E2.4 Analyse and solve diffraction and interference problems

Ontario Grade 12 Physics

The Wave Nature of Light

Using wave patterns, path differences, and screen geometry

In SPH3U, you studied waves as disturbances that transfer energy. Wavelength is the distance between matching points on consecutive cycles. Frequency is the number of cycles passing a point each second. This lesson considers light passing through one or two narrow openings and reaching a screen. The physical system is the opening or openings, the light, and the screen pattern. Use the screen as the reference frame. Set the central bright fringe as the origin and upward on the screen as positive. A screen position is a signed scalar: it has a magnitude and a sign, but no direction as a vector does. Path difference is also a scalar distance difference. The sign of a screen position indicates its location relative to the centre.

What you will learn

Explain how diffraction and interference produce bright and dark regions.
Use path difference to identify constructive and destructive interference.
Solve double-slit and single-slit problems with Grade 12 algebra and trigonometry.
Check units, direction, significant figures, and whether a result is reasonable.

1. Diffraction, interference, and path difference

Diffraction is the spreading of a wave as it passes through an opening or around an edge. The spreading is more noticeable when the opening is not much larger than the wavelength. Interference is the pattern formed when waves overlap. Their displacements combine at each point. This combining is called superposition.

A bright region forms when waves arrive in phase. Their crests line up with crests and their troughs with troughs. This is constructive interference. A dark region forms when waves arrive out of phase by half a cycle, so a crest lines up with a trough. This is destructive interference.

Path difference, written as

\Delta r

, compares the distances travelled by two waves to the same point. For two slits separated by distance

d

, the path difference at an angle

\theta

from the central axis is

d\sin\theta

. A whole-number wavelength difference produces constructive interference. A half-integer wavelength difference produces destructive interference. The central bright fringe has zero path difference.

\Delta r=d\sin\theta

Diffraction describes spreading; interference describes the pattern from overlapping waves.
Bright and dark regions depend on path difference.
The central bright fringe is at screen position $y=0$ .

2. Double-slit fringes and screen positions

For two narrow slits, let

d

be the slit separation, measured in metres. Let

\lambda

be the wavelength, also in metres. A bright fringe occurs when the path difference is an integer number of wavelengths. The integer

m

is the fringe order:

m=0

is the centre, and

m=1

and

m=-1

are the first bright fringes above and below it. A dark fringe occurs when the path difference is an odd number of half-wavelengths.

To relate angle to position, let

L

be the distance from the slits to the screen and

y

the signed position relative to the centre. For a distant screen and small displacement,

\sin\theta\approx\tan\theta=y/L

. This is the small-angle approximation. It gives the bright-fringe positions and the spacing between adjacent bright fringes. The spacing is a positive distance; an individual position can be positive or negative.

The model assumes that waves from the two slits overlap and have matching wavelengths. First identify whether the question asks for a bright or dark fringe. Then choose the condition that matches. Use metres for all lengths before substituting.

d\sin\theta=m\lambda,\qquad y_m\approx\frac{m\lambda L}{d},\qquad \Delta y\approx\frac{\lambda L}{d}

Use $d$ for the separation between two slits.
Positive $y$ means above the central maximum; negative $y$ means below it.
For a fixed wavelength and screen distance, larger slit separation gives smaller fringe spacing.

3. Single-slit diffraction minima

A single slit produces a diffraction pattern because waves from different parts of the opening reach the screen along different paths. The central bright region is wider than the bright regions beside it. Let

a

be the width of the slit. A dark minimum occurs when the path difference between waves from opposite sides of the opening is a whole number of wavelengths.

The integer

m

for these minima is nonzero. The first minima have

m=1

on either side of the centre. At small angles, their screen positions follow from

\sin\theta\approx y/L

. A smaller slit width produces a larger distance from the centre to the first minimum, so the central bright region is wider.

