Topics

Two-slit calculations

Covers AS91523

The fringe relationship

For two slits a distance $d$ apart, a screen a distance $L$ away, and fringes separated by $x$ :

$n\lambda = \frac{dx}{L}$

$n$ — order of the fringe ( $n = 1$ for the first bright fringe from the centre)
$\lambda$ — wavelength (m)
$d$ — slit separation, the distance between the two slits (m)
$x$ — distance from the central bright fringe to the $n$ th bright fringe (m)
$L$ — distance from the slits to the screen (m)
This relationship is on the resource sheet. It assumes the screen is far from the slits compared with their separation, which is always the case in practice ( $L$ in metres, $d$ in fractions of a millimetre).

Two ways the symbol $x$ is used

This is the single biggest source of error on this page, so read the question carefully:

$x$ as the distance to the $n$ th fringe — then use $n$ as that fringe's order.
$x$ as the fringe spacing (the gap between adjacent bright fringes) — then $n = 1$ , since adjacent fringes differ by one order.
The safest habit: write down which one the question gave you before substituting. If the question says "the fringes are $4.2$ mm apart", that is the spacing, so $n = 1$ . If it says "the third bright fringe is $12.6$ mm from the centre", then $x = 12.6$ mm and $n = 3$ .
Both give the same answer here ( $12.6/3 = 4.2$ ), which is a useful check.

What changes the fringe spacing

Rearranging for the spacing between adjacent fringes ( $n = 1$ ):

$x = \frac{\lambda L}{d}$

Change	Effect on fringe spacing
Longer wavelength (e.g. red instead of blue)	fringes further apart
Larger slit separation $d$	fringes closer together
Greater distance to the screen $L$	fringes further apart
Brighter source	no change in spacing (fringes just brighter)

Note the inverse relationship with $d$ — the counter-intuitive one. Moving the slits closer together spreads the pattern out.
Red light ( $\lambda \approx 700$ nm) produces fringes about $1.4$ times further apart than blue ( $\lambda \approx 450$ nm) in the same apparatus.

Getting the units right

$\lambda$ for visible light is in nanometres: $1$ nm $= 10^{-9}$ m.
$d$ is usually in millimetres or given as "slits per millimetre": $1$ mm $= 10^{-3}$ m.
$x$ is usually in millimetres.
$L$ is in metres.
Convert everything to metres before substituting. Mixing millimetres and metres in the same equation is the most common numerical error in this topic.

Worked ExampleFinding a wavelength from a fringe pattern

In a two-slit experiment, the slits are $0.25$ mm apart and the screen is $1.8$ m away. The bright fringes are measured to be $4.3$ mm apart. Find the wavelength of the light.

Step 1 — Convert everything to metres

$d = 0.25 \text{ mm} = 2.5 \times 10^{-4} \text{ m}$ $x = 4.3 \text{ mm} = 4.3 \times 10^{-3} \text{ m}$ $L = 1.8 \text{ m}$

Step 2 — Identify $n$

The question gives the spacing between adjacent fringes, so $n = 1$ .

Step 3 — Substitute

$n\lambda = \frac{dx}{L}$ $(1)\lambda = \frac{(2.5 \times 10^{-4})(4.3 \times 10^{-3})}{1.8}$ $\lambda = \frac{1.075 \times 10^{-6}}{1.8} = 5.97 \times 10^{-7} \text{ m}$

Step 4 — Express sensibly

$\lambda = 597 \approx 6.0 \times 10^{2} \text{ nm}$

$\lambda = 597 \text{ nm (orange-yellow light)}$

Worked ExampleUsing a higher-order fringe

Light of wavelength $633$ nm from a helium–neon laser passes through two slits $0.40$ mm apart. The fourth bright fringe from the centre is found $9.5$ mm from the central maximum. Find the distance from the slits to the screen.

Step 1 — Convert and identify

$\lambda = 6.33 \times 10^{-7} \text{ m}, \qquad d = 4.0 \times 10^{-4} \text{ m}$ $x = 9.5 \times 10^{-3} \text{ m}, \qquad n = 4$

Here $x$ is the distance from the centre to the fourth fringe, so $n = 4$ — not 1.

