Frequency vs Wavelength: What Is the Difference?

Frequency counts how many cycles pass each second, in hertz, and the source sets it. Wavelength is the distance between matching points on the wave, in metres. Wave speed locks them together. At a boundary the frequency cannot change, so the wavelength must.

AP Physics: Unit 14 (topics 14.2 Periodic Waves, 14.3 Boundary Behavior of Waves and Polarization). Both quantities are defined in AP Physics 2 Topic 14.2, Periodic Waves. Essential knowledge 14.2.A.1.ii gives frequency as the rate at which the wave repeats, with T = 1/f as the relevant equation, and 14.2.A.1.vi gives wavelength as the distance between successive corresponding positions, such as peaks or troughs, on a wave. Essential knowledge 14.2.A.3 connects them: for a periodic wave the wavelength is proportional to the wave's speed and inversely proportional to the wave's frequency, relevant equation lambda = v/f, printed on the AP Physics 2 sheet. Essential knowledge 14.2.A.2 keeps the time and space pictures apart with two example equations, x(t) = A cos(omega t) = A cos(2 pi f t) for displacement at a specific location as a function of time and y(x) = A cos(2 pi x / lambda) for displacement at a specific time as a function of position, both printed on the sheet. The separation between the two quantities is Topic 14.3: 14.3.A.1 says a wave travelling from one medium to another can be transmitted or reflected depending on the properties of the boundary, 14.3.A.1.i gives the low-mass to high-mass string example and says both reflected and transmitted waves result, 14.3.A.1.ii and 14.3.A.1.iii tie inversion of the reflected wave to whether the transmitted wave's speed decreases or increases, and 14.3.A.1.iv states that the frequency of a wave does not change when it travels from one medium to another. Supporting statements used here: 14.1.A.3.ii for the string speed, 14.1.A.3.iii for the temperature dependence of the speed of sound, 14.2.A.1.iii for amplitude being independent of period and frequency, 14.2.A.1.iv for energy increasing with frequency, 14.2.A.1.v for frequency and pitch, and 14.4.A.3 with 14.4.A.3.ii for electromagnetic waves being categorised by wavelength and visible light ordered from red to violet by decreasing wavelength. The optical version of the crossing is Topic 13.3, where 13.3.A.3 gives n = c/v. Neither Topic 14.2 nor Topic 14.3 carries a boundary statement. Unit 14 is weighted at 12 to 15 percent of the multiple-choice section over a suggested 14 to 23 class periods. The current AP Physics 1 framework has no waves unit, and its equation sheet prints T = 1/f with no lambda anywhere on it, checked by rendering the appendix page.

The distinction, stated once

Frequency is counted in time. Wavelength is measured in space. That is the whole difference, and everything else on this page follows from it.

AP Physics 2 essential knowledge 14.2.A.1.ii puts frequency as the rate at which the wave repeats, with the relevant equation $T = 1/f$ . Its unit is the hertz, one cycle per second. Stand at one point, hold a stopwatch, and count crests going past: that is $f$ .

Essential knowledge 14.2.A.1.vi puts wavelength as the distance between successive corresponding positions, such as peaks or troughs, on a wave. Its unit is the metre. Freeze the wave, hold a ruler along it, and measure from one crest to the next: that is $\lambda$ .

One quantity needs a clock and one needs a ruler, and no measurement gives you both. What connects them is the wave speed, through the relation printed on the AP Physics 2 equation sheet as the relevant equation for 14.2.A.3:

\lambda = \frac{v}{f}

The CED states the dependence in words as well: for a periodic wave, the wavelength is proportional to the wave's speed and inversely proportional to the wave's frequency. Read $v$ as the exchange rate between the clock and the ruler, in metres per second, and read the whole relation as saying that a wave repeating $f$ times a second while advancing $v$ metres a second must lay down its pattern every $v/f$ metres.

Which one is fixed depends on what you are changing. Stay in one medium and $v$ is fixed, so $f$ and $\lambda$ are locked in inverse proportion and either one determines the other. Cross into a new medium and $v$ changes, and now the two behave completely differently: essential knowledge 14.3.A.1.iv says the frequency of a wave does not change when it travels from one medium to another. The frequency belongs to the source, which is still shaking at the same rate. The wavelength has no such protection, so it absorbs the entire change.

That sentence is what refraction is built on, and it is the one students reverse.