Do not confuse this condition with the double-slit bright-fringe condition. A double-slit problem uses the separation

d

and may ask for bright or dark fringes. A single-slit problem uses the width

a

to locate dark minima.

a\sin\theta=m\lambda,\qquad m=1,2,3,\ldots

Use $a$ for the width of one slit and $d$ for separation between two slits.
Single-slit dark minima use nonzero integer orders.
A narrower slit produces a wider central bright region.

4. A reliable solving method

Sketch the opening or openings, central axis, screen, and fringe of interest. Label

a

or

d

,

L

, and

y

. State the chosen frame and positive direction. This prevents mixing up slit width and slit separation, or angle and screen position.

Choose the correct fringe condition and rearrange it before substituting. Convert nanometres and millimetres to metres. Keep units in the substitution, and report a sensible number of significant figures based on the given values. Use the small-angle relation only when the screen distance is large compared with the displacement.

Finally, check the result. Wavelength, slit width, and slit separation must be positive. The sign of

y

must match the chosen direction. A calculated position should have units of metres. Check the scale: increasing

L

or

\lambda

increases fringe spacing, while increasing

d

decreases double-slit spacing.

1\ \mathrm{nm}=1\times10^{-9}\ \mathrm{m}

Select the physical condition before selecting an equation.
Keep the sign convention and diagram consistent.
Check units, significant figures, direction, and scale.

Worked example

Finding double-slit fringe spacing

Light of wavelength

600\ \mathrm{nm}

passes through slits separated by

0.30\ \mathrm{mm}

. The screen is

2.0\ \mathrm{m}

away. Find the distance between adjacent bright fringes near the centre.

Identify the system and unknown
The system is the two slits and the screen pattern. Use the screen as the reference frame, with upward positive. The requested quantity is a distance between adjacent bright fringes, so it is positive.
$\Delta y=\frac{\lambda L}{d}$
Convert and substitute
Convert the wavelength and slit separation to metres. The remaining length is already in metres.
$\Delta y=\frac{(600\times10^{-9}\ \mathrm{m})(2.0\ \mathrm{m})}{0.30\times10^{-3}\ \mathrm{m}}$
Evaluate and check
The units reduce to metres. The result is a millimetre-scale spacing, reasonable for visible light and these slit and screen dimensions.
$\Delta y=4.0\times10^{-3}\ \mathrm{m}$

Answer: Adjacent bright fringes are

4.0\times10^{-3}\ \mathrm{m}

, or

4.0\ \mathrm{mm}

, apart.

Check: The calculation has two significant figures, matching the supplied values. The spacing is positive and has units of length.

Worked example

Finding wavelength from a bright fringe

A bright fringe of order

m=2

is at

y=+8.0\ \mathrm{mm}

on a screen

1.5\ \mathrm{m}

from slits separated by

0.25\ \mathrm{mm}

. Find the wavelength using the small-angle approximation.

Choose the bright-fringe relationship
The system is the two slits and screen. The screen frame uses upward as positive. Since the specified point is a bright fringe, use its order and signed position in the bright-fringe position relationship.
$y_m\approx\frac{m\lambda L}{d}$
Rearrange and convert
Solve for wavelength. Convert the position and slit separation to metres before substitution.
$\lambda=\frac{y_md}{mL}$
Substitute and evaluate
The positive position indicates a fringe above the centre. The units reduce to metres, as required for wavelength.
$\lambda=\frac{(8.0\times10^{-3}\ \mathrm{m})(0.25\times10^{-3}\ \mathrm{m})}{(2)(1.5\ \mathrm{m})}=6.7\times10^{-7}\ \mathrm{m}$

Answer: The wavelength is

6.7\times10^{-7}\ \mathrm{m}

, or

670\ \mathrm{nm}

.

Check: The wavelength is positive and in the visible-light range. The ratio

y/L

is about

0.0053

, so the small-angle approximation is appropriate.

Worked example

Locating a single-slit minimum

Light of wavelength

500\ \mathrm{nm}

passes through a slit

0.10\ \mathrm{mm}

wide. A screen is

2.0\ \mathrm{m}

away. Find the position of the first dark minimum above the centre.

