Wave Optics — JEE Mains Physics Formulas and Traps

Wavefronts and Light in a Medium

Speed, wavelength and frequency in a medium

Light in a medium

f = f_{0}, \qquad v = \frac{c}{\mu}, \qquad \lambda = \frac{\lambda_{0}}{\mu}, \qquad \frac{\lambda_{1}}{\lambda_{2}} = \frac{v_{1}}{v_{2}} = \frac{\mu_{2}}{\mu_{1}}

Wavefronts, rays and Huygens' principle

Source or situation	Shape of the wavefront	Rays
Point source nearby	Spherical	Spread out from the source
Line source, such as a lit slit or a tube light	Cylindrical	Spread out at right angles to the line
Very distant source, such as the Sun or a star	Plane	Parallel
Point source at the focus of a convex lens, after the lens	Plane	Parallel The lens turns a spherical wave into a plane one; this is how a parallel beam is made.
Plane wave after a convex lens	Spherical, shrinking onto the focus	Converge to the focus
Plane wave after a prism	Plane, turned through the deviation	Parallel, bent towards the base
Plane wave after a pinhole much smaller than the beam	Nearly spherical	Spread out from the hole A wider slit lets through a flatter, less curved wave.

The shape of the wavefront tells you the shape of the ray bundle: spherical spreads, plane stays parallel.

Common traps

The wave travels along the normal

The direction of travel is the normal to the wavefront, given by the coefficients of x, y and z, not a line lying in the wavefront.

A prism does not curve a plane wave

A prism only turns a plane wavefront. A lens curves it, and a tiny hole makes it nearly spherical by diffraction.

Photons, not waves

Interference, diffraction and polarisation are wave effects. The photoelectric and Compton effects are particle effects, and the wave theory fails on them.

Frequency never changes at a boundary

Only speed and wavelength change. An option that changes the frequency, or the colour, is wrong however neat its numbers.

The μ ratio is upside down

Wavelength and speed fall as μ rises, so λ₁/λ₂ = μ₂/μ₁. Writing μ₁/μ₂ gives the answer for going the other way.

Go through the vacuum value

Given a wavelength in one medium, multiply by its μ to get λ₀, then divide by the new μ. Skipping the step is where the ratio gets inverted.

Coherent Sources and Resultant Intensity

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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

Common traps

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.

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.

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.

Double Slit Fringe Width and Fringe Positions

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Fringe width and what it depends on

Fringe width

\beta = \frac{\lambda D}{d}, \qquad \theta = \frac{\lambda}{d}, \qquad \Delta\beta = \frac{\lambda\,\Delta D}{d}

The apparatus in a liquid

Fringes in a liquid

\beta' = \frac{\beta}{\mu} = \frac{\lambda D}{\mu d}, \qquad \theta' = \frac{\lambda}{\mu d}

Fringe positions and two wavelengths

Positions and coincidence

y_{\text{bright}} = \frac{n\lambda D}{d}, \qquad y_{\text{dark}} = \left(n - \tfrac{1}{2}\right)\frac{\lambda D}{d}, \qquad n_{1}\lambda_{1} = n_{2}\lambda_{2}

Common traps

Angular width ignores the screen

Moving the screen changes β but not λ/d. A statement that the angular separation grows as the screen moves away is false.

β falls as d rises

Doubling the slit gap halves the fringe width. Write the full ratio λ₂D₂d₁/(λ₁D₁d₂) and the d's cannot end up the wrong way up.

A percentage change is not a ratio

If β becomes 0.8 of its old value, the change is −20%, not 80%. Say which one the question asks for.

Divide by μ, do not multiply

The wavelength shrinks in a liquid, so the fringes get narrower. An option larger than the air value is the multiply-by-μ slip.

The given wavelength is the one in air

When a question gives λ in air and asks for β in the liquid, divide λ by μ before using λD/d.

Angles shrink too

The angular width λ/d also carries λ, so it falls by μ as well. It is only the screen distance that it ignores.

The ratio flips

n₁/n₂ = λ₂/λ₁: the longer wavelength needs the smaller order. Putting λ₁ on top gives a coincidence that does not exist.

Dark fringes are half a width short

The 3rd dark fringe is at 2.5β, not 3β and not 1.5β. Count the bright fringes, then step back half a width.

Reduce the ratio fully

The least distance needs the lowest-terms pair. 630 : 420 is 3 : 2; using 63 : 42 gives a coincidence 21 times too far out.

Double Slit Intensity and Slab Shifts

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Path difference from the geometry

Path difference

\Delta x = \frac{yd}{D}, \qquad \Delta x_{\text{opposite a slit}} = \frac{d^{2}}{2D}, \qquad \sqrt{D^{2} + d^{2}} - D \approx \frac{d^{2}}{2D}

Intensity from path and phase

Double-slit intensity

I = 4I_{0}\cos^{2}\frac{\phi}{2} = I_{\max}\cos^{2}\frac{\pi\,\Delta x}{\lambda}, \qquad \phi = \frac{2\pi}{\lambda}\Delta x

A thin sheet over one slit

Sheet shift

\Delta y = \frac{(\mu - 1)tD}{d} = \frac{(\mu - 1)t}{\lambda}\,\beta

Common traps

Opposite a slit is y = d/2

The centre of the screen is midway between the slits, so a slit is d/2 from it, not d. Using y = d doubles the path difference.

yd/D needs a far screen

When the screen is only a few slit gaps away, or a figure sets the point beside one slit, write each path with Pythagoras and subtract.

Know which I₀ the question means

Some stems call the maximum intensity I₀, others call the intensity of one slit I₀, which makes the maximum 4I₀. Read the definition before you answer.

Half the phase inside the cosine

I = I_max cos²(φ/2), not cos²φ. With the full phase, λ/4 would give zero instead of half.