Side by side

	Frequency	Wavelength
Symbol	$f$	$\lambda$
Unit	Hertz, one cycle per second	Metre
What you measure it with	A clock, at one fixed place	A ruler, at one fixed instant
CED definition	14.2.A.1.ii, the rate at which the wave repeats	14.2.A.1.vi, the distance between successive corresponding positions such as peaks or troughs
Read off which graph	Displacement against time, as the repeat interval $T = 1/f$	Displacement against position, as the repeat distance
Set by	The source	The source and the medium together, through $\lambda = v/f$
Changes when the wave enters a new medium	No, by 14.3.A.1.iv	Yes, by exactly the factor the speed changed
Changes when the source is retuned in one medium	Yes	Yes, inversely
Independent of amplitude	Yes, 14.2.A.1.iii	Yes, 14.2.A.1.iii
Physical property it tracks	Pitch of a sound, 14.2.A.1.v; energy of the wave, 14.2.A.1.iv	The category of an electromagnetic wave, 14.4.A.3
Appears in the interference relations	No	Yes, in every one of them, as $m\lambda$
For a longitudinal wave	Same definition, count compressions going past	Compression to compression, since there are no peaks

The row to reread is changes when the wave enters a new medium. It is the only row where one entry is a flat no and the other is a flat yes, and it is the row every refraction, thin film and fibre optic question is built on.

The last two rows explain a division of labour that is easy to miss. Pitch and energy are stated in terms of frequency. The interference and diffraction relations on the sheet, $\Delta D = m\lambda$ , $\Delta D = d\sin\theta$ and the rest, are stated in terms of wavelength and never in terms of frequency. So a question about what a sound sounds like wants $f$ , and a question about whether a pattern of fringes appears wants $\lambda$ .

Why the two graphs make this hard

The reason this pair gets swapped is that both are read off a curve that looks the same.

Displacement against time, taken at one location. The horizontal axis is seconds. The repeat interval is the period $T$ , and the frequency is $1/f$ inverted from it. Nothing on this graph is a wavelength, because the graph never leaves the one point in space.

Displacement against position, taken at one instant. The horizontal axis is metres. The repeat distance is the wavelength. Nothing on this graph is a frequency or a period, because the graph is frozen and no time passes on it.

The two curves have the same shape, the same amplitude and the same number of humps. Only the axis label distinguishes them, and axis labels are the first thing an eye skips. Essential knowledge 14.2.A.2 names both explicitly and keeps them apart with different variables: it gives the displacement from equilibrium at a specific location as a function of time,

x(t) = A\cos(\omega t) = A\cos(2\pi f t)

and the displacement at a specific time as a function of position,

y(x) = A\cos\!\left(\frac{2\pi x}{\lambda}\right)

Compare the two arguments. One has $f$ and a time in it, the other has $\lambda$ and a distance. The CED has already made the distinction for you in the shape of the two equations, and both are printed on the equation sheet.

The practical habit is to check the horizontal axis before reading anything else. If it says seconds you can get $T$ and $f$ . If it says metres you can get $\lambda$ . To get the other one you need $v$ , which the graph will not give you.

The case that separates them: light entering water

Send light of vacuum wavelength $500 \ \mathrm{nm}$ into water of index $n = 1.33$ . This is the situation where treating $f$ and $\lambda$ as interchangeable produces a wrong answer.

In vacuum, using $c = 3.00 \times 10^8$ m/s from the sheet's table of constants,

f = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{500 \times 10^{-9}} = 6.00 \times 10^{14} \ \mathrm{Hz}

Inside the water the speed drops, by $n = c/v$ from 13.3.A.3, to $v = (3.00 \times 10^8)/1.33 = 2.26 \times 10^8$ m/s.

Now the two quantities part company.

	In vacuum	In the water
Speed	$3.00 \times 10^8 \ \mathrm{m/s}$	$2.26 \times 10^8 \ \mathrm{m/s}$
Frequency	$6.00 \times 10^{14} \ \mathrm{Hz}$	$6.00 \times 10^{14} \ \mathrm{Hz}$ , unchanged
Wavelength	$500 \ \mathrm{nm}$	$376 \ \mathrm{nm}$

The frequency held because 14.3.A.1.iv says it must. The wavelength moved because $\lambda = v/f$ leaves it nowhere to hide: $v$ went down by the factor $1.33$ and $f$ did not move, so $\lambda$ went down by the factor $1.33$ as well.

