A full lesson: a cartoon movie to start, then why every wave bends round edges, why the dark bands fall at a sin θ = nλ, the width of the central maximum on a screen, how diffraction differs from interference, the Fresnel distance and the limit of resolution, NEET shortcuts, problems to solve yourself, and 12 MCQs.
Sound bends round a door because its wavelength is about as big as the door. Light bends too, but only through a gap a few wavelengths wide, like the gap between two fingers. Narrow slit, wide spread: the dark bands are at a sin θ = nλ.
Dadi calls from the kitchen: "Kabir, chai is ready!" Kabir is in his room, behind the wall. He cannot see Dadi, yet he hears her perfectly. Sound has bent round the edge of the doorway. The light from the kitchen bulb did not: it went straight through the door and lit only a patch on the opposite wall.
On the terrace that night, Meera holds two fingers very close together and looks at a distant street lamp through the tiny gap. The lamp is smeared sideways into a streak with dark lines across it. Squeeze the gap narrower and the streak becomes wider. That is light bending round edges too. The bending of waves round obstacles and through openings is called diffraction.
Try at home: look at a tube light or a far street lamp through the gap between two fingers (or two razor blades held edge to edge). Keep the gap parallel to the tube. Slowly close the gap and watch the bands spread.
"Every wave diffracts. Sound shows it all the time because its wavelength is about a metre. Light shows it only when the gap is just a few micrometres, because its wavelength is half a micrometre."
"Why can we hear round a corner but not see round it?"
The spreading angle is about λ/a. For sound through a door, λ ≈ a, so the wave spreads everywhere. For light through a door, λ/a ≈ 5 × 10⁻⁷, so it goes practically straight and casts a sharp shadow.
A plane wave meets a wall with a gap of width a. By Huygens' principle, every point of the gap is a source of wavelets. The picture is computed live by adding those wavelets. On the right is the pattern on a far screen. Change the gap and watch both.
Shadows of sharp edges show faint bands; a CD shows colours. Diffraction is the proof that light is a wave.
Noticeable only when a is a few λ or less. For light that means micrometre slits, thin hairs or fine lines.
Every lens is an opening, so even a perfect lens turns a star into a small blurred spot. More in Stage 6.
"Diffraction and interference are not two different physics. Diffraction is interference of the many wavelets coming from one opening."
Split the slit into many thin strips; all start in phase. Towards angle θ, the wavelet from the bottom edge travels an extra a sin θ compared with the top edge. Drag the yellow knob to tilt the direction. On the right, each strip's wave is a small arrow turned by its phase; the arrows are added head to tail (a phasor chain). The red arrow is the resultant.
Every wavelet arrives in step. The arrows lie in a straight line: biggest resultant.
Each strip in the top half has a partner a/2 below it in the bottom half. Partners differ by λ/2 and cancel. Every pair cancels: dark. The chain closes into a circle.
For a sin θ = 2λ, split into four quarters; quarters cancel in pairs. Note: n = 0 is NOT a minimum, it is the centre.
Split into three parts at 3λ/2: two cancel, one third survives, so it is much weaker. Exact value 1.43λ with only 4.7% of the central intensity.
"Students mix up the two formulas. In Young's double slit, d sin θ = nλ gives BRIGHT fringes. In a single slit, a sin θ = nλ gives DARK fringes. Same look, opposite meaning."
"What happens if the slit is narrower than one wavelength, a < λ?"
sin θ = λ/a would be more than 1, which is impossible. So there is no minimum at all: the central maximum fills the whole half-space in front of the slit.
Now put a screen at distance D. Angles are small, so sin θ ≈ tan θ ≈ θ = y/D. Move the sliders, drag the screen, or put the whole set-up under water.
Small-angle form, since a is many wavelengths in a real experiment.
The minima are equally spaced, λD/a apart, except across the centre.
Twice as wide as each secondary maximum (λD/a). Wider for longer λ, larger D and narrower a.
Under water (n = 4/3) the whole pattern shrinks to 3/4.
"Read the question carefully: 'width of the central maximum' or 'distance between the first minima on both sides' means 2λD/a. 'Position of the first minimum' or 'half width' means λD/a."
