Light of wavelength 500 nm falls at right angles on a barrier with two long parallel slits…

Light of wavelength $500\ \text{nm}$ falls at right angles on a barrier with two long parallel slits. Each slit is $0.10\ \text{mm}$ wide, and the distance between their centres is $0.50\ \text{mm}$ . A screen stands $2.00\ \text{m}$ behind the barrier, as the figure shows, and the pattern on it has narrow bright interference fringes inside a broad, bright diffraction band. The first minimum on each side of the broad band marks its edge, called the central maximum. A bright fringe whose position falls exactly on a dark point of the single-slit pattern is missing from the screen. Use the small-angle results $\sin\theta \approx \tan\theta$ .

A barrier with two long parallel slits seen from the front, with light of wavelength 500 nm falling on it from the left. Each slit is 0.10 mm wide, and the distance between the centres of the two slits is 0.50 mm, both marked with dimension arrows. A screen stands 2.00 m behind the barrier, with a double headed arrow for that distance and a dashed line along the axis between the slits.

What is the width of the central maximum of the diffraction pattern on the screen?
What is the fringe width of the interference fringes?
How many bright interference fringes lie inside the central maximum, counting the central one?
The slits are now widened to $0.20\ \text{mm}$ each, with the same centre to centre distance. How many bright fringes lie inside the central maximum?
In the original arrangement, what is the order of the first bright fringe that is missing?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The slits are wide enough to diffract, so the single-slit pattern is an envelope. The double-slit fringes show up inside it.

(a) Central maximum. Its half-width is $\dfrac{\lambda D}{a}$ , so the full width is $\dfrac{2\lambda D}{a} = \dfrac{2 \times 500\ \text{nm} \times 2.00\ \text{m}}{0.10\ \text{mm}} = 20.0\ \text{mm}$ .

(b) Fringe width. $\beta = \dfrac{\lambda D}{d} = \dfrac{500\ \text{nm} \times 2.00\ \text{m}}{0.50\ \text{mm}} = 2.0\ \text{mm}$ .

(c) Fringes inside. The $m$ th bright fringe lies inside if $\dfrac{m\lambda}{d} < \dfrac{\lambda}{a}$ , that is $m < \dfrac{d}{a} = 5.0$ . The order $m = 5$ falls exactly on the first diffraction minimum, so it is missing. The orders $0, \pm 1, \pm 2, \pm 3, \pm 4$ remain, which is $9$ fringes.

(d) Wider slits. Now $\dfrac{d}{a} = \dfrac{0.50}{0.20} = 2.5$ , so $m < 2.5$ . The orders $0, \pm 1, \pm 2$ remain, which is $5$ fringes.

(e) Missing order. It is the order $m = \dfrac{d}{a} = 5$ .

Answer: (a) $20.0\ \text{mm}$ . (b) $2.0\ \text{mm}$ . (c) $9$ . (d) $5$ . (e) $5$ .

Check. Count by listing: for each order $m$ from $-60$ to $60$ , find $\sin\theta = \dfrac{m\lambda}{d}$ and keep it if it is smaller than $\dfrac{\lambda}{a}$ . This keeps $9$ orders for the first set of slits and $5$ for the wider ones. The ratio $\dfrac{20.0\ \text{mm}}{2.0\ \text{mm}} = 10$ is twice $\dfrac{d}{a}$ .

Watch out. Dividing the central width by the fringe width gives $10$ , but the two edge positions hold missing fringes, so the count is $9$ . In general the number inside is $2\dfrac{d}{a} - 1$ only when $\dfrac{d}{a}$ is a whole number.

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