Q.In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Double Slit Interference
Double Slit Interference: From Ripples to Light
Imagine dropping two stones into a still pond at the same time, a short distance apart. Watch the ripples spread. Where a crest from one stone meets a crest from the other, the water rises higher. Where a crest meets a trough, the water flattens out. That is interference — waves adding or cancelling.
Now replace the water with light. Replace the stones with two narrow slits cut into a barrier. Shine a single colour of light (say, red laser light) onto the slits. On a screen behind the barrier, you do not see two bright spots. Instead, you see a pattern of alternating bright and dark bands — like a striped zebra crossing made of light.
That pattern is double slit interference. It is the single most convincing proof that light behaves as a wave.
The Core Idea
Light from a single source passes through two narrow slits. Each slit acts as a new source of waves. These two sets of waves spread out and overlap. At any point on the screen, the light you see is the sum of the waves from slit 1 and slit 2.
Whether they add (bright) or cancel (dark) depends on one thing: the path difference — how much farther one wave has travelled compared to the other.
For constructive interference (bright band): path difference = nλ (whole number of wavelengths)
For destructive interference (dark band): path difference = (n+21)λ (half-integer number of wavelengths)
Here λ is the wavelength of the light, and n=0,1,2,…
The Geometry
Let the slits be separated by distance d. The screen is far away at distance D (D≫d). For a point on the screen at angle θ from the centre:
- The path difference Δx=dsinθ
- Bright bands occur when dsinθ=nλ
- Dark bands occur when dsinθ=(n+21)λ
The position y of the n-th bright band on the screen (measured from the centre) is:
yn=dnλD
The spacing between consecutive bright bands (fringe width) is:
β=dλD
Fringe width β=dλD
What This Tells You
- Larger λ → wider fringes (red light spreads more than blue)
- Larger D → wider fringes (screen further away spreads the pattern)
- Smaller d → wider fringes (slits closer together spread the pattern more)
If you cover one slit, the pattern vanishes — you get a single blurry blob. The stripes only appear when both slits are open, proving that the light from the two slits is interfering.
Why It Matters
Double slit interference is not a classroom toy. It is the foundation of:
- Young's experiment (1801) — which settled the debate: light is a wave
- Diffraction gratings — used in spectrometers to identify elements by their light …
Why this formula?
Double Slit Interference: Why the Formula Holds
Let's build the understanding from first principles — not just memorise the formula, but see why it must be true.
1. The Core Idea: Path Difference Creates Phase Difference
Imagine two narrow slits S1 and S2, separated by distance d, illuminated by a single coherent source. Light from each slit travels to a point P on a screen at distance D (where D≫d).
- The two waves start in phase at the slits (same source).
- They travel different distances to reach P.
- This path difference Δx causes a phase difference Δϕ.
Key relation:
Δϕ=λ2π⋅Δx
Why? Because one full wavelength λ corresponds to a phase change of 2π radians.
2. Finding the Path Difference
From the geometry (see diagram in any textbook):
- For a point P at angle θ from the central axis, the extra distance travelled by the wave from the farther slit is approximately:
Δx=dsinθ
Why approximate? Because we assume D≫d, so the two paths are nearly parallel. This is the Fraunhofer (far-field) approximation — valid for most exam setups.
3. Condition for Constructive Interference (Bright Fringes)
Waves interfere constructively when they arrive in phase:
Δϕ=2πm(m=0,±1,±2,…)
Using Δϕ=λ2π⋅dsinθ, we get:
λ2π⋅dsinθ=2πm
Cancel 2π to obtain the bright fringe condition:
dsinθ=mλ
- m is called the order of the fringe.
- m=0 gives the central bright fringe (straight ahead).
4. Condition for Destructive Interference (Dark Fringes)
Waves interfere destructively when they arrive out of phase by π (half a cycle):
Δϕ=(2m+1)π
Substitute again:
λ2π⋅dsinθ=(2m+1)π
Cancel π to get the dark fringe condition:
dsinθ=(m+21)λ
5. From Angle to Position on Screen
For small angles (typical in exam problems), sinθ≈tanθ=Dy, where y is the distance from the central maximum on the screen.