I₀ for one slit means I_max = 4I₀

If I₀ is the intensity of one slit alone, the formula is 4I₀cos²(φ/2). Writing I₀cos²(φ/2) gives answers four times too small.

Path is not phase

A path of λ/3 is a phase of 2π/3. Put the path straight into the cosine and the answer is wrong by a factor of 2π/λ.

Towards the covered slit

The sheet lengthens that path, so the equal-path point moves to the side where the geometric path is shorter: the covered side.

Use μ − 1, not μ

The sheet replaces the same thickness of air, so it adds (μ − 1)t. Using μt gives a shift about three times too large for glass.

The fringes do not narrow

The sheet only moves the pattern. β = λD/d is unchanged, so an option that changes the fringe width is wrong.

Single-Slit Diffraction and Resolving Power

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Dark fringes of a single slit

Single-slit minima

a\sin\theta = n\lambda, \qquad y_{n} = \frac{n\lambda D}{a}, \qquad y_{3} - y_{1} = \frac{2\lambda D}{a}

Width of the central maximum

Central maximum

\theta_{\text{width}} = \frac{2\lambda}{a}, \qquad W = \frac{2\lambda D}{a}\ \left(\text{or } \frac{2\lambda f}{a}\right), \qquad N = \frac{2d}{a}

Resolving power of a telescope and a microscope

Resolution

\Delta\theta = \frac{1.22\lambda}{D}, \qquad \text{RP}_{\text{microscope}} = \frac{2\mu\sin\theta}{1.22\lambda}

Common traps

nλ is dark for a single slit

In the double slit, a path of nλ is bright. For one slit, a sin θ = nλ is a minimum. Mixing the two puts every fringe in the wrong place.

First to third is two steps

The first and third minima are 2λD/a apart, not 3λD/a. Subtract positions, do not read off the higher order.

Angle of one minimum or the angle between two?

"Angular divergence" can mean θ or 2θ. Check the options: they often contain both.

Twice λD/a, not λD/a

The central maximum spans from the first minimum on one side to the first on the other: 2λD/a. The single-step λD/a is the width of a secondary maximum.

a, not d

The single-slit width a sets the envelope; the gap d sets the double-slit fringes. Using λD/d for a central maximum mixes the two patterns.

Large angles need the sine

When λ/a is not small, as with microwaves on a slit a few wavelengths wide, find θ from sin θ = λ/a and double it.

Limit and power are opposites

The limit of resolution is an angle, 1.22λ/D; smaller is better. The resolving power is its reciprocal; larger is better.

Keep the 1.22

A round aperture gives 1.22λ/D, not λ/D. Dropping the factor lands on a distractor about 20% off.

Oil helps a microscope, not a telescope

The microscope's resolving power has μ in it, so oil raises it. A telescope's limit depends only on λ and the objective diameter.

Polarisation by Polaroids and by Reflection

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One polaroid after another: Malus' law

Malus' law

I_{1} = \frac{I_{0}}{2}, \qquad I_{2} = I_{1}\cos^{2}\theta

Chains of polaroids

A sheet between crossed polaroids

I = \frac{I_{0}}{2}\cos^{2}\theta\,\sin^{2}\theta = \frac{I_{0}}{8}\sin^{2}2\theta, \qquad I_{n} = \frac{I_{0}}{2^{n}}\ (45^{\circ}\ \text{steps})

Polarisation by reflection: Brewster's law

Brewster's law

\tan i_{B} = \frac{\mu_{2}}{\mu_{1}}, \qquad i_{B} + r = 90^{\circ}

Common traps

The first sheet halves, it does not use cos²

Unpolarised light has no single plane, so the first polaroid always passes half. Malus' law starts only from the second sheet.

Polarised light is not halved

If the light is already polarised, go straight to I cos²θ. Halving it first gives an answer half the right size.

Cos squared, not cos

Intensity goes as the square of the amplitude. At 60° the output is a quarter, not a half.

Angle to the previous sheet

Each cos² uses the angle from the sheet just before, not from the first sheet. A sheet at 37° to the first is 53° from the last of a crossed pair.

Count the halving once

Only the first sheet halves unpolarised light. With n sheets at 45°, the factor is ½ for the first and ½ for each of the other n − 1: I₀/2ⁿ in all.

Two angles can give the same output

Between crossed sheets, θ and 90° − θ pass the same intensity. If the question wants one angle, check which one it means.

Tangent, not sine

Brewster's law is tan i_B = μ. Snell's sine law gives the angle of refraction, but the Brewster angle itself comes from the tangent.

Refraction is the complement

At Brewster's angle r = 90° − i_B. There is no need to use Snell's law; the two angles add to a right angle.

Reverse direction, reverse ratio

From glass to air the ratio is 1/μ, giving 90° − i_B. Using tan⁻¹ μ again for the glass-to-air side is the usual slip.

Wave Optics formulas

Wavefronts and Light in a Medium

Speed, wavelength and frequency in a medium

Wavefronts, rays and Huygens' principle

Common traps

Coherent Sources and Resultant Intensity

Resultant intensity of two beams

Brightest and darkest fringes

Optical path and thin films

Common traps

Double Slit Fringe Width and Fringe Positions

Fringe width and what it depends on

The apparatus in a liquid

Fringe positions and two wavelengths

Common traps

Double Slit Intensity and Slab Shifts

Path difference from the geometry

Intensity from path and phase

A thin sheet over one slit

Common traps

Single-Slit Diffraction and Resolving Power

Dark fringes of a single slit

Width of the central maximum

Resolving power of a telescope and a microscope

Common traps

Polarisation by Polaroids and by Reflection

One polaroid after another: Malus' law

Chains of polaroids

Polarisation by reflection: Brewster's law

Common traps

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