CLASS 12 · CHAPTER 10 · OPTICS

Wave Optics

Slide the two slits closer together in this simulation. The fringe pattern on the screen spreads wider, not narrower, the exact opposite of what 'more room' intuition predicts.

01 · See it

Watch it happen

Slit separation d: 1.00 mm

Drag the slider toward smaller d and watch the two wave patterns fan out wider, not narrower, as the sources move apart in angle, that's the whole mechanism behind β = λD/d.

1.00 mm

Slit separation d

1.0 m

Screen distance D

600 nm

Wavelength λ

0.60 mm

Fringe width β

Drag the slit-separation slider down toward its smallest value. The bright-dark bands on the screen visibly widen out, even though the two slits themselves are moving closer together. Push the slider the other way, toward larger separation, and the bands pack in tighter until they’d be impossible to tell apart. Fringe spacing and slit spacing move in opposite directions.

02 · Derive it

Where the formula comes from

Two slits $S_1$ and $S_2$ , separated by a distance $d$ , are illuminated by the same coherent wavefront. A screen sits a distance $D$ away (with $D \gg d$ ). At a point on the screen a height $y$ above the centre, the path difference between the two waves arriving there is:

Path difference

\Delta = d\sin\theta

For the small angles relevant here ( $D \gg d, y$ ), $\sin\theta \approx \tan\theta = y/D$ , so:

Path difference (small angle)

\Delta \approx \dfrac{dy}{D}

Constructive interference (a bright fringe) occurs wherever this path difference is a whole number of wavelengths:

Bright fringe condition

\Delta = n\lambda, \quad n = 0, \pm1, \pm2, \ldots

Destructive interference (a dark fringe) occurs at odd half-wavelengths:

Dark fringe condition

\Delta = \left(n+\tfrac12\right)\lambda

Combining the bright-fringe condition with the small-angle path difference gives the position of the $n$ -th bright fringe:

Bright fringe position

y_n = \dfrac{n\lambda D}{d}

The spacing between consecutive bright fringes ( $y_{n+1} - y_n$ ) is the fringe width:

Fringe width

\beta = \dfrac{\lambda D}{d}

03 · Break it

Where the shortcut stops working

The whole chapter turns on a single ratio, $\beta = \lambda D/d$ , and on getting its direction right: $\beta$ is inversely proportional to $d$ , not directly proportional to it. The intuitive guess, “slits further apart means more room for the pattern to spread into,” describes the slits, not the fringes, and it points the wrong way.

Here’s why the real relationship runs backwards from that guess: increasing $d$ increases the path difference $\Delta \approx dy/D$ at any given point $y$ on the screen. Since successive bright fringes are separated by exactly one extra wavelength of path difference ( $\Delta\Delta = \lambda$ ), a bigger $d$ means you don’t have to move far in $y$ to rack up that one extra wavelength, so consecutive bright fringes end up closer together, not further apart.

Taken to its extreme, this is exactly why two separate ordinary light bulbs never show visible interference fringes: their effective $d$ is enormous (centimetres to metres) compared to $\lambda$ (hundreds of nanometres), so $\beta = \lambda D/d$ works out to a spacing far too small for any eye or detector to resolve, the fringes are there in principle, just packed far too tightly to see. (They also aren’t coherent with each other, which independently washes out any pattern, but even coherent bulbs at that separation would be hopeless.) Young’s experiment only works because $d$ is deliberately kept tiny, close to the scale of $\lambda$ itself, which is precisely what makes $\beta$ large enough to observe.

04 · Master it

Apply it under exam conditions

Q1. In a double-slit setup, slits 0.2 mm apart are illuminated by light of wavelength 500 nm, and the screen is 1.5 m away. Find the fringe width, in mm.

\beta = \dfrac{\lambda D}{d} = \dfrac{(500\times10^{-9})(1.5)}{0.2\times10^{-3}} = \dfrac{7.5\times10^{-7}}{2\times10^{-4}} = \textcolor{#e08a1e}{3.75\times10^{-3}\text{ m} = 3.75\text{ mm}}

Q2. For a fixed λ and D, the slit separation d is tripled. By what factor does the fringe width β change?

Since $\beta = \lambda D/d$ with $\lambda$ and $D$ held fixed, $\beta \propto 1/d$ . Tripling $d$ means:

\dfrac{\beta_{new}}{\beta_{old}} = \dfrac{d_{old}}{d_{new}} = \dfrac{d_{old}}{3d_{old}} = \textcolor{#e08a1e}{\dfrac{1}{3}}

The fringe width shrinks to one-third of its original value, the fringes pack three times closer together, confirming the inverse relationship the slider demonstrates above: more separation between the slits means a tighter, not wider, pattern.

05 · FAQs

Quick answers

Wouldn't moving the slits further apart spread the interference pattern out more, since there's more room between them?+

No, and this is the exact trap the chapter is built around. Fringe width β = λD/d is inversely proportional to the slit separation d, not proportional to it. Moving the slits apart (increasing d) packs the fringes closer together on the screen; moving them closer together (decreasing d) is what spreads the pattern out. 'More room between the slits' describes d, but what you see on the screen is governed by λD/d, and that ratio shrinks as d grows.

Why don't I see an interference pattern from two separate light bulbs in a room?+

Two ordinary bulbs fail on two counts: they aren't coherent sources (their light isn't in a fixed phase relationship, so any pattern washes out in a fraction of a second), and their separation d is enormous compared to a wavelength, so even if they were coherent, β = λD/d would be immeasurably small, far tighter than the eye or any detector could resolve. Young's experiment works because both slits are lit by the same wavefront (coherent) and are separated by a d small enough, comparable to a sub-millimetre scale, that β comes out large enough to actually see.

Does the central fringe sit exactly opposite the midpoint of the two slits?+

Yes. The central maximum is where the path difference from both slits is zero (Δ = 0), which by symmetry is the point directly opposite the midpoint of the two slits. All bright fringes are numbered relative to this central one, n = 0 at the centre, n = ±1, ±2, ... moving outward.

What happens to the fringe width if I switch to a shorter wavelength, say blue light instead of red?+

β = λD/d falls, since β is directly proportional to λ. Blue light (shorter λ) gives narrower, more tightly packed fringes than red light, for the same d and D. This is also why violet appears closest to the central maximum and red furthest out in a diffraction grating spectrum, longer wavelengths spread further.

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