Bright fringe position:
ym=dmλD
Dark fringe position:
ym=d(m+21)λD …
Concept: Young’s Double-Slit Interference — the position of bright fringes is given by yn=ndλD.
Reasoning:
-
For the n-th bright fringe (central is n=0), the distance from the centre is
yn=ndλD.
-
Here n=4, y4=1.2 cm=1.2×10−2 m,
d=0.28 mm=2.8×10−4 m,
D=1.4 m.
-
Rearranging: …
In Young’s double-slit interference, the bright fringe positions are given by yn=ndλD. Using the given values for the fourth bright fringe, the wavelength is found to be λ=600 nm.
The key to solving this problem is understanding that in Young’s double-slit experiment, bright fringes (constructive interference) occur at positions where the path difference from the two slits is an integer multiple of the wavelength. The formula yn=ndλD directly relates the fringe position to the wavelength, slit separation, and screen distance. Here, we know the distance to the fourth bright fringe (n=4), so we can solve for λ directly.
Let’s work through it step by step.
-
Identify the known quantities.
Slit separation: d=0.28 mm=0.28×10−3 m=2.8×10−4 m.
Screen distance: D=1.4 m.
Distance to the fourth bright fringe from the central maximum: y4=1.2 cm=1.2×10−2 m.
Fringe order: n=4.
-
Recall the formula for bright fringe positions.
For constructive interference in Young’s double-slit, the n-th bright fringe (where n=0,1,2,…) is located at a distance from the central maximum given by:
yn=ndλD
Here n=0 gives the central bright fringe, n=1 the first bright fringe, and so on. The problem states “the distance between the central bright fringe and the fourth bright fringe” — that is exactly y4.
- Substitute the known values into the formula.
1.2×10−2=4×2.8×10−4λ×1.4
- Solve for λ. First, simplify the right-hand side:
2.8×10−44×1.4λ=2.8×10−45.6λ=2×104λ
So the equation becomes:
1.2×10−2=2×104λ
Divide both sides by 2×104:
λ=2×1041.2×10−2=0.6×10−6 m=6×10−7 m …
Method: Fringe Width Method for Young’s Double-Slit Experiment
This method uses the relationship between fringe spacing, slit separation, screen distance, and wavelength.
Step-by-step solution
Step 1: Identify the given data
- Slit separation, d=0.28 mm=0.28×10−3 m
- Screen distance, D=1.4 m
- Distance from central bright to fourth bright fringe, y4=1.2 cm=1.2×10−2 m
- Order of fringe, n=4
Step 2: Recall the formula for bright fringe position
For nth bright fringe from centre:
yn=dnλD
Step 3: Substitute for the fourth bright fringe
y4=d4λD
Step 4: Solve for wavelength λ
λ=4Dy4⋅d
Step 5: Plug in the values
λ=4×1.4(1.2×10−2)×(0.28×10−3)
λ=5.63.36×10−6
λ=6.0×10−7 m …
Common Mistakes in Young's Double-Slit Problems
Students often lose marks on this exact type of question. Here are the most frequent errors and how to avoid each:
1. Confusing fringe order with fringe number
The Mistake:
Students take n=4 for the "fourth bright fringe" but incorrectly use the formula for the first fringe.
Why it happens:
The central bright fringe is n=0, so the fourth bright fringe corresponds to n=4, not n=3.
How to avoid:
Always remember:
- Central bright fringe → n=0
- First bright fringe → n=1
- Fourth bright fringe → n=4
So here, n=4 is correct.
2. Unit conversion errors
The Mistake:
Using d=0.28 (in mm) or D=1.4 (in m) without converting to consistent units.
Why it happens:
The slit separation is given in mm, the screen distance in m, and the fringe distance in cm — three different units.