Step 2 — Rearrange for $L$

$n\lambda = \frac{dx}{L} \;\Rightarrow\; L = \frac{dx}{n\lambda}$

Step 3 — Substitute

$L = \frac{(4.0 \times 10^{-4})(9.5 \times 10^{-3})}{(4)(6.33 \times 10^{-7})}$ $L = \frac{3.80 \times 10^{-6}}{2.532 \times 10^{-6}} = 1.50 \text{ m}$

$L = 1.5 \text{ m}$

Worked ExamplePredicting the effect of a change

A two-slit pattern is produced with blue light of wavelength $450$ nm, giving fringes $3.0$ mm apart. Predict the new fringe spacing if (a) the light is changed to red at $700$ nm, and (b) the slit separation is doubled instead.

Step 1 — Write the relationship for the spacing

With $n = 1$ :

$x = \frac{\lambda L}{d}$

$L$ and $d$ are unchanged in part (a), so $x \propto \lambda$ .

(a) Changing to red light

$\frac{x_{\text{red}}}{x_{\text{blue}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}} = \frac{700}{450} = 1.556$ $x_{\text{red}} = (3.0)(1.556) = 4.7 \text{ mm}$

The fringes move further apart, because a longer wavelength needs a larger path difference — and therefore a larger angle — to reach each order.

(b) Doubling the slit separation

Now $\lambda$ and $L$ are unchanged, so $x \propto \dfrac{1}{d}$ :

$x_{\text{new}} = \frac{3.0}{2} = 1.5 \text{ mm}$

The fringes move closer together.

$\text{(a) } 4.7 \text{ mm}, \qquad \text{(b) } 1.5 \text{ mm}$

Two-slit calculations

Covers AS91523

The fringe relationship

For two slits a distance $d$ apart, a screen a distance $L$ away, and fringes separated by $x$ :

$n\lambda = \frac{dx}{L}$

$n$ — order of the fringe ( $n = 1$ for the first bright fringe from the centre)
$\lambda$ — wavelength (m)
$d$ — slit separation, the distance between the two slits (m)
$x$ — distance from the central bright fringe to the $n$ th bright fringe (m)
$L$ — distance from the slits to the screen (m)
This relationship is on the resource sheet. It assumes the screen is far from the slits compared with their separation, which is always the case in practice ( $L$ in metres, $d$ in fractions of a millimetre).

Two ways the symbol $x$ is used

This is the single biggest source of error on this page, so read the question carefully:

$x$ as the distance to the $n$ th fringe — then use $n$ as that fringe's order.
$x$ as the fringe spacing (the gap between adjacent bright fringes) — then $n = 1$ , since adjacent fringes differ by one order.
The safest habit: write down which one the question gave you before substituting. If the question says "the fringes are $4.2$ mm apart", that is the spacing, so $n = 1$ . If it says "the third bright fringe is $12.6$ mm from the centre", then $x = 12.6$ mm and $n = 3$ .
Both give the same answer here ( $12.6/3 = 4.2$ ), which is a useful check.

What changes the fringe spacing

Rearranging for the spacing between adjacent fringes ( $n = 1$ ):

$x = \frac{\lambda L}{d}$

Change	Effect on fringe spacing
Longer wavelength (e.g. red instead of blue)	fringes further apart
Larger slit separation $d$	fringes closer together
Greater distance to the screen $L$	fringes further apart
Brighter source	no change in spacing (fringes just brighter)

Note the inverse relationship with $d$ — the counter-intuitive one. Moving the slits closer together spreads the pattern out.
Red light ( $\lambda \approx 700$ nm) produces fringes about $1.4$ times further apart than blue ( $\lambda \approx 450$ nm) in the same apparatus.

Getting the units right

$\lambda$ for visible light is in nanometres: $1$ nm $= 10^{-9}$ m.
$d$ is usually in millimetres or given as "slits per millimetre": $1$ mm $= 10^{-3}$ m.
$x$ is usually in millimetres.
$L$ is in metres.
Convert everything to metres before substituting. Mixing millimetres and metres in the same equation is the most common numerical error in this topic.

