Young’s Double Slit Experiment

Q: What is fringe width in Young's double slit experiment?

Fringe width is the distance between two consecutive bright or dark fringes: beta = lambda D / d. It grows with wavelength and screen distance and shrinks as the slits move apart. The angular fringe width lambda/d does not depend on D. In a liquid of index mu the fringe width becomes beta/mu.

Q: Why are two slits needed instead of two bulbs in YDSE?

Two bulbs are independent sources whose phases change randomly, so their interference averages out. Two slits lit by the same source take their light from one wavefront, so they keep a constant phase difference. They act as coherent sources and give a stationary fringe pattern on the screen.

Q: What happens when a glass slab is placed in front of one slit?

The slab adds an optical path (mu - 1) t to that slit's light, so the zero path difference point moves towards the covered slit. The whole pattern shifts by (mu - 1) t D / d, which equals (mu - 1) t / lambda fringe widths. The fringe width itself does not change.

Q: Why is the central fringe white when white light is used in YDSE?

At the centre of the screen the path difference is zero for every wavelength, so all colours interfere constructively there and combine into white. Away from the centre each colour has a different fringe width, so a few coloured fringes appear with violet nearest the centre, and then the colours overlap into uniform light.

Q: How many maxima can be seen in a double slit experiment?

The path difference cannot exceed d, so the orders satisfy n lambda less than or equal to d. The highest order is the integer part of d/lambda. If d/lambda is a whole number, the highest orders lie at 90 degrees and do not reach a flat screen, so the visible count is 2(d/lambda) - 1.

Q: Why is the edge fringe dark in Lloyd's mirror experiment?

Light reflected from the mirror, which is optically denser than air, suffers a phase change of pi, equal to an extra half wavelength of path. At the mirror edge the geometrical path difference is zero, so the net difference is half a wavelength and the fringe there is dark instead of bright.

Q: Which YDSE questions are common in NEET?

NEET usually asks for the fringe width lambda D / d and how it changes with lambda, D, d or a liquid medium, the position of a bright or dark fringe, the intensity at a given path difference, the shift due to a thin sheet and the white central fringe. Most are single-formula questions with unit conversions.

Q: How is YDSE tested in JEE Main and Advanced?

JEE combines YDSE with other ideas: coincidence of two wavelengths, slabs on both slits, oblique incidence, a displaced source, large-angle maxima where the small-angle formula fails, intensity with unequal slits, and Lloyd's mirror or biprism variants. Drawing the path difference carefully before using any formula is the key skill.

PhysicsWave OpticsFor NEET aspirants

Young's double slit experiment (YDSE) splits one wavefront into two coherent sources, $S_{1}$ and $S_{2}$ , and records their interference as equally spaced bright and dark fringes. The whole experiment rests on one quantity, the path difference $Δ x = \frac{y d}{D}$ , and on one result, the fringe width $β = \frac{λ D}{d}$ . Young's double slit experiment is among the most-asked topics in JEE Main and NEET optics, with questions on fringe width, intensity, slab shift and white light.

On this page1The set-up2Path difference3Fringe positions4Fringe width5Intensity6Large angles7Slab shift8Oblique light9Fringe shapes10White light11Lloyd's mirror and biprism

Key Formulas - Quick Reference

Path difference: $Δ x = d sin θ \approx \frac{y d}{D}$ (for $D ≫ d$ and $y ≪ D$ )
Bright fringes: $y_{n} = n \frac{λ D}{d}$ , $n = 0, \pm 1, \pm 2, \dots$
Dark fringes: $y_{n} = (2 n - 1) \frac{λ D}{2 d}$ , $n = \pm 1, \pm 2, \dots$
Fringe width $β = \frac{λ D}{d}$ ; angular fringe width $θ_{β} = \frac{λ}{d}$ ; in a medium $β^{'} = \frac{β}{μ}$
Intensity on the screen: $I = I_{ma x} cos^{2} (\frac{π y d}{λ D})$ , with $I_{ma x} = 4 I_{0}$ for equal slits
Exact maxima when $d \sim λ$ : $d sin θ = nλ$ ; a flat screen shows only orders with $∣ n ∣ < \frac{d}{λ}$
Coincidence of two wavelengths: $n_{1} λ_{1} = n_{2} λ_{2}$ (first coincidence at the LCM of $β_{1}$ and $β_{2}$ )
Slab of thickness $t$ over one slit: shift $= \frac{( μ - 1 ) t D}{d} = \frac{( μ - 1 ) t}{λ} β$ towards that slit
Oblique incidence at $θ_{0}$ : $Δ x = d sin θ_{0} + d sin θ$ ; central maximum at $θ = θ_{0}$ on the other side
Lloyd's mirror: $d = 2 a$ , centre of the edge fringe dark; biprism: $d = 2 (μ - 1) A l_{1}$ , $D = l_{1} + l_{2}$

1. The Experiment

In 1801 Thomas Young produced a stationary interference pattern of light for the first time. Two ordinary lamps cannot do this because they are not coherent. Young's idea was to take one wavefront and divide it into two.

Figure 1: Young's arrangement. One wavefront from

S_{0}

is divided by

S_{1}

and

S_{2}

, so the two slits act as coherent sources; where their waves overlap, a stationary pattern of bright and dark fringes appears on the screen (fringes drawn from

I \propto cos^{2}

Monochromatic light falls on a narrow slit $S_{0}$ (screen A), which acts as a single source.
The wavefront from $S_{0}$ reaches two narrow slits $S_{1}$ and $S_{2}$ (screen B), a distance $d$ apart and equidistant from $S_{0}$ . Points on one wavefront are in phase, so $S_{1}$ and $S_{2}$ act as coherent sources. This is division of wavefront.
Waves diffracted from $S_{1}$ and $S_{2}$ overlap and interfere on a screen C at distance $D$ ( $D ≫ d$ ).
Where crests meet crests the screen is bright; where crests meet troughs it is dark. The result is a set of equally spaced straight fringes with a bright fringe at the centre.