The direction of the reasoning matters more than the number. The frequency is not "also unchanged", as though the two were both conserved and one happened to move. The frequency is the fixed one, imposed by the source at the far end, and the wavelength is what the relation has left to adjust. Get that ordering the wrong way round and every boundary question comes out backwards.

A note on colour, since it is where students test this idea. Essential knowledge 14.4.A.3 says categories of electromagnetic waves are characterized by their wavelengths, and 14.4.A.3.ii orders visible light by decreasing wavelength: red, orange, yellow, green, blue, violet. The CED does not say whether those are vacuum wavelengths. The boundary rule is what tells you they must be, because the same light has a shorter wavelength in water while it is unmistakably the same light. Whenever a problem quotes a wavelength for light inside a material, check which of the two it means; they differ by the factor $n$ .

The same logic on a rope, where you can see it

The light case is the one that gets examined and the rope case is the one that makes it obvious, because the CED uses the rope as its own example.

Essential knowledge 14.3.A.1 says a wave that travels from one medium to another can be transmitted or reflected, depending on the properties of the boundary separating the two media, and 14.3.A.1.i names the example directly: a wave traveling between low-mass and high-mass strings will result in reflected and transmitted waves.

Tie a light string to a heavy one and shake the free end. Your hand is the source and it sets the frequency, once, for the whole system. The crests arriving at the knot must leave it at the same rate, or crests would pile up at the knot without limit. So $f$ is the same on both sides, which is 14.3.A.1.iv seen mechanically rather than quoted.

What differs is the speed, from 14.1.A.3.ii,

v_{\text{string}} = \sqrt{\frac{F_T}{m/\ell}}

The tension is the same throughout a string pulled taut, so the heavier string, with its larger mass per length, carries a slower wave. With $f$ pinned and $v$ smaller, $\lambda = v/f$ makes the wavelength shorter. The pattern visibly bunches up as it crosses onto the heavy rope, and the bunching is the wavelength change you cannot see happening to light in glass.

The boundary also does something to the reflected wave, and the CED ties it to the same speed change. Essential knowledge 14.3.A.1.ii says a reflected wave is inverted if the transmitted wave travels into a medium in which the speed of the wave decreases, and 14.3.A.1.iii says it is not inverted if the speed increases. Light rope to heavy rope, the transmitted wave slows, so the reflected pulse comes back upside down. That is a phase statement rather than a frequency or wavelength statement, and it does not disturb either of them.

When it costs a mark

Saying the frequency changes when light enters glass. Essential knowledge 14.3.A.1.iv says it does not. This is the single error the whole topic is built to catch, and it usually shows up as a student computing a new frequency from the new speed and the old wavelength.

Saying the wavelength is unchanged. The mirror image of the same error. If $v$ changed and $f$ did not, then $\lambda$ changed, by arithmetic and with no room for judgement.

Quoting a wavelength for light in a medium without saying which medium. The number $500 \ \mathrm{nm}$ means one thing in vacuum and describes different light in water. A frequency is unambiguous everywhere, which is one practical reason to convert to $f$ early.

Reading a wavelength off a time axis. A displacement-against-time graph has a period on it, not a wavelength. Getting a length out of it requires the wave speed, which the graph does not show.

Treating $f$ and $\lambda$ as inversely proportional in every situation. They are, but only at fixed $v$ , which means only within one medium. Across a boundary $f$ is fixed and $\lambda$ tracks $v$ directly, which is a proportionality with the opposite sense. Two different rules, and which one applies depends on whether the medium changed.

Using frequency in an interference relation. Every fringe relation on the AP Physics 2 sheet is written with $\lambda$ : $\Delta D = m\lambda$ , $\Delta D = d\sin\theta$ , $d(y_{\text{max}}/L) \approx m\lambda$ . Substituting a frequency in place of a wavelength is a units error that a quick dimension check catches.

Assuming the speed of sound is a number you already know. The AP Physics 2 table of constants prints the speed of light and does not print a speed of sound, so a sound question has to give you one, or give you enough to get it. Essential knowledge 14.1.A.3.iii tells you only the direction of the dependence: in a given medium, the speed of sound waves increases with the temperature of the medium.