How to avoid:
Convert everything to metres before plugging into the formula:
- d=0.28 mm=0.28×10−3 m=2.8×10−4 m
- D=1.4 m
- y4=1.2 cm=1.2×10−2 m
3. Using the wrong formula for fringe position
The Mistake:
Using yn=dnλD when the problem gives the distance from the central fringe.
Why it happens:
Students sometimes use the formula for fringe spacing (β=dλD) and multiply by n, which is actually correct — but they forget that yn is measured from the centre.
How to avoid:
For bright fringes, the distance from the central fringe to the nth bright fringe is:
yn=dnλD
Here, y4=1.2×10−2 m, n=4, D=1.4 m, d=2.8×10−4 m.
4. Algebraic rearrangement errors
The Mistake:
Solving for λ incorrectly — e.g., multiplying instead of dividing.
How to avoid:
Rearrange step by step:
yn=dnλD
Multiply both sides by d:
ynd=nλD …
Showing the 12 most recent of 13 on this concept.
- GUJCET 2026Set x1 markMCQQ.In a Young's double slit experiment, the slits are separated by 0.2 mm and placed 2.0 m away. The distance between the central bright fringe and the first bright fringe (fringe width) is measured to be 1.5 cm. Determine the wavelength of light used in the experiment. (A) 4200 A˚ (B) 5000 A˚ (C) 4600 A˚ (D) 15000 A˚ (Calculated result)
›Reveal solutionSolution
[!TLDR] Substituting the given numbers into λ=βd/D yields 1.5×10−6 m =15000A˚, matching option (D).
Concept
In Young's double-slit experiment the fringe width is β=dλD, where d is the slit separation and D the slit-to-screen distance. Rearranged, λ=Dβd.
Solution
Given d=0.2mm=0.2×10−3 m, D=2.0 m, β=1.5cm=1.5×10−2 m. …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.If path difference between two waves is 3 lambda then phase difference related with them is ___.(a) 4 pi rad(b) 3 pi rad(c) 2 pi rad(d) 6 pi rad
›Reveal solutionSolution
Path difference and phase difference are related by delta(phi) = (2 pi/lambda) x (path difference), because one full wavelength of path corresponds to a full 2 pi cycle of phase.
…
- GUJCET 2025Set 031 markMCQQ.Two waves of same intensity I0 emitted from two sources having same phase difference (ϕ). Due to superposition of two waves, the intensity of resultant wave is directly proportional to ______. (A) sin2(2ϕ) (B) sin2ϕ (C) cos2(2ϕ) (D) cos2ϕ
›Reveal solutionSolution
Superposition of two equal-intensity coherent waves gives I=4I0cos2(ϕ/2).
Concept — interference. Resultant intensity of two waves of intensities I1,I2 with phase difference ϕ is I=I1+I2+2I1I2cosϕ.
Steps.
- With I1=I2=I0: I=2I0+2I0cosϕ=2I0(1+cosϕ). …
- GUJCET 2025Set 031 markMCQQ.In Young's double slit experiment, the slits are separated by 0.54 mm and the screen is placed 1.8 m away. The distance between central bright fringe and sixth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. (A) 5000 Å (B) 600 nm (C) 8000 nm (D) 800 nm
›Reveal solutionSolution
[!TLDR]
From λ=xnd/(nD) the wavelength is 600nm.
Concept
In Young's double-slit experiment the distance of the n-th bright fringe from the centre is xn=dnλD, so λ=nDxnd.
Solution
Given d=0.54mm=0.54×10−3m, D=1.8m, x6=1.2cm=1.2×10−2m, n=6: …
- GUJCET 2024Set 131 markMCQQ.In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Then the wavelength of light used in the experiment is ________. (A) 500 nm (B) 660 nm (C) 600 nm (D) 550 nm
›Reveal solutionSolution
Fourth bright fringe: y4=d4λD, solve for λ.