Conditions for a sustained, clear pattern: coherent sources; same (or nearly same) amplitude for good contrast; narrow slits and a narrow source slit; small $d$ and large $D$ so that fringes are wide enough to see; and monochromatic light (white light gives only a few coloured fringes).

2. Path Difference at a Point on the Screen

Let P be a point at height $y$ above the centre O of the screen. Waves leave $S_{1}$ and $S_{2}$ in phase, so the brightness at P is decided only by the path difference $Δ x = S_{2} P - S_{1} P$ .

Figure 2: For

D ≫ d

the rays

r_{1}

and

r_{2}

are nearly parallel, so the path difference is

S_{2} N = d sin θ

; for small

θ

sin θ \approx tan θ = \frac{y}{D}

gives

Δ x = \frac{y d}{D}

2.1 Exact expression

Δ x = S_{2} P - S_{1} P = D^{2} + (y + \frac{d}{2})^{2} - D^{2} + (y - \frac{d}{2})^{2}

2.2 Approximation I: $D ≫ d$

The rays $r_{1}$ and $r_{2}$ are almost parallel, both at angle $θ$ to the axis. Drop a perpendicular $S_{1} N$ on $S_{2} P$ ; then $N P \approx S_{1} P$ and

Δ x = S_{2} N = d sin θ

2.3 Approximation II: $y ≪ D$ as well

For small $θ$ , $sin θ \approx tan θ = \frac{y}{D}$ , so

★ Must learn

Δ x = \frac{y d}{D}, Δ ϕ = \frac{2 π}{λ} \cdot \frac{y d}{D}

When to stop using $\frac{y d}{D}$ : if the order $n$ is not much smaller than $\frac{d}{λ}$ , or if $y$ is comparable to $D$ , the small-angle form fails. Use $d sin θ = nλ$ and $y = D tan θ$ instead (Solved Examples 3 and 4).

3. Positions of Bright and Dark Fringes

3.1 Bright fringes (maxima)

Constructive interference needs $Δ x = nλ$ :

\frac{y d}{D} = nλ \Rightarrow y_{n} = \frac{nλ D}{d}, n = 0, \pm 1, \pm 2, \dots

$n = 0$ is the central maximum at O (zero path difference), $n = \pm 1$ are the first maxima, and so on.

3.2 Dark fringes (minima)

Destructive interference needs $Δ x = \pm \frac{λ}{2}, \pm \frac{3 λ}{2}, \pm \frac{5 λ}{2}, \dots$ :

y_{n} = (2 n - 1) \frac{λ D}{2 d}, n = 1, 2, 3, \dots above O (and the same distances below)

$n = 1$ is the first minimum (at $\frac{β}{2}$ ), $n = 2$ the second (at $\frac{3 β}{2}$ ), and so on. There is no "zeroth" minimum.

Figure 3: Equally spaced fringes. Bright fringes

B_{n}

sit at

y = n β

, dark fringes

D_{n}

halfway between, and the intensity follows

I = 4 I_{0} cos^{2} (\frac{π y}{β})

4. Fringe Width

★ Must learnFringe width

β

is the distance between two consecutive bright fringes (or two consecutive dark fringes) on the same side of the centre:

β = y_{n + 1} - y_{n} = \frac{λ D}{d}

The angular fringe width is

θ_{β} = \frac{β}{D} = \frac{λ}{d}

(independent of

D

All fringes (bright and dark) have the same width $β$ , a bright and the next dark fringe are $\frac{β}{2}$ apart, and the $n$ th bright fringe is simply at $n β$ .

Figure 4: Same slits, same screen, three colours. Fringe width

β = \frac{λ D}{d}

grows with wavelength, so red fringes are about

1.5

times as wide as violet ones. All colours are bright at the centre (dashed line), so white light gives a white central fringe, a few coloured fringes, then a washed-out whitish glow.

4.1 What changes $β$ ?

Change	Effect on $β = \frac{λ D}{d}$	Reason
Increase $λ$ (violet to red)	Increases	$β \propto λ$
Move the screen away (increase $D$ )	Increases	$β \propto D$ ; angular width $\frac{λ}{d}$ unchanged
Bring slits closer (decrease $d$ )	Increases	$β \propto \frac{1}{d}$
Immerse the whole set-up in a liquid of index $μ$	Decreases to $\frac{β}{μ}$	$λ$ becomes $\frac{λ}{μ}$
Place a thin slab over one slit	No change	Pattern shifts as a whole (Section 7)
Make one slit brighter	No change	Only contrast changes (Figure 5)
Widen the source slit too much	Fringes fade	Condition $\frac{s}{S} < \frac{λ}{d}$ fails ( $s$ = source width, $S$ = source-to-slit distance)

Exam Trick

"Ratio questions" in one line. Write $β \propto \frac{λ D}{μ d}$ and change only what the question changes. Example: $d$ halved and $D$ doubled makes $β$ four times; the whole set-up in water ( $μ = \frac{4}{3}$ ) makes it $\frac{3}{4}$ of the value in air.

Key idea

Everything scales with

β = \frac{λ D}{d}

: bright fringes at

n β

, dark fringes halfway between, all equally spaced as long as the angles are small.

5. Intensity Distribution on the Screen

With slit intensities $I_{1}$ and $I_{2}$ and $Δ ϕ = \frac{2 π}{λ} \cdot \frac{y d}{D}$ :

I = I_{1} + I_{2} + 2 I_{1} I_{2} cos (\frac{2 π y d}{λ D})

For identical slits ( $I_{1} = I_{2} = I_{0}$ ):

★ Must learn

I = 4 I_{0} cos^{2} (\frac{π y d}{λ D}) = I_{ma x} cos^{2} (\frac{π y}{β})

Bright fringes have

4 I_{0}

, dark fringes zero, and at a quarter-fringe (

Δ x = \frac{λ}{4}

) the intensity is

\frac{I _{ma x}}{2}

Figure 5: Fringe positions do not depend on slit intensities, but contrast does. Equal slits give

I_{min} = 0

; slits of

4 I_{0}

and

I_{0}

give

I_{ma x} = 9 I_{0}

and

I_{min} = I_{0}

(poorer contrast).

