Young’s Double Slit Fringe Width Calculator
Young's Double Slit: Fringe Width, and Whether You Could See It
NCERT gives the fringe POSITIONS and leaves the fringe width to be read off the spacing. This page does both, converts the three quantities out of the three different units they always arrive in, and answers the question a number alone does not: at this slit separation and this screen distance, is the pattern coarse enough for anyone to see?
Fringe width, fringe positions and visibility
Sodium light at 589 nm, slits 0.5 mm apart, screen 1.00 m away, first-order fringe
A spacing NCERT leaves you to read off, and the three units it arrives in
- β = λD/d
- the fringe width: the distance from one bright fringe to the next, which is also the distance from one dark fringe to the next. NCERT does not print this formula. Its Chapter 10 gives the fringe POSITIONS — Equation 10.13, bright where xd/D = mλ — and states that the fringes are equally spaced, leaving λD/d to be read off as the gap between consecutive orders. Worth knowing if you are asked to derive rather than quote it
- d
- the SEPARATION of the two slits, not the width of either. The width controls the single-slit envelope that decides how bright each fringe is and how far the pattern extends before fading; it does not appear in this formula anywhere. Students who substitute the slit width get an answer out by a factor of ten or more, and nothing in the arithmetic complains
- λ/d, the angular fringe width
- the property of the slits alone. The fringe width in millimetres depends on where you put the screen; the angle between fringes does not. That makes λ/d the quantity to quote when comparing two arrangements, and it is also the one that tells you when the formula is about to fail: above about 0.1 radian the small-angle step behind λD/d is no longer safe
- the three units
- λ in nanometres, d in millimetres, D in metres — and that is not this page being awkward, it is how all three are genuinely quoted. The conversion is the actual difficulty of the question: 589 nm is 5.89×10−7 m and 0.5 mm is 5×10−4 m, a ratio of 1.178×10−3, which multiplied by a 1 m screen distance is 1.178 mm. Get the powers of ten right and the question is finished
- d sinθ = mλ
- the exact condition, of which λD/d is the small-angle limit. It matters in two places. When d is comparable with λ the fringes are widely spaced and visibly unevenly spaced on a flat screen. And when d is smaller than λ there is no first-order maximum at all, because sinθ would have to exceed 1
- immersion
- put the whole apparatus in a medium of index n and every wavelength shortens to λ/n, so the fringe width shrinks by the same factor. Under water at n = 1.33 the pattern is 1.33 times finer. The frequency is unchanged, as it always is at a boundary, which is why the wavelength is the quantity that has to move
Worked example
Sodium light at 589 nm, slits 0.5 mm apart, screen 1.00 m away, first-order fringe
THE UNITS FIRST, because that is the whole difficulty. λ = 589 nm = 589 × 10−9 = 5.89×10−7 m. d = 0.5 mm = 5×10−4 m. D = 1.00 m. Write all three in metres before substituting anything; every wrong answer to this question that is not a wrong formula is a power of ten.
THE FRINGE WIDTH. β = λD/d = (5.89×10−7 × 1.00) / 5×10−4 = 1.178×10−3 m = 1.1780 mm. Just over a millimetre between consecutive bright fringes, which is a comfortable pattern: about 8.5 fringes to the centimetre, countable by eye and easy on a travelling microscope.
WHERE THE FRINGES ACTUALLY ARE. The first-order bright fringe sits at x = mβ = 1.1780 mm from the centre, and the first-order dark fringe at (m − ½)β = 0.5890 mm — exactly halfway in, which is what makes the dark fringes interleave the bright ones. The path difference at that bright fringe is mλ = 589 nm, one whole wavelength, which is the condition for constructive interference and the only thing the geometry is really doing.
THE ANGLE, which is the part that travels. The angular fringe width is λ/d = 5.89×10−7/5×10−4 = 1.178×10−3 rad, which is 0.0675° or 4.05 arcminutes. That number belongs to the slits and not to the bench: move the screen and the millimetres change while the arcminutes do not. It is also comfortably below 0.1 rad, so the small-angle step behind λD/d is safe here to better than a part in a million.
AND TWO SANITY CHECKS WORTH DOING. Put the apparatus under water and every wavelength shortens by n = 1.33, so the fringe width becomes 1.1780/1.33 = 0.8857 mm — finer, not coarser, which catches the common error of multiplying. Then ask what slit separation would give a round 1 mm fringe at this screen distance: d = λD/β = 5.89×10−7/10−3 = 0.589 mm. Both are one rearrangement of the same formula, and being able to run it in any direction is what the question is actually testing.
