Width of the central maximum (qualitative)
The central maximum stretches between the first minima on either side. It widens when the slit is narrower or the wavelength is longer.
Where the central maximum ends
The central maximum lies between the first minimum on each side, at θ = +λ/a and θ = −λ/a. So its angular width is 2λ/a. On a screen at distance D, its linear width is 2λD/a. We use these results without derivation. Each secondary maximum is only about half as wide. The key idea is simple: the narrower the slit, the more the light spreads.
How the central maximum changes
| Change | Effect on the central maximum | Reason |
|---|---|---|
| Slit made narrower (smaller a) | Wider and fainter | Width 2λD/a ∝ 1/a, and less light passes |
| Slit made much wider than λ | Very narrow; light goes almost straight | Spreading becomes negligible, so ray optics holds |
| Red light in place of blue light | Wider | Red light has the longer wavelength |
| Screen moved farther away (larger D) | Wider on the screen; angular width unchanged | Linear width 2λD/a ∝ D |
| Whole set-up dipped in water | Narrower | The wavelength in water is λ/μ |
| White light used | White centre with coloured edges | Each colour has its own width; red spreads the most |
Interference vs diffraction
In Young's experiment each slit also diffracts, so the interference fringes are seen inside the broad single slit pattern.
Worked example: width of the central maximum
Light of wavelength 600 nm falls on a slit of width 0.2 mm. The screen is 1 m away. Find (a) the angular width and (b) the linear width of the central maximum. (c) What is the linear width if the slit width is doubled?
- Given: λ = 6 × 10⁻⁷ m, a = 2 × 10⁻⁴ m, D = 1 m.
- (a) Angular width = 2λ/a = 2 × (6 × 10⁻⁷ m) ÷ (2 × 10⁻⁴ m) = 6 × 10⁻³ rad.
- (b) Linear width = 2λD/a = 6 × 10⁻³ × 1 m = 6 × 10⁻³ m = 6 mm.
- (c) Width ∝ 1/a, so doubling a halves the width: 6 mm ÷ 2 = 3 mm.
(a) 6 × 10⁻³ rad; (b) 6 mm; (c) 3 mm
In a pinhole camera made from a shoebox in school, making the hole too small blurs the image, because light spreads out by diffraction. The same spreading limits how much detail cameras and telescopes can show, which is why their lenses are made wide.
2 marks: “How does the width of the central maximum change when (a) the slit width is halved, (b) blue light is replaced by red light?” (a) It doubles, because width ∝ 1/a. (b) It increases, because width ∝ λ and red light has a longer wavelength.
2 marks: “Give two differences between the interference pattern of Young's experiment and the diffraction pattern of a single slit.” Interference: equal fringe widths, bright fringes about equally bright, two coherent sources. Diffraction: central maximum twice as wide, brightness falls quickly, wavelets from one wavefront.
3 marks: “Light of wavelength 500 nm falls on a slit of width 0.25 mm, and the screen is 1.5 m away. Find the width of the central maximum.” Width = 2λD/a = 2 × (5 × 10⁻⁷) × 1.5 ÷ (2.5 × 10⁻⁴) m = 6 × 10⁻³ m = 6 mm.
Narrow slit, wide centre. Width = 2λD/a, with the slit width a downstairs.