Exam Trick

Path to intensity in one step: $\frac{I}{I _{ma x}} = cos^{2} (\frac{π Δ x}{λ})$ . So $Δ x = \frac{λ}{3}$ gives $cos^{2} 6 0^{\circ} = \frac{1}{4}$ , $Δ x = \frac{λ}{4}$ gives $\frac{1}{2}$ , $Δ x = \frac{λ}{6}$ gives $\frac{3}{4}$ .

Quick Recall: tap to check

d

is halved and

D

doubled. What happens to

β

It becomes

4 β

, since

β \propto \frac{D}{d}

The whole apparatus is put in water (

μ = \frac{4}{3}

). New fringe width?

\frac{3}{4} β

: the wavelength in water is

\frac{λ}{μ}

Equal slits. Intensity where

Δ x = \frac{λ}{3}

I_{ma x} cos^{2} 6 0^{\circ} = \frac{I _{ma x}}{4}

Distance between the 3rd bright fringe and the 2nd dark fringe on the same side?

3 β - 1.5 β = 1.5 β

6. Large Angles and the Number of Fringes

The path difference can never exceed $d$ (its value at $θ = 9 0^{\circ}$ ). So the orders that exist obey $∣ n ∣ λ \leq d$ :

Highest order of maximum: $n_{ma x} = ⌊ \frac{d}{λ} ⌋$ ; total maxima in all directions $2 n_{ma x} + 1$ . If $\frac{d}{λ}$ is a whole number, the orders $\pm n_{ma x}$ lie at $θ = 9 0^{\circ}$ and never reach a flat screen.
Highest order of minimum: largest $n$ with $(2 n - 1) \frac{λ}{2} < d$ ; total minima $2 n_{min}$ .
Position of an order on a flat screen: $sin θ = \frac{nλ}{d}$ , then $y = D tan θ$ .

Figure 6: When

d

is only a few wavelengths the small-angle formula fails. Use

d sin θ = nλ

: for

d = 2 λ

the orders are at

0^{\circ}

\pm 3 0^{\circ}

and

\pm 9 0^{\circ}

(the last run along the slit plane).

6.1 Two wavelengths together

If light has two wavelengths $λ_{1}$ and $λ_{2}$ , each makes its own pattern. Bright fringes coincide where $y = \frac{n _{1} λ _{1} D}{d} = \frac{n _{2} λ _{2} D}{d}$ , that is

n_{1} λ_{1} = n_{2} λ_{2} \Rightarrow \frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} in lowest terms

Figure 7: Bright fringes of

6000 \overset{˚}{A}

(

β_{1} = 0.6 mm

) and

4500 \overset{˚}{A}

(

β_{2} = 0.45 mm

) first coincide at the LCM,

1.8 mm

, where

n_{1} = 3

and

n_{2} = 4

(Solved Example 5).

7. Optical Path and Fringe Shift by a Thin Slab

7.1 Geometrical path and optical path

A light wave $E = E_{0} sin (ω t - k x + ϕ)$ changes phase by $k x$ over a distance $x$ . In a medium of index $μ$ the speed is $\frac{c}{μ}$ , so

Δ ϕ = \frac{ω}{v} x = \frac{ω}{c} (μx) = \frac{2 π}{λ _{0}} (μx)

★ Must learnOptical path

= μx

. A distance

x

in a medium causes the same phase change as a distance

μx

in vacuum. Always compute path differences as optical path differences and use the vacuum wavelength

λ_{0}

Δ ϕ = \frac{2 π}{λ _{0}} Δ (optical path)

. Equivalently

Δ ϕ = \frac{2 π}{λ} x

with

λ = \frac{λ _{0}}{μ}

in the medium.

7.2 Slab in front of one slit

Figure 8: A slab over

S_{1}

adds optical path

(μ - 1) t

, so zero path difference now occurs nearer

S_{1}

: the whole pattern shifts towards the covered slit by

\frac{( μ - 1 ) t D}{d}

without changing

β

A slab of thickness $t$ replaces a length $t$ of air, so the optical path $S_{1} P$ increases by $μ t - t = (μ - 1) t$ .
Path difference at P: $Δ x = S_{2} P - [S_{1} P + (μ - 1) t] = \frac{y d}{D} - (μ - 1) t$ .
Central maximum ( $Δ x = 0$ ):
$y_{0} = \frac{( μ - 1 ) t D}{d} = \frac{( μ - 1 ) t}{λ} β$
Every fringe moves by the same amount, so $β$ is unchanged. Number of fringes that cross O: $N = \frac{( μ - 1 ) t}{λ}$ .

Exam Trick

The pattern chases the slab. The slab slows light from the covered slit, so the other slit must travel a longer geometrical path to "catch up": the central fringe moves towards the covered slit. With slabs on both slits, use the net extra optical path $(μ_{1} - 1) t_{1} - (μ_{2} - 1) t_{2}$ .

Slab over one slit

Only one path gets extra optical length $(μ - 1) t$ . The whole pattern shifts by $\frac{( μ - 1 ) t D}{d}$ towards the covered slit; $β$ is unchanged.

Whole set-up in a liquid

Both paths are in the liquid, so no extra path difference appears: no shift. The wavelength becomes $\frac{λ}{μ}$ , so $β^{'} = \frac{β}{μ}$ (fringes squeeze).

Key idea

A slab moves the pattern towards the covered slit by

\frac{( μ - 1 ) t D}{d}

but never changes the fringe width.