When they move together, and why that lulls you

Nearly every problem a student meets first happens inside a single medium, and inside a single medium the pair really does behave like one quantity.

With $v$ fixed, $\lambda = v/f$ is a strict inverse proportion. Double the frequency and the wavelength halves. Quote either one and you have quoted both. A whole course of practice can go by without ever needing to know which is which, because the arithmetic works from either end and the answers come out right.

The habit that builds is: frequency and wavelength are two ways of saying the same thing. It is a good habit inside one medium and a wrong one the moment a boundary appears, which is exactly where the exam puts the question.

Two more overlaps keep the disguise up.

Both are independent of the amplitude. Essential knowledge 14.2.A.1.iii says the amplitude of a wave is independent of the period and the frequency of that wave, so changing the loudness of a sound leaves both $f$ and $\lambda$ where they were. They fail to respond to the same input, which makes them look like the same input.

Both increase the energy in the same direction, sort of. Essential knowledge 14.2.A.1.iv says the energy of a wave increases with increasing frequency. Since a shorter wavelength is a higher frequency in fixed $v$ , it is tempting to read that as an energy statement about wavelength. The CED writes it about frequency, so write it about frequency.

The question that breaks the tie, every time: did the medium change? If no, the two are interchangeable and inversely proportional. If yes, $f$ is the survivor and $\lambda$ is the casualty.

Where this sits on the AP exam

Both quantities are defined in Topic 14.2, Periodic Waves, inside Unit 14, Waves, Sound, and Physical Optics, which the CED weights at 12 to 15 percent of the multiple-choice section over a suggested 14 to 23 class periods. The boundary rule that separates them is in Topic 14.3, and the optical version of the same crossing is Topic 13.3, Refraction.

The current AP Physics 1 framework has no waves unit, so neither quantity is examined there in a wave context. Its Unit 7, Oscillations uses frequency and period for an oscillating object and never uses a wavelength, and the AP Physics 1 equation sheet prints $T = 1/f$ with no $\lambda$ anywhere on it. That was checked by rendering the appendix page rather than inferred.

The suggested skills for Topic 14.2 are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Topic 14.3 lists the same four. Neither topic carries a boundary statement, checked page by page.

Skill 2.C is again the shape of the question: compare the same wave in two media, or the same medium at two frequencies, and say which quantities moved. The unit's own guidance for building the science practices names functional dependence, skill 2.D, and gives an example that is a wavelength question in disguise, asking students to determine the new distance between bright fringes if the frequency of the light through a single slit is doubled.

For the arithmetic itself, rearranging $\lambda = v/f$ in every direction and reading each quantity off a graph, the wave speed, frequency and wavelength guide owns that procedure and works it in both directions. For the definitions alone see frequency and wavelength. For the quantity that is genuinely fixed by the medium rather than by the source, see wave speed vs particle speed, and for the time-domain twin of frequency see period vs frequency.

Light of 500 nm entering water: which number moves

Monochromatic light with a wavelength of $500 \ \mathrm{nm}$ in vacuum enters water, index of refraction $n = 1.33$ . Take $c = 3.00 \times 10^8$ m/s. (a) Find the frequency of the light in vacuum. (b) Find the speed of the light in the water. (c) Find the frequency and the wavelength of the light in the water. (d) State which quantity the boundary was not allowed to change, and why.

(a) Rearranging the sheet relation $\lambda = v/f$ with $v = c$ in vacuum gives $f = c/\lambda$ . Convert first: $500 \ \mathrm{nm} = 500 \times 10^{-9} \ \mathrm{m} = 5.00 \times 10^{-7} \ \mathrm{m}$ .
$f = (3.00 \times 10^8)/(5.00 \times 10^{-7}) = 6.00 \times 10^{14} \ \mathrm{Hz}$ .
(b) Essential knowledge 13.3.A.3 gives $n = c/v$ , so $v = c/n = (3.00 \times 10^8)/1.33 = 2.26 \times 10^8 \ \mathrm{m/s}$ .
(c) The frequency in the water is still $6.00 \times 10^{14} \ \mathrm{Hz}$ , by 14.3.A.1.iv. Then $\lambda = v/f = (2.26 \times 10^8)/(6.00 \times 10^{14}) = 3.76 \times 10^{-7} \ \mathrm{m} = 376 \ \mathrm{nm}$ .
Shortcut check: with $f$ fixed, $\lambda$ is proportional to $v$ , and $v$ was divided by $1.33$ , so $\lambda$ is too. $500/1.33 = 376 \ \mathrm{nm}$ to three significant figures, which agrees with the long route.
(d) The frequency. Essential knowledge 14.3.A.1.iv states that the frequency of a wave does not change when it travels from one medium to another, because the source on the far side of the boundary is still emitting at the same rate. With $f$ pinned and $v$ reduced, $\lambda = v/f$ has only the wavelength left to change.

