JEE Mains Physics · Wave Optics

Coherent Sources and Resultant Intensity

Coherent waves add as amplitudes, so the intensity carries a cross term 2√(I₁I₂) cos φ; incoherent waves just add their intensities.

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Why this matters

Twenty-three PYQs, fifteen of them multiple choice, and none yet from 2026. Six ask for the resultant of two beams at a given phase difference, coherent or not. Thirteen ask for the ratio of the brightest to the darkest fringe, five of them from slit widths rather than intensities. Four turn a layer of a medium, a column of air or a thin film into an extra optical path. Every one of them starts from amplitudes, not intensities.

Concept 1 of 3: Resultant intensity of two beams

Two coherent waves keep a fixed phase difference, so their amplitudes add like two arrows at that angle. Squaring the resultant arrow gives the intensity, and the cross term 2√(I₁I₂) cos φ is the interference. For incoherent waves the phase difference jumps about so fast that the cross term averages to zero, and only the intensities add.

Definition

Coherent sources keep a constant phase difference (in practice, both come from one source).
Amplitudes add as phasors: $A^{2} = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos\phi$ .
Since $I \propto A^{2}$ : coherent, $I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\phi$ ; incoherent, $I = I_{1} + I_{2}$ .
Two equal beams $I_{0}$ each: $I = 2I_{0}(1 + \cos\phi) = 4I_{0}\cos^{2}\dfrac{\phi}{2}$ .
Beyond $90^{\circ}$ the cosine is negative, so the coherent sum is less than the incoherent one.

Two coherent beams

I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\phi, \qquad A^{2} = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos\phi

Worked example

Two coherent beams of intensities 3I and 12I meet with a phase difference of

120^{\circ}

. Find the resultant intensity, and compare it with the intensity if the beams were incoherent.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q26Moderate

Example 1 · Wave Optics · Coherent Sources and Resultant Intensity

Two waves of intensity ratio

1:9

cross each other at a point. The resultant intensities at the point, when (a) Waves are incoherent is

I_{1}

(b) Waves are coherent is

I_{2}

and differ in phase by

60^{\circ}

. If

\frac{I_{1}}{I_{2}}=\frac{10}{x}

then

x =

______ .

Incoherent means no cross term

For incoherent beams the intensities simply add. Putting in 2√(I₁I₂) cos φ for them double-counts interference that averages away.

Add amplitudes, not intensities

The cross term uses √(I₁I₂), the product of the amplitudes. Writing 2I₁I₂ cos φ is dimensionally wrong and always among the options.

Watch the sign of cos φ

Past 90° the cosine is negative. At 120° the cross term subtracts, so the coherent sum is smaller than the incoherent one.

Drill 5 more on resultant intensity of two beams

Concept 2 of 3: Brightest and darkest fringes

The brightest fringe is where the two amplitudes add straight on; the darkest is where they subtract. So everything comes from the amplitude ratio, the square root of the intensity ratio. Equal beams cancel completely and give perfectly dark fringes; unequal beams never quite cancel.

Definition

$I_{\max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$ , $I_{\min} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}$ .
With the amplitude ratio $r = A_{1}/A_{2} = \sqrt{I_{1}/I_{2}}$ : $\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{r + 1}{r - 1}\right)^{2}$ .
$I_{\max} - I_{\min} = 4\sqrt{I_{1}I_{2}}$ and $I_{\max} + I_{\min} = 2(I_{1} + I_{2})$ , so $\dfrac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}} = \dfrac{r^{2} + 1}{2r}$ .
Equal beams $I_{0}$ : $I_{\max} = 4I_{0}$ , $I_{\min} = 0$ .
Slit widths: by default the intensity through a slit is proportional to its width, so widths $w_{1} : w_{2}$ give $r = \sqrt{w_{1}/w_{2}}$ . If the question says the amplitude is proportional to the width, use the width ratio as r directly.

Fringe contrast

\frac{I_{\max}}{I_{\min}} = \left(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}\right)^{2} = \left(\frac{r + 1}{r - 1}\right)^{2}

Worked example

Two coherent beams have intensities in the ratio 16 : 1. Find

I_{\max} : I_{\min}

and

\dfrac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q81Moderate

Example 2 · Wave Optics · Coherent Sources and Resultant Intensity

Two monochromatic light beams have intensities in the ratio

1:9

. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is.