8. YDSE with Oblique Incidence

If parallel light falls on the slits at angle $θ_{0}$ to the axis, the wave reaches one slit earlier. Before the slits there is already a path difference $d sin θ_{0}$ .

Figure 9: Oblique incidence at

θ_{0}

. The light reaching

S_{2}

has already travelled

d sin θ_{0}

more, so zero net path difference occurs at

θ = θ_{0}

below the axis: the pattern shifts by

D tan θ_{0}

Point P on the screen	Path difference $Δ x$ (for $d ≪ D$ )
On the side of $S_{1}$ (above O), at angle $θ_{1}$	$d sin θ_{0} + d sin θ_{1}$
At O	$d sin θ_{0}$
Between O and O', at angle $θ_{2}$	$d sin θ_{0} - d sin θ_{2}$
Beyond O', at angle $θ_{2}$	$d sin θ_{2} - d sin θ_{0}$

The central maximum is where $Δ x = 0$ : $θ_{2} = θ_{0}$ , on the line of the incident light. The pattern shifts by $D tan θ_{0}$ and the fringe width is unchanged.

JEE Advanced

Source moved off the axis. If the point source $S_{0}$ is moved a distance $b$ above the axis, at distance $L$ from the slits, the waves reach $S_{1}$ and $S_{2}$ with path difference $\frac{b d}{L}$ . The central maximum moves to $\frac{b D}{L}$ below O (opposite to the source), again with unchanged $β$ . A wide source is many such points; their patterns wash out unless $\frac{s}{L} < \frac{λ}{d}$ .

9. Shapes of Fringes

A fringe is the set of points on the screen with the same path difference. Its shape therefore depends on the sources.

Figure 10: Fringe shape = the curve of constant path difference on the screen. Slits give straight lines; point sources along the axis give circles; point sources side by side give hyperbolas (nearly straight near the centre).

Sources	Fringe shape on the screen
Two long parallel slits (normal YDSE)	Straight lines parallel to the slits
Two point sources on a line perpendicular to the screen	Concentric circles (path difference $d cos θ$ ); the centre has the highest order
Two point sources on a line parallel to the screen	Hyperbolas, nearly straight near the centre

10. YDSE with White Light

At the centre $Δ x = 0$ for every wavelength, so every colour is bright there: the central fringe is white. Elsewhere each colour has its own $β = \frac{λ D}{d}$ . Violet has the smallest $β$ , so the first minimum and the first coloured edge on either side of the centre are violet-side, with red on the outer edge. After a few fringes the colours overlap and the screen looks uniformly whitish (bottom strip of Figure 4).

Finding the zero order in practice: with monochromatic light all fringes look alike, so the central fringe cannot be found. Switch to white light: the one white fringe marks $Δ x = 0$ . This is how the zero-order fringe (and the slab shift) is located in practice.

11. Lloyd's Mirror

Light from a narrow source S reaches the screen in two ways: directly, and after reflection at grazing incidence from a long plane mirror. The reflected light appears to come from the image $S^{'}$ , so $S$ and $S^{'}$ act as coherent sources a distance $d = 2 a$ apart ( $a$ = height of S above the mirror).

Figure 11: Lloyd's mirror. Direct light from

S

and reflected light (which appears to come from the image

S^{'}

) overlap in the shaded region;

d = 2 a

. Reflection adds a phase of

π

, so the fringe at the mirror plane is dark.

When the screen touches the end of the mirror, the edge of the pattern (where the geometrical path difference is zero) is dark, not bright. The direct beam has no phase change, so the reflected beam must suffer a phase change of $π$ (path $\frac{λ}{2}$ ) on reflection from the denser glass. Hence the conditions are reversed:

S^{'} P - S P = nλ (dark), S^{'} P - S P = (n + \frac{1}{2}) λ (bright)

The fringe width is still $β = \frac{λ D}{2 a}$ .

12. Fresnel's Biprism

A biprism is two thin prisms (angle $A$ , index $μ$ ) joined base to base. Each half bends light from the slit S towards the axis by $δ = (μ - 1) A$ , so the two beams appear to come from virtual sources $S_{1}$ and $S_{2}$ .

Figure 12: Fresnel's biprism (angles exaggerated). Each half deviates light by

δ = (μ - 1) A

, creating virtual sources

S_{1}

S_{2}

a distance

d = 2 (μ - 1) A l_{1}

apart; fringes form where the beams overlap.

With $l_{1}$ = slit-to-biprism distance and $l_{2}$ = biprism-to-screen distance:

d = 2 l_{1} δ = 2 (μ - 1) A l_{1}, D = l_{1} + l_{2}, β = \frac{λ ( l _{1} + l _{2} )}{2 ( μ - 1 ) A l _{1}}

Fringes are straight lines, seen only in the overlap region BC.

JEE Advanced

Displacement method for $d$ . A convex lens placed between the biprism and the eyepiece gives sharp images of $S_{1}$ , $S_{2}$ at two lens positions, $d_{1}$ and $d_{2}$ apart. By the lens conjugate property the magnifications are reciprocal, so $d = d_{1} d_{2}$ (Solved Example 13).

Key idea

Lloyd's mirror and the biprism are YDSE in disguise: find

d

(the gap between the real or virtual sources) and

D

, then use

β = \frac{λ D}{d}

. Lloyd's mirror adds a

π

shift.

Quick Recall: tap to check

A thin sheet covers the upper slit. Which way does the pattern move?

Upwards, towards the covered slit; the fringe width does not change.

Why is the central fringe white when white light is used?

Δ x = 0

every wavelength is bright, so all colours add to white.

What is the fringe at the mirror edge in Lloyd's mirror?

Dark: reflection adds a phase of

π

How many maxima exist in all directions for

d = 3.5 λ

n_{ma x} = 3

, so

2 (3) + 1 = 7

maxima.

