JEE Mains Physics · Wave Optics

Double Slit Fringe Width and Fringe Positions

In Young's double slit the bright fringes sit at whole multiples of the fringe width β = λD/d and the dark ones halfway between, so every position question is a count of fringe widths.

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Why this matters

Twenty-seven PYQs, sixteen of them multiple choice, and four from 2026: the largest page in the chapter. Twelve ask how the fringe width changes with the wavelength, the slit gap or the screen distance, several as true-or-false statements. Five put the whole apparatus in a liquid. Ten locate a fringe, and half of those ask where the bright fringes of two wavelengths first fall on top of each other.

Concept 1 of 3: Fringe width and what it depends on

Moving one fringe along the screen changes the path difference by one wavelength. The path difference grows as yd/D, so the step from one bright fringe to the next is λD/d. Longer waves, a farther screen or closer slits all spread the fringes out. The angle between fringes, λ/d, does not depend on the screen at all.

Definition

Fringe width (bright to bright, or dark to dark): $\beta = \dfrac{\lambda D}{d}$ .
Angular fringe width: $\theta = \dfrac{\lambda}{d}$ , independent of D.
Scaling: $\dfrac{\beta_{2}}{\beta_{1}} = \dfrac{\lambda_{2}}{\lambda_{1}} \cdot \dfrac{D_{2}}{D_{1}} \cdot \dfrac{d_{1}}{d_{2}}$ .
Moving the screen by $\Delta D$ changes the fringe width by $\Delta\beta = \dfrac{\lambda\,\Delta D}{d}$ .
Shorter wavelength (red to violet, orange to blue) gives narrower fringes; the central fringe stays bright.
If d varies with time, the widest fringes come with the smallest gap and the narrowest with the largest.

Fringe width

\beta = \frac{\lambda D}{d}, \qquad \theta = \frac{\lambda}{d}, \qquad \Delta\beta = \frac{\lambda\,\Delta D}{d}

Worked example

In a double-slit set-up, d = 0.5 mm, D = 1.2 m and λ = 600 nm. Find the fringe width and the angular fringe width. The screen is then moved 30 cm closer to the slits. Find the new fringe width.

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The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q19Moderate

Example 1 · Wave Optics · Double Slit Fringe Width and Fringe Positions

Using Young's double slit experiment, a monochromatic light of wavelength

5000\text{Å}

produces fringes of fringe width

0.5\text{ }mm

. If another monochromatic light of wavelength

6000\text{Å}

is used and the separation between the slits is doubled, then the new fringe width will be:

Angular width ignores the screen

Moving the screen changes β but not λ/d. A statement that the angular separation grows as the screen moves away is false.

β falls as d rises

Doubling the slit gap halves the fringe width. Write the full ratio λ₂D₂d₁/(λ₁D₁d₂) and the d's cannot end up the wrong way up.

A percentage change is not a ratio

If β becomes 0.8 of its old value, the change is −20%, not 80%. Say which one the question asks for.

Drill 11 more on fringe width and what it depends on

Concept 2 of 3: The apparatus in a liquid

Fill the space between the slits and the screen with a liquid and the light slows down by μ. Its frequency is fixed, so its wavelength shrinks by μ. Every fringe width and every angle carries λ, so all of them shrink by the same factor μ. Nothing else in the set-up changes.

Definition

In a liquid of index μ: $\lambda' = \dfrac{\lambda}{\mu}$ .
So $\beta' = \dfrac{\beta}{\mu} = \dfrac{\lambda D}{\mu d}$ and the angular width becomes $\dfrac{\lambda}{\mu d}$ .
The central fringe is still bright; the fringes simply crowd closer together.
To get the old fringe width back, the screen must go $\mu$ times farther away.

Fringes in a liquid

\beta' = \frac{\beta}{\mu} = \frac{\lambda D}{\mu d}, \qquad \theta' = \frac{\lambda}{\mu d}

Worked example

In air the fringe width is 1.6 mm. The whole apparatus is put in water of refractive index 4/3. Find the new fringe width, and the factor by which D must change to restore 1.6 mm.

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The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q76Moderate

Example 2 · Wave Optics · Double Slit Fringe Width and Fringe Positions

Young's double slit interference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be :

Divide by μ, do not multiply

The wavelength shrinks in a liquid, so the fringes get narrower. An option larger than the air value is the multiply-by-μ slip.