Each slit of Young's experiment is itself a single slit, so the real double-slit pattern is the double-slit fringes inside a single-slit envelope. Switch the view and change the separation d (in units of the slit width a).
| Feature | Interference (double slit) | Diffraction (single slit) |
|---|---|---|
| Waves added | two coherent sources | a continuous set of wavelets from one slit |
| Bright at | d sin θ = nλ | θ = 0, and ≈ (n + ½)λ/a |
| Dark at | d sin θ = (n + ½)λ | a sin θ = nλ, n ≠ 0 |
| Fringe width | all equal, λD/d | central 2λD/a, others λD/a |
| Intensity | all bright fringes equal | falls fast away from the centre |
"NCERT's way to remember it: interference is the superposition of a few waves, diffraction is the superposition of a very large number of wavelets from one wavefront."
Syllabus note: Fresnel distance and resolving power were removed from the rationalised NCERT and are beyond the current NEET syllabus. Treat this stage, Problem 4 and MCQs 10 and 12 as enrichment.
A beam through a hole of width a spreads by about λ/a. After a distance z the spread is zλ/a. When that equals the hole size a, diffraction takes over: this is the Fresnel distance z_F = a²/λ. Lenses are holes too, so two stars very close together blur into one: the limit of resolution.
a = 3 mm, λ = 500 nm: 18 m. Closer than z_F, rays work; beyond it, the beam spreads by diffraction.
D is the diameter of the lens. Two points are just resolved when one spot's centre falls on the other's first dark ring (Rayleigh).
Big telescope mirrors and blue light give sharper images. Your eye (pupil 2 mm): about 1 arc-minute.
Single slit gives DARK at nλ. Double slit gives BRIGHT at nλ. Never swap them.
Half-width (one side) is λD/a. Secondary maxima are λD/a wide: half the central one.
Double a → half width. Blue instead of red → narrower. Under water → ÷ n.
Number of double-slit fringe widths inside the central diffraction maximum.
Square the aperture (in metres!) and divide by λ.
Smaller Δθ is better. Increase D or decrease λ. Distance resolved = separation ÷ Δθ.
"Nine out of ten NEET questions on this lesson are just 2λD/a with a twist: one side or both sides, air or water, red or blue. Write the formula first, then put the numbers."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.
A slit 0.1 mm wide is lit by light of wavelength 600 nm. The screen is 2 m away. Find the position of the first and second minima and the width of the central maximum.
yₙ = nλD/a; central width = 2λD/a.
Red light of wavelength 650 nm gives the first minimum at 30°. Find the slit width. Is a second minimum seen?
a sin θ = nλ, and sin θ can never exceed 1.
In a double-slit experiment with λ = 500 nm and slit separation 1 mm, what should be the width of each slit so that 10 maxima of the double-slit pattern lie within the central maximum of the single-slit pattern?
Central maximum spans 2λ/a; one fringe is λ/d.
For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?
z_F = a²/λ, all in metres.
Take the pupil of the eye as 2 mm wide and λ = 500 nm. Find the limit of resolution. Up to what distance can the two headlights of a car, 1.2 m apart, be seen as separate?
Δθ = 1.22 λ/D, then distance = separation / Δθ.
Write the formula first, then click. Watch for "one side" versus "both sides".
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "first minimum", "first dark fringe" | a sin θ = λ, y₁ = λD/a | Using a sin θ = λ/2 (that is a double-slit idea) |
| "width of central maximum" | 2λD/a (angular 2λ/a) | Giving only λD/a, the half width |
| "distance between first minima on either side" | 2λD/a | Same trap: one side is half of it |
| "secondary maximum" | a sin θ ≈ (n + ½)λ | Putting the bright fringes at nλ as in Young's |
| "apparatus immersed in liquid" | λ → λ/n, width ÷ n | Forgetting that the pattern shrinks |
| "slit width is doubled" | width ∝ 1/a → halves | Thinking a wider slit gives a wider pattern |
| "fringes within central maximum" | 2d/a | Writing d/a |
| "ray optics valid up to" | z_F = a²/λ | Writing λ/a² or forgetting mm → m |
| "limit of resolution", "resolving power" | Δθ = 1.22λ/D, RP = 1/Δθ | Saying a larger λ resolves better |
Lesson 6 asks a different question: in which direction does the light wave vibrate? Polaroids, Malus' law and Brewster's angle show that light is a transverse wave.
Values used: visible λ ≈ 400–700 nm, n(water) = 4/3, pupil of the eye ≈ 2 mm. Secondary maxima: NCERT gives (n + ½)λ/a; the exact first one is at 1.43λ/a.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.