Figure 13: YDSE in three moves: choose small-angle (

β = \frac{λ D}{d}

) or exact (

d sin θ = nλ

), correct for a slab, a liquid or oblique light, then

I = I_{ma x} cos^{2} \frac{π Δ x}{λ}

Figure 14: Revision map of YDSE: set-up, fringe positions,

β = \frac{λ D}{d}

, intensity, pattern shifts and the variants (white light, Lloyd's mirror, biprism).

13. Solved Examples

Solved Example 1

In a YDSE performed with wavelength

λ = 5890 \overset{˚}{A}

the angular fringe width is

0.4 0^{\circ}

. What is the angular fringe width if the entire set-up is immersed in water (

μ = \frac{4}{3}

Solution:

Angular fringe width $θ_{β} = \frac{λ}{d}$ . In water $λ_{w} = \frac{λ}{μ}$ , so $\frac{θ _{w}}{θ _{ai r}} = \frac{λ _{w}}{λ} = \frac{1}{μ} = \frac{3}{4}$ .

$θ_{w} = 0.4 0^{\circ} \times \frac{3}{4} = 0.3 0^{\circ}$ .

Answer: $0.3 0^{\circ}$ .

Solved Example 2

A beam of light of two wavelengths,

6500 \overset{˚}{A}

and

5200 \overset{˚}{A}

, is used in a YDSE. The slit separation is

2 mm

and the screen is

120 cm

away. Find the least distance from the central maximum where bright fringes of both wavelengths coincide.

Solution:

Let the $m$ th bright of $6500 \overset{˚}{A}$ coincide with the $n$ th bright of $5200 \overset{˚}{A}$ : $m \times 6500 = n \times 5200 \Rightarrow \frac{m}{n} = \frac{5200}{6500} = \frac{4}{5}$ .

Smallest values: $m = 4$ , $n = 5$ .

$y = \frac{4 \times 6500 \times 1 0 ^{- 10} \times 1.2}{2 \times 1 0 ^{- 3}} = 1.56 \times 1 0^{- 3} m$ .

Answer: $1.56 mm$ (4th bright of $6500 \overset{˚}{A}$ = 5th bright of $5200 \overset{˚}{A}$ ).

Solved Example 3

In a YDSE,

D = 1 m

d = 1 mm

and

λ = \frac{1}{2} mm

. (i) Find the distance between the first and the central maxima on the screen. (ii) Find the number of maxima and minima.

Solution:

Here $\frac{d}{λ} = 2$ , so $n$ is not much smaller than $\frac{d}{λ}$ : the formula $\frac{y d}{D}$ cannot be used. Use $d sin θ = nλ$ (Figure 6).

(i) First maximum: $sin θ = \frac{λ}{d} = \frac{1}{2} \Rightarrow θ = 3 0^{\circ}$ , so $y = D tan 3 0^{\circ} = \frac{1}{3} m \approx 0.577 m$ .

(ii) $Δ x_{ma x} = d = 2 λ$ . Maxima: $n = 0, \pm 1, \pm 2$ , i.e. 5 in all directions, but $n = \pm 2$ are at $θ = 9 0^{\circ}$ and never reach a flat screen, so 3 maxima appear on the screen. Minima: $(2 n - 1) \frac{λ}{2} < 2 λ$ gives $n = 1, 2$ on each side ( $θ = 14. 5^{\circ}$ and $48. 6^{\circ}$ ), so 4 minima.

Answer: (i) $\frac{1}{3} m$ ; (ii) 3 maxima on the screen (5 counting the grazing $\pm 9 0^{\circ}$ orders) and 4 minima.

Solved Example 4

Monochromatic light of wavelength

5000 \overset{˚}{A}

is used in a YDSE with

d = 1 mm

and

D = 1 m

. The intensities at the two slits are

I_{1} = 4 I_{0}

and

I_{2} = I_{0}

. Find (i) the fringe width, (ii) the distance of the 5th minimum from the central maximum, (iii) the intensity at

y = \frac{1}{3} mm

, (iv) the distance of the 1000th maximum, (v) the distance of the 5000th maximum.

Solution:

(i) $β = \frac{λ D}{d} = \frac{5 \times 1 0 ^{- 7} \times 1}{1 0 ^{- 3}} = 0.5 mm$ .

(ii) $y = (2 n - 1) \frac{β}{2}$ with $n = 5$ : $y = 9 \times 0.25 = 2.25 mm$ .

(iii) $y ≪ D$ , so $Δ x = \frac{y d}{D} = \frac{( 1/3 ) \times 1 0 ^{- 3} \times 1 0 ^{- 3}}{1} = \frac{1}{3} μ m = \frac{2 λ}{3}$ , and $Δ ϕ = \frac{4 π}{3}$ .

$I = 4 I_{0} + I_{0} + 2 4 I_{0} \cdot I_{0} cos \frac{4 π}{3} = 5 I_{0} - 2 I_{0} = 3 I_{0}$ .

(iv) $\frac{d}{λ} = 2000$ , and $n = 1000$ is not much smaller, so use $d sin θ = 1000 λ$ : $sin θ = \frac{1}{2}$ , $θ = 3 0^{\circ}$ , $y = D tan 3 0^{\circ} = \frac{1}{3} m$ .

(v) $n_{ma x} \leq \frac{d}{λ} = 2000$ , so a 5000th maximum does not exist.

Answer: $0.5 mm$ ; $2.25 mm$ ; $3 I_{0}$ ; $\frac{1}{3} m \approx 0.577 m$ ; not possible.

Solved Example 5

Light of wavelengths

6000 \overset{˚}{A}

and

4500 \overset{˚}{A}

is used in a YDSE with

D = 1 m

and

d = 1 mm

. Find the least distance from the central maximum where bright fringes of the two wavelengths coincide.

Solution:

$β_{1} = \frac{6000 \times 1 0 ^{- 10} \times 1}{1 0 ^{- 3}} = 0.6 mm$ and $β_{2} = \frac{4500 \times 1 0 ^{- 10} \times 1}{1 0 ^{- 3}} = 0.45 mm$ .