The given wavelength is the one in air

When a question gives λ in air and asks for β in the liquid, divide λ by μ before using λD/d.

Angles shrink too

The angular width λ/d also carries λ, so it falls by μ as well. It is only the screen distance that it ignores.

Drill 4 more on the apparatus in a liquid

Concept 3 of 3: Fringe positions and two wavelengths

Count in fringe widths. The nth bright fringe is n widths from the centre; the nth dark fringe is half a width short of that. With two colours, each has its own fringe width, and their bright fringes coincide where a whole number of one wavelength equals a whole number of the other.

Definition

nth bright fringe: $y_{n} = n\beta = \dfrac{n\lambda D}{d}$ (the centre is n = 0).
nth dark fringe: $y_{n} = \left(n - \tfrac{1}{2}\right)\beta$ .
Same side: subtract positions. Opposite sides: add them. The nth bright on both sides are $2n\beta$ apart.
Two wavelengths: bright fringes coincide where $n_{1}\lambda_{1} = n_{2}\lambda_{2}$ . Reduce $\dfrac{n_{1}}{n_{2}} = \dfrac{\lambda_{2}}{\lambda_{1}}$ to lowest terms; the least distance from the centre is $\dfrac{n_{1}\lambda_{1}D}{d}$ .
A frequency question works the same way: find λ from the position, then $f = c/\lambda$ .

Positions and coincidence

y_{\text{bright}} = \frac{n\lambda D}{d}, \qquad y_{\text{dark}} = \left(n - \tfrac{1}{2}\right)\frac{\lambda D}{d}, \qquad n_{1}\lambda_{1} = n_{2}\lambda_{2}

Worked example

Light of wavelengths 630 nm and 420 nm falls on slits 0.9 mm apart, with the screen 1.5 m away. Find the least distance from the central fringe where bright fringes of both wavelengths coincide.

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The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q117Moderate

Example 3 · Wave Optics · Double Slit Fringe Width and Fringe Positions

A beam of light consisting of two wavelengths

7000\text{Å}

and

5500\text{Å}

is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is

2.5\text{ }mm

and the distance between the plane of slits and the screen is

150\text{ }cm

. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is

n \times10^{- 5}\text{ }m

. The value of

n

is___

The ratio flips

n₁/n₂ = λ₂/λ₁: the longer wavelength needs the smaller order. Putting λ₁ on top gives a coincidence that does not exist.

Dark fringes are half a width short

The 3rd dark fringe is at 2.5β, not 3β and not 1.5β. Count the bright fringes, then step back half a width.

Reduce the ratio fully

The least distance needs the lowest-terms pair. 630 : 420 is 3 : 2; using 63 : 42 gives a coincidence 21 times too far out.

Drill 9 more on fringe positions and two wavelengths

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Fringe width and what it depends on
Fringe width
$\beta = \frac{\lambda D}{d}, \qquad \theta = \frac{\lambda}{d}, \qquad \Delta\beta = \frac{\lambda\,\Delta D}{d}$
The apparatus in a liquid
Fringes in a liquid
$\beta' = \frac{\beta}{\mu} = \frac{\lambda D}{\mu d}, \qquad \theta' = \frac{\lambda}{\mu d}$
Fringe positions and two wavelengths
Positions and coincidence
$y_{\text{bright}} = \frac{n\lambda D}{d}, \qquad y_{\text{dark}} = \left(n - \tfrac{1}{2}\right)\frac{\lambda D}{d}, \qquad n_{1}\lambda_{1} = n_{2}\lambda_{2}$

Watch out for (9)

Angular width ignores the screen
→ Fringe width and what it depends on
β falls as d rises
→ Fringe width and what it depends on
A percentage change is not a ratio
→ Fringe width and what it depends on
Divide by μ, do not multiply
→ The apparatus in a liquid
The given wavelength is the one in air
→ The apparatus in a liquid
Angles shrink too
→ The apparatus in a liquid
The ratio flips
→ Fringe positions and two wavelengths
Dark fringes are half a width short
→ Fringe positions and two wavelengths
Reduce the ratio fully
→ Fringe positions and two wavelengths

Test yourself on Wave Optics

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