and their superposition creates a pattern.

Feature	Interference (YDSE)	Single Slit Diffraction
Sources	Two separate slits	One slit (many Huygens sources)
Central maximum	Same width as others	Twice the width of other maxima
Fringe spacing	Equal	Unequal (central is widest)
Intensity	All bright fringes equally bright	Intensity falls off from centre
Minima condition	$d\sin\theta = (n+\frac{1}{2})\lambda$	$a\sin\theta = n\lambda$
Fringe width	$\beta = \lambda D/d$	Central width $= 2\lambda D/a$

In double slit interference, the path difference changes linearly with position — giving equally spaced fringes. In single slit diffraction, the condition for minima involves $\sin\theta$ , which is approximately linear only near the centre. The central maximum is special because all wavelets arrive nearly in phase — giving double the width.

Why This Works

Both phenomena arise from the wave nature of light, but the source geometry creates fundamentally different patterns.

graph TD
    A["Wave Optics Pattern"] --> B{"How many sources?"}
    B -->|"Two coherent slits"| C["Interference"]
    B -->|"Single slit"| D["Diffraction"]
    C --> C1["Equal fringes<br/>β = λD/d"]
    C --> C2["All maxima same intensity"]
    D --> D1["Central max is widest<br/>Width = 2λD/a"]
    D --> D2["Intensity decreases<br/>away from centre"]
    A --> E{"In practice?"}
    E --> F["YDSE shows BOTH:<br/>interference modulated<br/>by diffraction envelope"]

In a real double slit experiment, you see the interference pattern modulated by the single slit diffraction envelope. The diffraction pattern acts as an “envelope” that reduces the intensity of the interference fringes away from the centre. JEE Advanced loves testing this combined pattern.

Alternative Method — Compare Minima Conditions

A quick way to distinguish: for interference, dark fringes occur at half-integer multiples of wavelength. For diffraction, minima occur at integer multiples. This reversal is because interference combines two sources while diffraction integrates over a continuous aperture.

For JEE: the missing order concept combines both phenomena. If the $n$ th interference maximum coincides with the $m$ th diffraction minimum, that bright fringe is missing. The condition: $d/a = n/m$ (where $d$ = slit separation, $a$ = slit width). This appears in JEE Advanced regularly.

Common Mistake

Students treat interference and diffraction as completely separate phenomena. In reality, every double slit experiment shows both simultaneously — the interference fringes are contained within the diffraction envelope. If you calculate fringe visibility without accounting for the diffraction envelope, your intensity predictions will be wrong for fringes far from the centre.

Question

Solution — Step by Step

Why This Works

Alternative Method — Compare Minima Conditions

Common Mistake

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