Coincidence needs $y = n_{1} β_{1} = n_{2} β_{2}$ ; the smallest such $y$ is the LCM of $0.6 mm$ and $0.45 mm$ , which is $1.8 mm$ .

Answer: $1.8 mm$ , where the 3rd maximum of $6000 \overset{˚}{A}$ meets the 4th maximum of $4500 \overset{˚}{A}$ (Figure 7).

Solved Example 6

In a YDSE with

d = 1 mm

and

D = 1 m

, slabs (

t = 1 μ m

μ = 3

) and (

t = 0.5 μ m

μ = 2

) are placed in front of the upper and the lower slit respectively. Find the shift of the fringe pattern.

Solution:

Optical path from the upper slit: $S_{1} P + (3 - 1) (1 μ m) = S_{1} P + 2 μ m$ .

Optical path from the lower slit: $S_{2} P + (2 - 1) (0.5 μ m) = S_{2} P + 0.5 μ m$ .

$Δ x = (S_{2} P + 0.5) - (S_{1} P + 2) = \frac{y d}{D} - 1.5 μ m$ . Central maximum: $Δ x = 0 \Rightarrow y = \frac{1.5 \times 1 0 ^{- 6} \times 1}{1 0 ^{- 3}} = 1.5 mm$ .

Answer: the whole pattern shifts $1.5 mm$ upwards (towards the slit with the larger extra optical path).

Solved Example 7

Interference fringes are produced by Young's method with light of wavelength

6000 \overset{˚}{A}

. The slit separation is

2 mm

and the slit-screen distance is

10 cm

. When a transparent plate

0.5 mm

thick is placed over one slit, the pattern shifts by

5 mm

. Find the refractive index of the plate.

Solution:

Given: $d = 2 \times 1 0^{- 3} m$ , $D = 0.10 m$ , $t = 0.5 \times 1 0^{- 3} m$ , shift $x_{0} = 5 \times 1 0^{- 3} m$ .

$x_{0} = \frac{( μ - 1 ) t D}{d} \Rightarrow μ - 1 = \frac{x _{0} d}{t D} = \frac{5 \times 1 0 ^{- 3} \times 2 \times 1 0 ^{- 3}}{0.5 \times 1 0 ^{- 3} \times 0.10} = 0.2$ .

Answer: $μ = 1.2$ . (The wavelength is not needed: the shift of the central fringe does not depend on $λ$ .)

Solved Example 8

In a YDSE, light of wavelength

5000 \overset{˚}{A}

emerges in phase from two slits

3 \times 1 0^{- 7} m

apart. A transparent sheet of thickness

1.5 \times 1 0^{- 7} m

and refractive index

1.17

is placed over one slit. Where does the central maximum now appear?

Solution:

The sheet adds optical path $(μ - 1) t$ ; the central maximum moves to the direction where $d sin θ = (μ - 1) t$ .

$sin θ = \frac{( 1.17 - 1 ) \times 1.5 \times 1 0 ^{- 7}}{3 \times 1 0 ^{- 7}} = 0.085 \Rightarrow θ = sin^{- 1} (0.085) = 4.8 8^{\circ}$ .

Here $d$ is comparable to $λ$ , so the answer is given as an angle. If $D$ were given, the shift for small angles would be $Y = \frac{D ( μ - 1 ) t}{d}$ .

Answer: at $4.8 8^{\circ}$ from the axis, on the side of the covered slit.

Solved Example 9

In a YDSE with

D = 1 m

and

d = 1 mm

, light of wavelength

500 nm

is incident at

0.5 7^{\circ}

to the axis of symmetry, whose centre on the screen is O. Find (i) the position of the central maximum, (ii) the intensity at O in terms of the central-maximum intensity

I_{0}

, (iii) the number of maxima between O and the central maximum.

Solution:

$θ_{0} = 0.5 7^{\circ} \approx \frac{0.57}{57} rad = 1 0^{- 2} rad$ .

(i) Central maximum at $θ = θ_{0}$ on the far side: $y = - D tan θ_{0} \approx - 1 \times 1 0^{- 2} m = - 1 cm$ (1 cm below O, Figure 9).

(ii) At O, $Δ x = d sin θ_{0} \approx 1 0^{- 3} \times 1 0^{- 2} = 1 0^{- 5} m = 10000 nm = 20 λ$ . So O is the 20th maximum and its intensity is $I_{0}$ .

(iii) Between the central maximum ( $n = 0$ ) and O ( $n = 20$ ) lie orders $1$ to $19$ .

Answer: 1 cm below O; $I_{0}$ ; 19 maxima.

Solved Example 10

In Lloyd's mirror experiment, 10 fringes occupy

1.5 mm

. The source-screen distance is

1.25 m

and

λ = 6000 \overset{˚}{A}

. Find the distance of the source from the plane of the mirror.

Solution:

$β = \frac{1.5}{10} = 0.15 mm$ . From $β = \frac{λ D}{d}$ : $d = \frac{λ D}{β} = \frac{6 \times 1 0 ^{- 7} \times 1.25}{0.15 \times 1 0 ^{- 3}} = 5 \times 1 0^{- 3} m = 5 mm$ .

The source and its image are $d = 2 a$ apart, so $a = \frac{d}{2}$ .

Answer: $2.5 mm$ above the mirror.

Solved Example 11

In a biprism experiment the slit is lit with light of wavelength

4800 \overset{˚}{A}

. The slit-biprism distance is

20 cm

and the biprism-eyepiece distance is

80 cm

. If the virtual sources are

0.3 cm

apart, find the distance between the 5th bright band on one side of the central band and the 4th dark band on the other side.