Steps. d=0.28×10−3 m, D=1.4 m, y4=1.2×10−2 m, n=4. …
- GUJCET 2023Set 091 markMCQQ.Two slits are made 10 millimeter apart and the screen is placed 1.5 metre away. What is the fringe separation when a wavelength of 7000A˚ is used? (A) 105μm (B) 1.05μm (C) 10.5μm (D) 0.105μm
›Reveal solutionSolution
[!TLDR]
Using β=λD/d gives a fringe width of 105μm.
Concept
In Young's double-slit experiment the spacing between adjacent bright (or dark) fringes is β=dλD, where D is the slit-to-screen distance and d the slit separation.
Solution
Convert units: λ=7000A˚=7×10−7m, D=1.5m, d=10mm=1×10−2m. …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.In Young's double experiment the distance between two slits is 0.2 mm and the distance between slit and screen is 1.5 m. The wavelength of light used is 600 nm. The distance between any two consecutive bright fringes is ___ mm.(a) 0.8(b) 4.5(c) 0.5(d) 2.0
›Reveal solutionSolution
The spacing of consecutive bright fringes is beta = lambda D/d = 4.5 mm.
Fringe width in Young's double-slit experiment:
beta = lambda D / d.
Substitute lambda = 600 nm = 600 x 10^-9 m, D = 1.5 m, d = 0.2 mm = 0.2 x 10^-3 m: …
- GUJCET 2022Set 171 markMCQQ.Two slits are made 3 millimetre (3 mm) apart and the screen is placed 2 m away. What is the fringe separation when blue-green light of wavelength 600 nm is used? (A) 0.4 mm (B) 0.6 mm (C) 0.5 mm (D) 0.7 mm
›Reveal solutionSolution
β=dλD.
Steps.
- λ=600 nm=6×10−7 m, D=2 m, d=3 mm=3×10−3 m. …
- GUJCET 2021Set 151 markMCQQ.The wavelength of light 500 nm is used in a Young's double-slit experiment. The distance between the slits and screen is 100 cm and the slits are separated by 1 mm. Then find distance between fifth (5th) and third (3rd) bright fringes. (A) 1 mm (B) 3 mm (C) 2 mm (D) 4 mm
›Reveal solutionSolution
Consecutive bright fringes are beta = lambda D/d apart; the 5th and 3rd differ by 2 beta.
Concept. β=dλD, and x5−x3=(5−3)β=2β. …
- GUJCET 2020Set 071 markMCQQ.The distance between two slits is 3 mm & screen is placed at 2 m distance. When blue-green light of wavelength 500 nm is used then distance between two fringes will be? (A) 0.5 mm (B) 0.43 mm (C) 0.33 mm (D) 0.4 mm
›Reveal solutionSolution
Fringe spacing in YDSE is β=dλD.
Concept: In Young's double-slit experiment the distance between adjacent bright (or dark) fringes is β=dλD. …
- GUJCET 2019Set 131 markMCQQ.In Young's experiment fourth bright fringe produced by light of 5000A˚ superposes on the fifth bright fringe of an unknown wavelength. the unknown wavelength is ............ A˚ . (A) 8000 (B) 5000 (C) 6000 (D) 4000
›Reveal solutionSolution
Overlapping bright fringes have equal positions, so n1λ1=n2λ2.
Concept first: The position of the n-th bright fringe is yn=nλD/d. If the 4th bright fringe of 5000A˚ coincides with the 5th bright fringe of the unknown wavelength: …
- GUJCET 2015Set C1 markMCQQ.Light of wave length λ is incident on slit of width d. The resulting diffraction pattern is observed on a screen placed at distance D. The linear width of central maximum is equal to width of the slit, then D = _____ (A) d2λ2 (B) 2λd2 (C) λd (D) d2λ
›Reveal solutionSolution
[!TLDR] Setting the central-maximum width equal to the slit width gives D=2λd2, option (B).
Concept …
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