Define the system and direction
The system is one slit and the screen. Use the screen as the reference frame and upward as positive. The first minimum above the centre has order $m=1$ and positive position.
$a\sin\theta=m\lambda$
Relate angle to position
For a small angle, $\sin\theta\approx y/L$ . Combining this with the single-slit minimum condition gives the screen position.
$y\approx\frac{m\lambda L}{a}$
Substitute and evaluate
Convert wavelength and slit width to metres. The positive result places the minimum above the central maximum.
$y\approx\frac{(1)(500\times10^{-9}\ \mathrm{m})(2.0\ \mathrm{m})}{0.10\times10^{-3}\ \mathrm{m}}=1.0\times10^{-2}\ \mathrm{m}$

Answer: The first dark minimum above the centre is at

y=+1.0\times10^{-2}\ \mathrm{m}

, or

+10\ \mathrm{mm}

.

Check: The units reduce to metres. The sign agrees with upward as positive. The displacement is small compared with the

2.0\ \mathrm{m}

screen distance, consistent with the small-angle model.

Common mistakes and how to avoid them

Using slit width for a double-slit pattern, or slit separation for a single-slit minimum.

Correction: Use

d

for the distance between two slits and

a

for the width of one slit.

Using a whole-wavelength difference for a double-slit dark fringe.

Correction: A bright fringe has a whole-number wavelength path difference. A dark fringe has an odd half-number wavelength path difference.

Substituting nanometres or millimetres without converting units.

Correction: Convert all lengths to metres before using the equations.

Treating every screen position as a positive distance.

Correction: A distance between fringes is positive, but an individual position has a sign based on the stated direction.

Lesson summary

Diffraction is wave spreading, while interference is the pattern formed by overlapping waves.
For two slits, path difference determines whether a fringe is bright or dark.
For one slit, use the slit-width condition to locate dark minima.
Use screen geometry and the small-angle approximation when appropriate.
Convert to SI units and check units, sign, and scale.

Check your understanding

Question 1

For a double slit, what path difference gives the first bright fringe beside the centre?

Zero
One wavelength
Half a wavelength
One and a half wavelengths

Show answer and explanation

One wavelength

The first bright fringe has order

m=1

. Its path difference is one wavelength.

Question 2

If slit separation doubles while wavelength and screen distance stay fixed, what happens to the spacing between adjacent bright fringes?

It doubles.
It is cut in half.
It stays the same.
It becomes zero.

Show answer and explanation

It is cut in half.

The fringe spacing is inversely proportional to slit separation. Doubling the separation halves the spacing.

Question 3

A single slit has width

0.10\ \mathrm{mm}

. Light of wavelength

500\ \mathrm{nm}

reaches a screen

2.0\ \mathrm{m}

away. Where is the first minimum above the centre?

$+1.0\ \mathrm{mm}$
$+5.0\ \mathrm{mm}$
$-5.0\ \mathrm{mm}$
$+10\ \mathrm{mm}$

Show answer and explanation

+10\ \mathrm{mm}

Using

y\approx\lambda L/a

gives

0.010\ \mathrm{m}

, or

+10\ \mathrm{mm}

. The sign is positive because the point is above the centre.

Key terms

Diffraction: The spreading of a wave as it passes through an opening or around an edge.
Interference: The pattern that forms when overlapping waves combine.
Path difference: The difference between the distances travelled by two waves to the same point.
Constructive interference: Wave overlap that reinforces displacement and produces a bright region in the ideal light pattern.
Destructive interference: Wave overlap that reduces displacement and produces a dark region in the ideal light pattern.
Fringe: A bright or dark band in an interference or diffraction pattern.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation E2.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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E2.4 · Analyse and solve diffraction and interference problems

Ontario Grade 12 Physics

Using wave patterns, path differences, and screen geometry

What you will learn

1. Diffraction, interference, and path difference

2. Double-slit fringes and screen positions

3. Single-slit diffraction minima

4. A reliable solving method

Finding double-slit fringe spacing

Finding wavelength from a bright fringe

Locating a single-slit minimum

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review