Solution:

$D = 20 + 80 = 100 cm = 1 m$ , $d = 3 \times 1 0^{- 3} m$ , so $β = \frac{4.8 \times 1 0 ^{- 7} \times 1}{3 \times 1 0 ^{- 3}} = 0.16 mm$ .

5th bright on one side: $5 β = 0.80 mm$ . 4th dark on the other side: $(2 \times 4 - 1) \frac{β}{2} = 3.5 β = 0.56 mm$ .

They are on opposite sides, so the distances add: $0.80 + 0.56 = 1.36 mm$ .

Answer: $1.36 mm$ ( $8.5 β$ ).

Solved Example 12

In a biprism experiment the fringe width is

0.4 mm

. When the eyepiece is moved

30 cm

away from the biprism, the fringe width increases by 50%. If the virtual sources are

0.6 mm

apart, find the wavelength.

Solution:

$β_{1} = \frac{λ D _{1}}{d}$ and $β_{2} = 1.5 β_{1} = \frac{λ ( D _{1} + 0.3 )}{d}$ . Dividing: $\frac{D _{1} + 0.3}{D _{1}} = 1.5 \Rightarrow D_{1} = 0.6 m$ .

$λ = \frac{β _{1} d}{D _{1}} = \frac{0.4 \times 1 0 ^{- 3} \times 0.6 \times 1 0 ^{- 3}}{0.6} = 4 \times 1 0^{- 7} m$ .

Answer: $4000 \overset{˚}{A}$ .

Solved Example 13

Interference fringes from a Fresnel biprism are viewed in the focal plane of a reading microscope

100 cm

from the slit. A lens placed between the biprism and the microscope gives two images of the slit pair at two positions,

4.05 mm

apart in one case and

2.90 mm

in the other. With

λ = 5893 \overset{˚}{A}

, find the fringe width.

Solution:

Displacement method: $d = d_{1} d_{2} = 4.05 \times 2.90 mm = 11.745 mm = 3.43 mm$ .

$β = \frac{λ D}{d} = \frac{5893 \times 1 0 ^{- 10} \times 1}{3.43 \times 1 0 ^{- 3}} = 1.72 \times 1 0^{- 4} m$ .

Answer: $β \approx 0.172 mm$ ( $0.0172 cm$ ).

Solved Example 14

In a YDSE the slit separation is halved and the screen distance is doubled. The fringe width becomes
(A) the same
(B) double
(C) four times
(D) half

Solution:

$β = \frac{λ D}{d} \Rightarrow β^{'} = \frac{λ ( 2 D )}{d /2} = 4 β$ .

Answer: (C). Four times.

Solved Example 15

In a YDSE with identical slits the maximum intensity is

I_{ma x}

. Find the intensity at a point where the path difference is

\frac{λ}{4}

Solution:

$Δ ϕ = \frac{2 π}{λ} \cdot \frac{λ}{4} = \frac{π}{2}$ , and $I = I_{ma x} cos^{2} \frac{Δ ϕ}{2} = I_{ma x} cos^{2} 4 5^{\circ}$ .

Answer: $\frac{I _{ma x}}{2}$ .

Solved Example 16

A glass sheet (

μ = 1.5

) of thickness

6 μ m

is placed over one slit of a YDSE using light of

600 nm

. How many fringes shift across the centre? If

β = 0.4 mm

, find the shift.

Solution:

$N = \frac{( μ - 1 ) t}{λ} = \frac{0.5 \times 6 \times 1 0 ^{- 6}}{6 \times 1 0 ^{- 7}} = 5$ .

Shift $= N β = 5 \times 0.4 = 2 mm$ towards the covered slit.

Answer: 5 fringes; $2 mm$ .

Solved Example 17

In a YDSE (

d = 1 mm

D = 1 m

) the point source is moved

1 mm

above the axis. It is

20 cm

from the slits. Where is the central maximum now?

Solution:

Path difference before the slits: $\frac{b d}{L} = \frac{1 0 ^{- 3} \times 1 0 ^{- 3}}{0.2} = 5 \times 1 0^{- 6} m$ (the upper slit is nearer the source).

After the slits the lower slit must make up this difference: $\frac{y d}{D} = \frac{b d}{L} \Rightarrow y = \frac{b D}{L} = \frac{1 0 ^{- 3} \times 1}{0.2} = 5 \times 1 0^{- 3} m$ .

Answer: $5 mm$ below O, on the opposite side to the source displacement.

Solved Example 18

In a YDSE a thin transparent sheet is placed in front of one slit. Then
(A) the fringe width increases
(B) the fringe width decreases
(C) the pattern shifts but the fringe width is unchanged
(D) the fringes disappear

Solution:

The sheet adds the same extra optical path $(μ - 1) t$ to every point of the screen, so the condition for each fringe is met at a shifted position: $y_{n} = \frac{nλ D}{d} + \frac{( μ - 1 ) t D}{d}$ . The spacing between neighbours is still $\frac{λ D}{d}$ .

Answer: (C). The whole pattern shifts towards the covered slit; $β$ stays $\frac{λ D}{d}$ . (Fringes vanish only if the sheet is so thick that coherence is lost, which is not the case for thin sheets.)

Practice Questions

In a YDSE, $λ = 600 nm$ , $d = 0.3 mm$ , $D = 1.5 m$ . Find $β$ and the position of the 3rd dark fringe.Answer: $β = 3 mm$ ; $y = 2.5 β = 7.5 mm$ .
The fringe width in air is $0.8 mm$ . Find it when the apparatus is in a liquid of $μ = 1.6$ .Answer: $0.5 mm$ .
The 10th bright fringe of $λ_{1} = 600 nm$ coincides with the 12th bright fringe of $λ_{2}$ . Find $λ_{2}$ .Answer: $500 nm$ .
A YDSE with identical slits has $I_{ma x} = I$ . Find the intensity at a point where the path difference is $\frac{λ}{3}$ .Answer: $I cos^{2} 6 0^{\circ} = \frac{I}{4}$ .
How thick a mica sheet ( $μ = 1.6$ ) shifts the central fringe by 3 fringes for $λ = 600 nm$ ?Answer: $t = \frac{3 λ}{μ - 1} = 3 μ m$ .
In a YDSE $d = 5 λ$ . How many maxima appear on a large flat screen?Answer: $n = 0, \pm 1, \dots, \pm 4$ : 9 maxima ( $n = \pm 5$ are at $9 0^{\circ}$ ).
Why is the central fringe white but the others coloured when white light is used?Answer: At the centre $Δ x = 0$ for every wavelength; elsewhere each colour has its own $β \propto λ$ .