Worked ExampleFinding a wavelength from a fringe pattern

In a two-slit experiment, the slits are $0.25$ mm apart and the screen is $1.8$ m away. The bright fringes are measured to be $4.3$ mm apart. Find the wavelength of the light.

Step 1 — Convert everything to metres

$d = 0.25 \text{ mm} = 2.5 \times 10^{-4} \text{ m}$ $x = 4.3 \text{ mm} = 4.3 \times 10^{-3} \text{ m}$ $L = 1.8 \text{ m}$

Step 2 — Identify $n$

The question gives the spacing between adjacent fringes, so $n = 1$ .

Step 3 — Substitute

$n\lambda = \frac{dx}{L}$ $(1)\lambda = \frac{(2.5 \times 10^{-4})(4.3 \times 10^{-3})}{1.8}$ $\lambda = \frac{1.075 \times 10^{-6}}{1.8} = 5.97 \times 10^{-7} \text{ m}$

Step 4 — Express sensibly

$\lambda = 597 \approx 6.0 \times 10^{2} \text{ nm}$

$\lambda = 597 \text{ nm (orange-yellow light)}$

Worked ExampleUsing a higher-order fringe

Step 1 — Convert and identify

$\lambda = 6.33 \times 10^{-7} \text{ m}, \qquad d = 4.0 \times 10^{-4} \text{ m}$ $x = 9.5 \times 10^{-3} \text{ m}, \qquad n = 4$

Here $x$ is the distance from the centre to the fourth fringe, so $n = 4$ — not 1.

Step 2 — Rearrange for $L$

$n\lambda = \frac{dx}{L} \;\Rightarrow\; L = \frac{dx}{n\lambda}$

Step 3 — Substitute

$L = \frac{(4.0 \times 10^{-4})(9.5 \times 10^{-3})}{(4)(6.33 \times 10^{-7})}$ $L = \frac{3.80 \times 10^{-6}}{2.532 \times 10^{-6}} = 1.50 \text{ m}$

$L = 1.5 \text{ m}$

Worked ExamplePredicting the effect of a change

Step 1 — Write the relationship for the spacing

With $n = 1$ :

$x = \frac{\lambda L}{d}$

$L$ and $d$ are unchanged in part (a), so $x \propto \lambda$ .

(a) Changing to red light

$\frac{x_{\text{red}}}{x_{\text{blue}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}} = \frac{700}{450} = 1.556$ $x_{\text{red}} = (3.0)(1.556) = 4.7 \text{ mm}$

The fringes move further apart, because a longer wavelength needs a larger path difference — and therefore a larger angle — to reach each order.

(b) Doubling the slit separation

Now $\lambda$ and $L$ are unchanged, so $x \propto \dfrac{1}{d}$ :

$x_{\text{new}} = \frac{3.0}{2} = 1.5 \text{ mm}$

The fringes move closer together.

$\text{(a) } 4.7 \text{ mm}, \qquad \text{(b) } 1.5 \text{ mm}$

Two-slit calculations

The fringe relationship

Two ways the symbol xxx is used

What changes the fringe spacing

Getting the units right

Step 1 — Convert everything to metres

Step 2 — Identify nnn

Step 3 — Substitute

Step 4 — Express sensibly

Step 1 — Convert and identify

Step 2 — Rearrange for LLL

Step 3 — Substitute

Step 1 — Write the relationship for the spacing

(a) Changing to red light

(b) Doubling the slit separation

Two-slit calculations

The fringe relationship

Two ways the symbol xxx is used

What changes the fringe spacing

Getting the units right

Step 1 — Convert everything to metres

Step 2 — Identify nnn

Step 3 — Substitute

Step 4 — Express sensibly

Step 1 — Convert and identify

Step 2 — Rearrange for LLL

Step 3 — Substitute

Step 1 — Write the relationship for the spacing

(a) Changing to red light

(b) Doubling the slit separation

Two ways the symbol $x$ is used

Step 2 — Identify $n$

Step 2 — Rearrange for $L$

Two ways the symbol $x$ is used

Step 2 — Identify $n$

Step 2 — Rearrange for $L$