Take the square root first

An intensity ratio of 1 : 9 is an amplitude ratio of 1 : 3. Putting 1 and 9 into (r + 1)/(r − 1) squares the ratio twice.

Width ratio: intensity or amplitude?

By default intensity is proportional to slit width. If the stem says amplitude is proportional to width, the widths are the amplitude ratio. Read which one the question states before you start.

Only equal beams give true darkness

I_min = 0 needs equal amplitudes. With unequal beams the dark fringes still carry (√I₁ − √I₂)².

Drill 12 more on brightest and darkest fringes

Concept 3 of 3: Optical path and thin films

Inside glass the wavelength is shorter, so more waves fit into the same thickness. A thickness t of index μ holds as many waves as a thickness μt of vacuum: that is its optical path. In a thin film, two reflected beams travel different optical paths, and a reflection off a denser medium adds half a wave more. Counting those half-waves is most of the work.

Definition

Optical path of thickness t in index μ: $\mu t$ . Extra over the same thickness of vacuum (or air): $(\mu - 1)t$ .
Phase difference from it: $\Delta\phi = \dfrac{2\pi}{\lambda}(\mu - 1)t$ ; extra number of waves $\dfrac{(\mu - 1)t}{\lambda}$ .
Equal thicknesses of two media: path difference $(\mu_{2} - \mu_{1})t$ , because the wavelengths differ.
Reflection off a denser medium reverses the phase ( $\pi$ , half a wave); off a rarer medium it does not.
Thin film at normal incidence, path difference $2\mu t$ . One reversal: reflected dark at $2\mu t = n\lambda$ , bright at $2\mu t = (n - \tfrac{1}{2})\lambda$ . None or two: the other way round.
Transmitted light is brightest where reflected light is darkest. As a film thins, the pattern repeats every $\Delta t = \dfrac{\lambda}{2\mu}$ .

Optical path and films

\Delta = (\mu - 1)t, \qquad \Delta\phi = \frac{2\pi}{\lambda}(\mu - 1)t, \qquad 2\mu t = n\lambda \ \text{or}\ \left(n - \tfrac{1}{2}\right)\lambda

Worked example

A film of oil (μ = 1.25) floats on water (μ = 1.33). Light of wavelength 600 nm falls on it normally. Find the least thickness of oil that gives a reflected minimum.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q92Moderate

Example 3 · Wave Optics · Coherent Sources and Resultant Intensity

A transparent film of refractive index, 2.0 is coated on a glass slab of refractive index, 1.45. What is the minimum thickness of transparent film to be coated for the maximum transmission of Green light of wavelength 550 nm . [Assume that the light is incident nearly perpendicular to the glass surface.]

Count the reversals first

Mark each surface: into a denser medium reverses, into a rarer one does not. One reversal and two reversals give opposite conditions for the same thickness.

Transmission is the opposite of reflection

A question about maximum transmission is a question about minimum reflection. Solve for the reflected minimum.

The extra path is (μ − 1)t

Compared with air, the plate adds (μ − 1)t, not μt. Using μt forgets that the same thickness of air already had a path t.

Drill 3 more on optical path and thin films

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 (3)

Resultant intensity of two beams
Two coherent beams
$I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\phi, \qquad A^{2} = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos\phi$
Brightest and darkest fringes
Fringe contrast
$\frac{I_{\max}}{I_{\min}} = \left(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}\right)^{2} = \left(\frac{r + 1}{r - 1}\right)^{2}$
Optical path and thin films
Optical path and films
$\Delta = (\mu - 1)t, \qquad \Delta\phi = \frac{2\pi}{\lambda}(\mu - 1)t, \qquad 2\mu t = n\lambda \ \text{or}\ \left(n - \tfrac{1}{2}\right)\lambda$

Watch out for (9)

Incoherent means no cross term
→ Resultant intensity of two beams
Add amplitudes, not intensities
→ Resultant intensity of two beams
Watch the sign of cos φ
→ Resultant intensity of two beams
Take the square root first
→ Brightest and darkest fringes
Width ratio: intensity or amplitude?
→ Brightest and darkest fringes
Only equal beams give true darkness
→ Brightest and darkest fringes
Count the reversals first
→ Optical path and thin films
Transmission is the opposite of reflection
→ Optical path and thin films
The extra path is (μ − 1)t
→ Optical path and thin films

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