Fringe width in millimetres, for a 1.00 m screen distance
| Slit separation d | Violet, 400 nm | Sodium, 589 nm | Red HeNe, 632.8 nm | Visible to the eye at 589 nm |
|---|---|---|---|---|
| 0.05 mm | 8.000 | 11.780 | 12.656 | Yes, very wide |
| 0.10 mm | 4.000 | 5.890 | 6.328 | Yes, wide |
| 0.25 mm | 1.600 | 2.356 | 2.531 | Yes, comfortable |
| 0.50 mm | 0.800 | 1.178 | 1.266 | Yes, comfortable |
| 1.00 mm | 0.400 | 0.589 | 0.633 | Only just; use a magnifier |
| 2.00 mm | 0.200 | 0.295 | 0.316 | No, too fine unaided |
| 5.00 mm | 0.080 | 0.118 | 0.127 | No, needs a microscope |
The formula NCERT does not print, and the three units that decide the answer
β = λD/d is the answer to almost every double-slit question, and it is worth knowing that NCERT does not print it. Chapter 10 of the Class XII Part II textbook gives the fringe POSITIONS — Equation 10.13 puts bright fringes where xd/D = mλ, Equation 10.14 puts dark ones where xd/D = (m + ½)λ — and then says the fringes are equally spaced, leaving the spacing itself to be read off as the gap between consecutive orders. If a question asks you to DERIVE the fringe width rather than quote it, that is the derivation it wants: subtract one position from the next and the order index cancels.
The genuine difficulty of the question is units, not physics. Wavelengths arrive in nanometres, slit separations in millimetres and screen distances in metres, because that is how each is naturally quoted, and the formula wants all three in the same unit. 589 nm is 5.89×10−7 m; 0.5 mm is 5×10−4 m; the ratio is 1.178×10−3; times a 1 m screen distance that is 1.178 mm. Every incorrect answer to this question that is not a wrong formula is a power of ten, so convert everything to metres on a separate line before substituting. This page keeps the three fields in their natural units deliberately, for the same reason.
Then there is d, which is the separation and not the width. A double slit has two numbers: how far apart the slits are, and how wide each one is. The separation sets the fringe SPACING and is the d in this formula. The width sets the single-slit diffraction envelope, which decides how bright each fringe is and how far out the pattern persists before fading away, and it does not appear in λD/d at all. Substituting the width gives an answer out by a factor of ten or more that looks perfectly reasonable. One practical consequence of the envelope is worth carrying: at high orders a fringe can be exactly where the arithmetic puts it and have no light in it, because the envelope has already gone to zero there.
The angular fringe width is the number that travels. λ/d is a property of the slits alone: move the screen and the millimetres change while the angle does not. For the default settings it is 1.178×10−3 rad, which is 4.05 arcminutes — and it is also the health check on the formula, because λD/d rests on setting sinθ and tanθ both equal to θ. Below about 0.1 rad that is good to roughly a tenth of a per cent. Above it the fringes on a flat screen stop being evenly spaced and you need the exact d sinθ = mλ. You reach that regime by making the slit separation comparable with the wavelength, and when d drops below λ there is no first-order fringe at all.
And the verdict on the bar is there because a correct number can describe an impossible experiment. A fringe width of 0.03 mm is arithmetically right and invisible: the unaided eye resolves about a tenth of a millimetre at reading distance, so the pattern is a uniform grey smear. The remedies are all in the formula — push the screen further away, use a smaller slit separation, or put an eyepiece on it, which is what the laboratory apparatus does and why it carries a micrometer. The last of those changes nothing about the physics and everything about whether the experiment works, which is a distinction a bare number cannot make and a verdict can.
Frequently asked questions
What is the formula for fringe width in Young’s double slit experiment?
β = λD/d, where λ is the wavelength, D the distance from the slits to the screen and d the separation BETWEEN the slits. All three must be in the same unit before you substitute. It is worth knowing that NCERT does not print this formula directly: Chapter 10 of Class XII Part II gives the fringe positions as xd/D = mλ and notes that the fringes are equally spaced, so λD/d is the gap between consecutive orders. If you are asked to derive it, subtract one position from the next.
Is d the slit separation or the slit width?
The separation between the two slits. The width of each slit is a different quantity that does not appear in this formula at all — it controls the single-slit diffraction envelope, which decides how bright the fringes are and how far the pattern extends, not how far apart they are. Substituting the width typically gives an answer out by a factor of ten or more, and nothing in the arithmetic will warn you.
What happens to the fringes if the experiment is done under water?