Common Mistakes to Avoid

Watch out

Using $\frac{y d}{D}$ when the order is comparable to $\frac{d}{λ}$ or $y$ is comparable to $D$ ; switch to $d sin θ = nλ$ and $y = D tan θ$ .
Placing the first dark fringe at $β$ . Dark fringes lie halfway between bright ones: the first at $\frac{β}{2}$ , the $n$ th at $(n - \frac{1}{2}) β$ .
Moving the pattern away from the slab, or using $μ t$ instead of $(μ - 1) t$ for its extra optical path in air. The central fringe shifts towards the covered slit by $\frac{( μ - 1 ) t D}{d}$ .
Expecting a slab (or unequal slit intensities) to change $β$ . Only $λ$ , $D$ , $d$ and the medium change it.
Adding distances of fringes on the same side when the question says 'other side' (or vice versa). Draw the centre first.
Counting the maxima at exactly $9 0^{\circ}$ as visible on a flat screen.
Taking the edge fringe in Lloyd's mirror as bright. Reflection adds a phase of $π$ , so it is dark.
Forgetting to convert units: Å to m ( $1 0^{- 10}$ ), mm to m, cm to m before substituting in $\frac{λ D}{d}$ .

Frequently Asked Questions

What is fringe width in Young's double slit experiment?

Fringe width is the distance between two consecutive bright (or dark) fringes: $β = λ D / d$ . It grows with wavelength and screen distance and shrinks as the slits move apart. The angular fringe width $λ / d$ does not depend on $D$ . In a liquid of index $μ$ the fringe width becomes $β / μ$ .

Why are two slits needed instead of two bulbs in YDSE?

Two bulbs are independent sources whose phases change randomly, so their interference averages out. Two slits lit by the same source take their light from one wavefront, so they keep a constant phase difference. They act as coherent sources and give a stationary fringe pattern on the screen.

What happens when a glass slab is placed in front of one slit?

The slab adds an optical path $(μ - 1) t$ to that slit's light, so the zero path difference point moves towards the covered slit. The whole pattern shifts by $(μ - 1) t D / d$ , which equals $(μ - 1) t / λ$ fringe widths. The fringe width itself does not change.

Why is the central fringe white when white light is used in YDSE?

At the centre of the screen the path difference is zero for every wavelength, so all colours interfere constructively there and combine into white. Away from the centre each colour has a different fringe width, so a few coloured fringes appear with violet nearest the centre, and then the colours overlap into uniform light.

How many maxima can be seen in a double slit experiment?

The path difference cannot exceed $d$ , so the orders satisfy $nλ \leq d$ . The highest order is the integer part of $d / λ$ . If $d / λ$ is a whole number, the highest orders lie at 90 degrees and do not reach a flat screen, so the visible count is $2 (d / λ) - 1$ .

Why is the edge fringe dark in Lloyd's mirror experiment?

Light reflected from the mirror, which is optically denser than air, suffers a phase change of pi, equal to an extra half wavelength of path. At the mirror edge the geometrical path difference is zero, so the net difference is half a wavelength and the fringe there is dark instead of bright.

Which YDSE questions are common in NEET?

NEET usually asks for the fringe width $λ D / d$ and how it changes with $λ$ , $D$ , $d$ or a liquid medium, the position of a bright or dark fringe, the intensity at a given path difference, the shift due to a thin sheet and the white central fringe. Most are single-formula questions with unit conversions.

How is YDSE tested in JEE Main and Advanced?

JEE combines YDSE with other ideas: coincidence of two wavelengths, slabs on both slits, oblique incidence, a displaced source, large-angle maxima where the small-angle formula fails, intensity with unequal slits, and Lloyd's mirror or biprism variants. Drawing the path difference carefully before using any formula is the key skill.

Previous year questions on Young’s Double Slit Experiment

35 questions from past papers, each with a step-by-step solution.

Show all 35 questions

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1. The Experiment

2. Path Difference at a Point on the Screen

2.1 Exact expression

2.2 Approximation I: D≫d

2.3 Approximation II: y≪D as well

3. Positions of Bright and Dark Fringes

3.1 Bright fringes (maxima)

3.2 Dark fringes (minima)

4. Fringe Width

4.1 What changes β?

5. Intensity Distribution on the Screen

6. Large Angles and the Number of Fringes

6.1 Two wavelengths together

7. Optical Path and Fringe Shift by a Thin Slab

7.1 Geometrical path and optical path

7.2 Slab in front of one slit

8. YDSE with Oblique Incidence

9. Shapes of Fringes

10. YDSE with White Light

11. Lloyd's Mirror

12. Fresnel's Biprism

13. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is fringe width in Young's double slit experiment?

Why are two slits needed instead of two bulbs in YDSE?

What happens when a glass slab is placed in front of one slit?

Why is the central fringe white when white light is used in YDSE?

How many maxima can be seen in a double slit experiment?

Why is the edge fringe dark in Lloyd's mirror experiment?

Which YDSE questions are common in NEET?

How is YDSE tested in JEE Main and Advanced?

Previous year questions on Young’s Double Slit Experiment

Ready to master Wave Optics?

2.2 Approximation I: $D ≫ d$

2.3 Approximation II: $y ≪ D$ as well

4.1 What changes $β$ ?