$f = 6.00 \times 10^{14} \ \mathrm{Hz}$ in both media. In the water the speed is $2.26 \times 10^8 \ \mathrm{m/s}$ and the wavelength is $376 \ \mathrm{nm}$ . Frequency is the quantity the boundary cannot change, so the wavelength absorbs the whole factor of $1.33$ .

A wave crossing from a light string to a heavy one

A light string with a mass per length of $0.0025$ kg/m is knotted to a heavy string with a mass per length of $0.010$ kg/m. The whole arrangement is pulled taut with a tension of $36$ N, and the free end of the light string is shaken at $30 \ \mathrm{Hz}$ . (a) Find the wave speed on each string. (b) Find the frequency on each string. (c) Find the wavelength on each string. (d) Say whether the pulse reflected at the knot is inverted.

(a) Use $v_{\text{string}} = \sqrt{F_T/(m/\ell)}$ from 14.1.A.3.ii, printed on the AP Physics 2 sheet. The tension is the same throughout, $36$ N.
Light string: $v_1 = \sqrt{36/0.0025} = \sqrt{14400} = 120 \ \mathrm{m/s}$ . Heavy string: $v_2 = \sqrt{36/0.010} = \sqrt{3600} = 60 \ \mathrm{m/s}$ . Four times the mass per length gives half the speed, since the mass per length sits under a square root.
(b) $30 \ \mathrm{Hz}$ on both, by 14.3.A.1.iv. The hand at the free end is the only source, and the knot cannot change how often crests arrive at it.
(c) $\lambda_1 = v_1/f = 120/30 = 4.0 \ \mathrm{m}$ and $\lambda_2 = v_2/f = 60/30 = 2.0 \ \mathrm{m}$ . The wavelength halved, tracking the speed exactly, because the frequency was not free to move.
(d) Essential knowledge 14.3.A.1.ii says a reflected wave is inverted if the transmitted wave travels into a medium in which the speed of the wave decreases. The transmitted wave went from $120 \ \mathrm{m/s}$ to $60 \ \mathrm{m/s}$ , so the speed decreased and the reflected pulse is inverted.
Consistency check on the ratios: $v_2/v_1 = 60/120 = 0.50$ and $\lambda_2/\lambda_1 = 2.0/4.0 = 0.50$ , while $f_2/f_1 = 1$ . Speed and wavelength moved by the same factor and frequency did not move at all, which is $\lambda = v/f$ with $f$ held fixed.

Speeds $120 \ \mathrm{m/s}$ and $60 \ \mathrm{m/s}$ ; frequency $30 \ \mathrm{Hz}$ on both strings; wavelengths $4.0 \ \mathrm{m}$ and $2.0 \ \mathrm{m}$ . The reflected pulse is inverted, because the transmitted wave slowed down.

Two notes an octave apart in the same air

In a room where the speed of sound is given as $340 \ \mathrm{m/s}$ , one tuning fork sounds at $220 \ \mathrm{Hz}$ and another at $440 \ \mathrm{Hz}$ . (a) Find the wavelength of each note. (b) State the ratio of the wavelengths and the ratio of the frequencies. (c) Explain why this proportionality is the opposite in character to the one in the two examples above.

(a) Both notes travel in the same air, so both use the same $v = 340 \ \mathrm{m/s}$ . This value has to be given: the AP Physics 2 table of constants prints the speed of light and no speed of sound.
$\lambda_{220} = v/f = 340/220 = 1.55 \ \mathrm{m}$ to three significant figures. $\lambda_{440} = 340/440 = 0.773 \ \mathrm{m}$ .
(b) $f_{440}/f_{220} = 2.00$ . For the wavelengths, divide before rounding: $\lambda_{440}/\lambda_{220} = (340/440)/(340/220) = 220/440 = 0.500$ exactly. Doubling the frequency exactly halved the wavelength.
(c) Here the medium never changed, so $v$ was the fixed quantity and the two variables traded off inversely: $\lambda \propto 1/f$ . In the light and string examples the medium did change, so $f$ was the fixed quantity and the wavelength moved directly with the speed: $\lambda \propto v$ . Same relation $\lambda = v/f$ , two different quantities held still, two opposite behaviours.
Worth noting what does not enter either calculation. The loudness of the forks is irrelevant, because 14.2.A.1.iii says the amplitude of a wave is independent of the period and the frequency of that wave.

