MHT-CET Physics · Wave Optics

Intensity in an Interference Pattern

Two coherent waves of equal intensity I₀ combine to 4I₀cos²(φ/2), where the phase difference is φ = 2π × (path difference)/λ; with unequal intensities the maxima and minima are (√I₁ ± √I₂)², so the dark fringes are no longer dark.

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Why this matters

29 PYQs, six HARD. Two shapes: the intensity at a point with a given path or phase difference (and which fringe sits there), and the maximum-to-minimum ratio when the two sources are unequal — including light reflected from the two faces of a glass plate.

Concept 1 of 2: Intensity From the Phase Difference

Convert the path difference into a phase difference, halve it, and square the cosine. A whole number of wavelengths gives the full 4I₀ (a bright fringe); an odd number of half-wavelengths gives zero (a dark fringe); everything else lies between.

Definition

$\phi = \dfrac{2\pi}{\lambda}\Delta x$ ; $I = 4I_0\cos^2\dfrac{\phi}{2} = I_{\max}\cos^2\dfrac{\phi}{2}$ for equal sources.
Path $\dfrac{\lambda}{4}$ : $\dfrac{I_{\max}}{2}$ . $\dfrac{\lambda}{6}$ : $\dfrac{3I_{\max}}{4}$ . $\dfrac{\lambda}{3}$ : $\dfrac{I_{\max}}{4}$ .
Dark fringes: $\phi = (2n - 1)\pi$ ; bright: $\phi = 2n\pi$ . A path difference of $n\lambda$ is the $n$ th bright band.
At a height $y$ on the screen: $\Delta x = \dfrac{yd}{D}$ .
$y_1 = a\sin(\omega t - kx)$ and $y_2 = a\cos(\omega t - kx + \phi)$ differ in phase by $\phi + \dfrac{\pi}{2}$ .

Two equal coherent sources

I = 4I_0\cos^2\frac{\phi}{2}, \qquad \phi = \frac{2\pi}{\lambda}\,\Delta x

Worked example

The maximum intensity is

I_{\max}

. What is the intensity where the path difference is

\dfrac{\lambda}{3}

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q44Moderate

Example 1 · Wave Optics · Interference Intensity and Coherent Sources

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is

\lambda

x

units,

\lambda

being the wavelength of light used. The intensity at a point where the path difference is

\lambda/4

will be (

\cos 2\pi = 1,\, \cos \frac{\pi}{2} = 0

)

Forgetting to halve the phase

I \propto \cos^2\frac{\phi}{2}

. At

\frac{\lambda}{4}

\phi = 90^\circ

and the answer is

\frac{1}{2}

, not

\cos^2 90^\circ = 0

Drill 17 more on intensity from the phase difference

Concept 2 of 2: Unequal Sources: Maxima, Minima and Contrast

Amplitudes add at a bright fringe and subtract at a dark one, and intensity goes as amplitude squared. With unequal sources the subtraction never reaches zero, so the dark fringes glow and the contrast drops. Work with square roots of intensities throughout.

Definition

$I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ , $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$ . General point: $I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi$ .
$I_1 : I_2 = 9 : 1$ ⇒ $\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{3 + 1}{3 - 1}\right)^2 = 4$ . Reverse: $\dfrac{I_{\max}}{I_{\min}} = 9$ ⇒ $I_1 : I_2 = 4 : 1$ .
Amplitude ratio at bright and dark fringes $= \sqrt{\dfrac{I_{\max}}{I_{\min}}}$ .
One slit made wider (or brighter): both maxima and minima brighten. One slit dimmed: maxima dim, minima brighten.
Reflections from the two faces of a plate: each carries its share of the energy — compute both, then use the same formula.

Maxima and minima

\frac{I_{\max}}{I_{\min}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2

Worked example

Two coherent sources have intensities in the ratio 16 : 1. Ratio of maximum to minimum intensity?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q29Moderate

Example 2 · Wave Optics · Interference Intensity and Coherent Sources

The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is

9:1

. The intensities of the light sources used are in the ratio

Taking the intensity ratio as the max–min ratio

Sources in the ratio 9 : 1 do NOT give fringes in the ratio 9 : 1. Take square roots (3 and 1), add and subtract, then square: 4 : 1.

Drill 10 more on unequal sources: maxima, minima and contrast

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Intensity From the Phase Difference
Two equal coherent sources
$I = 4I_0\cos^2\frac{\phi}{2}, \qquad \phi = \frac{2\pi}{\lambda}\,\Delta x$
Unequal Sources: Maxima, Minima and Contrast
Maxima and minima
$\frac{I_{\max}}{I_{\min}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2$

Watch out for (2)

Forgetting to halve the phase
→ Intensity From the Phase Difference
Taking the intensity ratio as the max–min ratio
→ Unequal Sources: Maxima, Minima and Contrast

Test yourself on Wave Optics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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Related notes

Wavefronts and Coherent Sources →