They get closer together, by the refractive index. The frequency of the light cannot change at a boundary, so the wavelength shortens to λ/n, and since β is proportional to λ the fringe width shrinks by the same factor. At n = 1.33 a 1.178 mm pattern becomes 0.886 mm. The common error is to multiply rather than divide; the check is that a shorter wavelength must always give a finer pattern.
Why is the central fringe white in white light?
Because it is the one place where the path difference is zero, and zero is a whole number of wavelengths for every colour at once. Every wavelength therefore has a bright fringe at the centre and they overlap into white. Away from the centre the orders depend on wavelength — a red fringe is 58 per cent wider than a violet one — so the colours separate and the higher orders smear into overlapping coloured bands. The white centre is how you identify order zero, and it was part of Young’s own evidence that the effect was interference.
When does β = λD/d stop being accurate?
When the angles get large, because the formula comes from setting sinθ and tanθ both equal to θ. The practical gauge is the angular fringe width λ/d: below about 0.1 radian the approximation is good to roughly a tenth of a per cent, and above it the fringes on a flat screen are no longer evenly spaced and you need the exact condition d sinθ = mλ. You get there by making the slit separation comparable with the wavelength, and if d is smaller than λ there is no first-order maximum at all because sinθ cannot exceed 1.
How do I make the fringes easier to see?
Three handles, all of them in β = λD/d. Move the screen further away, which scales the fringe width proportionally and is the easiest change to make, at the cost of a dimmer pattern. Use a slide with a smaller slit separation, which also scales it but is fixed by the apparatus you have. Or do not try to see it with the naked eye at all: view it through a travelling microscope or an eyepiece, which is what the laboratory version does and why the apparatus has a micrometer on it. The unaided eye resolves roughly a tenth of a millimetre at reading distance, which is the threshold this page’s bar uses.
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References
- NCERT, Physics Textbook for Class XII, Part II, Chapter 10, Wave Optics, section 10.5 on Young’s experiment (read 11 November 2026). Equation 10.13 places the bright fringes where xd/D = nλ with n = 0, ±1, ±2, …, and Equation 10.14 places the dark fringes where xd/D = (n + ½)λ. The chapter gives no explicit fringe-width formula; it states that the bright and dark fringes are equally spaced, and λD/d follows as the difference between consecutive positions. That is worth recording precisely because every secondary source presents β = λD/d as though it were quoted from the textbook. Citing the chapter is not a claim that any board has approved this page.
- The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019. The metre is defined by fixing the speed of light in vacuum at c = 299 792 458 m/s exactly, so c carries no uncertainty at all: it is a definition, not a measurement, and every refractive index on these pages divides that exact number. The same brochure fixes the Boltzmann and Avogadro constants exactly, which makes the molar gas constant R = 8.314 462 618 153 24 J mol−1 K−1 exact as well — and that is what the speed-of-sound figures here are derived from rather than copied out of a table.
- NCERT, Physics Textbook for Class XII, Part II, Chapter 9, Ray Optics and Optical Instruments (read 11 November 2026). Section 9.2.1 states the Cartesian sign convention: all distances measured from the pole of the mirror or the optical centre of the lens, distances measured in the same direction as the incident light taken as positive, heights above the axis positive. Equation 9.7 is the mirror equation 1/v + 1/u = 1/f and Equation 9.9 gives m = −v/u; Equation 9.23 is the thin lens formula 1/v − 1/u = 1/f and Equation 9.24 gives m = v/u. Table 9.1 lists refractive indices against air with their critical angles: water 1.33 and 48.75°, crown glass 1.52 and 41.14°, diamond 2.42 and 24.41°.
- The wavelengths used in the table are laboratory conventions rather than derived quantities: 589 nm for the sodium doublet (which is really two lines at 588.995 and 589.592 nm, unresolved by this experiment and conventionally quoted as 589), 632.8 nm for the helium-neon laser red line, 532 nm for a frequency-doubled green laser, and 400 nm as a round figure for the violet end of the visible band. The visible range itself is given here as roughly 380 to 750 nm, which is a convention rather than a measurement of anybody’s eye: sensitivity falls away gradually at both ends and the figures quoted in different places differ by tens of nanometres.
- The visibility threshold on this page’s bar is a rule of thumb, and is marked as one. A tenth of a millimetre at normal reading distance corresponds to roughly one arcminute of visual angle, which is the conventional figure for the resolution limit of a good unaided eye. This batch did not read a primary psychophysics source for it, and the real figure depends on contrast, on illumination and on the observer. The band edges at 0.1, 1 and 5 mm are therefore practical guidance about whether an experiment will work rather than measured thresholds, and nothing in the arithmetic above depends on them.