$1.55 \ \mathrm{m}$ at $220 \ \mathrm{Hz}$ and $0.773 \ \mathrm{m}$ at $440 \ \mathrm{Hz}$ . Doubling the frequency halves the wavelength, because in one medium $v$ is fixed. Across a boundary the fixed quantity is the frequency instead, and the wavelength follows the speed.

Frequently asked questions

What is the difference between frequency and wavelength?

Frequency is how many complete cycles pass a fixed point each second, measured in hertz with a clock. Wavelength is the distance between two successive corresponding positions on the wave, such as peak to peak, measured in metres with a ruler. AP Physics 2 defines them at essential knowledge 14.2.A.1.ii and 14.2.A.1.vi. Wave speed links them through lambda = v / f, printed on the AP Physics 2 equation sheet, so inside one medium each determines the other, and at a boundary they behave completely differently.

Does frequency or wavelength change when a wave enters a new medium?

The wavelength changes and the frequency does not. Essential knowledge 14.3.A.1.iv in AP Physics 2 states that the frequency of a wave does not change when it travels from one medium to another, because the source setting the rate has not changed. The speed does change in the new medium, and since wavelength equals speed divided by frequency, the wavelength changes by exactly the factor the speed changed by. Light of vacuum wavelength 500 nanometres entering water of index 1.33 keeps its frequency of 6.00 times ten to the fourteen hertz and shortens to 376 nanometres.

Are frequency and wavelength inversely proportional?

Only at a fixed wave speed, which means only within one medium. There, lambda = v / f with v constant makes them a strict inverse proportion, so doubling the frequency halves the wavelength. Across a boundary the rule is different: the frequency is the fixed quantity by essential knowledge 14.3.A.1.iv, the speed changes, and the wavelength becomes directly proportional to the speed. Applying the inverse-proportion habit to a boundary problem is the standard way this topic goes wrong.

Why does light change wavelength but not colour when it enters glass?

Because the frequency is what survives the crossing. Essential knowledge 14.3.A.1.iv holds the frequency fixed at a boundary, while the speed drops to c / n and the wavelength drops by the same factor. AP Physics 2 categorises electromagnetic waves by wavelength in 14.4.A.3, listing visible light from red down to violet by decreasing wavelength, and those bands only make sense as vacuum wavelengths, since the same light has a shorter wavelength inside glass. So when a problem names a wavelength for light in a material, check whether it means the vacuum value or the in-medium one.

How do I find wavelength from frequency?

Use lambda = v / f, the relevant equation for essential knowledge 14.2.A.3 and printed on the AP Physics 2 sheet, and note that you need the wave speed in that medium as well as the frequency. For light in vacuum the speed is the constant c = 3.00 times ten to the eighth metres per second, printed in the AP Physics 2 table of constants. For sound there is no printed constant, so the question has to supply a speed. The step-by-step rearrangements in each direction are worked in the wave speed, frequency and wavelength guide.

Which one do interference and diffraction problems use?

Wavelength, always. Every path-difference and fringe relation on the AP Physics 2 equation sheet is written in terms of lambda: the path difference equals m lambda, and the slit relations use a sin theta and d sin theta against m lambda. Frequency never appears in them. If a question gives you a frequency and asks about fringe spacing, the first step is to convert to a wavelength with lambda = v / f, and for light in vacuum that means dividing c by the frequency.

Can two waves have the same frequency but different wavelengths?

Yes, and it happens at every boundary. The same light has a frequency of 6.00 times ten to the fourteen hertz and a wavelength of 500 nanometres in vacuum, and the same frequency with a wavelength of 376 nanometres inside water of index 1.33. It also happens between two media side by side: a 30 hertz wave on a taut light string can have a wavelength of 4.0 metres while the same 30 hertz wave on the heavier string it is knotted to has a wavelength of 2.0 metres.