The experimental arrangement
Two ordinary, independent light sources cannot interfere. NCERT notes that the light emitted from a source such as a sodium lamp undergoes abrupt phase changes in times of the order of $10^{-10}$ seconds, so the waves from two separate lamps carry no fixed phase relationship. Such sources are incoherent, and their intensities simply add up — no fringes appear on the screen.
Thomas Young's insight was to derive both interfering beams from a single source. He made a pinhole $S$ illuminated by a bright source, and placed two closely spaced pinholes $S_1$ and $S_2$ on a second opaque screen. Because the light reaching $S_1$ and $S_2$ comes from the same original wavefront, any abrupt phase change at $S$ appears identically in both — the two are locked in phase and behave as coherent sources.
The spherical waves spreading from $S_1$ and $S_2$ overlap on the observation screen $GG'$, producing alternate bright and dark bands called fringes. NIOS records that this pattern was observed experimentally by Young in 1802 and provided direct proof of the wave nature of light.
Path difference and geometry
Whether a point $P$ on the screen is bright or dark depends on the path difference between the two waves reaching it. Since the waves leave $S_1$ and $S_2$ in phase, any phase difference at $P$ arises purely from the difference in path lengths. NIOS writes this path difference as
$$\Delta = S_2P - S_1P = d\sin\theta$$
In the standard geometry the slit separation $d$ is far smaller than the screen distance $D$, so the angle $\theta$ is small. For small angles, $\sin\theta \approx \tan\theta = x/D$, where $x$ is the distance of $P$ from the central point $O$. The path difference therefore reduces to the working form:
$$\Delta = d\sin\theta \approx \frac{d\,x}{D}$$
Bright and dark fringe positions
A bright fringe (constructive interference) forms wherever the path difference is a whole number of wavelengths, $\Delta = n\lambda$. A dark fringe (destructive interference) forms wherever the path difference is an odd half-integer of wavelengths, $\Delta = (n + \tfrac{1}{2})\lambda$. Substituting $\Delta = dx/D$ gives the fringe positions used directly in NEET problems.
Setting $\dfrac{dx}{D} = n\lambda$ yields the bright-fringe positions (NCERT Eq. 10.13), and $\dfrac{dx}{D} = (n+\tfrac{1}{2})\lambda$ yields the dark-fringe positions (NCERT Eq. 10.14):
$$x_n^{\text{bright}} = \frac{n\lambda D}{d}, \quad n = 0, \pm 1, \pm 2,\dots$$ $$x_n^{\text{dark}} = \left(n+\tfrac{1}{2}\right)\frac{\lambda D}{d}, \quad n = 0, \pm 1, \pm 2,\dots$$
| Quantity | Condition | Position from centre |
|---|---|---|
| Bright fringe (max) | Δ = nλ | x = nλD/d |
| Dark fringe (min) | Δ = (n + ½)λ | x = (n + ½)λD/d |
| Central fringe | Δ = 0 (n = 0) | x = 0 — bright |
Because the positions step up in equal increments of $\lambda D/d$, NCERT concludes from Eqs. 10.13 and 10.14 that the dark and bright fringes are equally spaced.
Fringe width β = λD/d
The fringe width $\beta$ is the spacing between two consecutive bright (or dark) fringes. Taking the third and second bright fringes from the position formula, NIOS subtracts:
$$\beta = x_3^{\text{bright}} - x_2^{\text{bright}} = \frac{3\lambda D}{d} - \frac{2\lambda D}{d} = \frac{\lambda D}{d}$$
This is the single most examined relation in the topic. Fringe width is directly proportional to the wavelength $\lambda$ and the screen distance $D$, and inversely proportional to the slit separation $d$.
The fringe pattern only forms because two wave amplitudes superpose. Review coherent and incoherent addition of waves to see exactly where the $4I_0\cos^2(\phi/2)$ result comes from.
How β responds to λ, D, d and a liquid
Read $\beta = \lambda D/d$ as three separate proportionalities and one immersion rule. Increasing $\lambda$ or $D$ widens the fringes; increasing $d$ narrows them. When the whole apparatus is dipped in a liquid of refractive index $n$, the wavelength becomes $\lambda/n$, so $\beta$ is divided by $n$ — the pattern shrinks. The central fringe stays bright regardless, since the path difference there remains zero.
$\beta \propto \lambda,\ \beta \propto D,\ \beta \propto 1/d$; in a medium $\beta' = \beta/n$; central fringe is always bright.
Intensity distribution
The brightness across the screen is not a simple on/off pattern; it varies smoothly. If each slit alone produces intensity $I_0$ and the phase difference at a point is $\phi$, the resultant displacement has amplitude $2a\cos(\phi/2)$, so the intensity is
$$I = 4I_0\cos^2\!\left(\frac{\phi}{2}\right)$$
When $\phi = 0, \pm 2\pi, \pm 4\pi,\dots$ the cosine is maximal and $I = 4I_0$ — a bright fringe. When $\phi = \pm\pi, \pm 3\pi,\dots$ the cosine is zero and $I = 0$ — a dark fringe. NIOS stresses that no energy is destroyed at the dark fringes: the energy missing from the minima reappears at the maxima, so the pattern simply redistributes energy between $4I_0$ and zero.
Conditions for sustained interference
A fringe pattern is only stable — "sustained" — if the phase difference between the two waves at every point stays fixed in time. NCERT explains that if the sources keep a constant phase difference, the positions of maxima and minima do not change and a stable pattern results. If the phase difference fluctuates rapidly, the maxima and minima shift so quickly that the eye records only a uniform time-averaged intensity $I = 2I_0$, and no fringes are seen.
| Requirement | Why it matters |
|---|---|
| Coherent sources (constant phase difference) | Keeps maxima and minima fixed; from a single source via two slits |
| Same wavelength / monochromatic | Fixes one fringe width $\beta = \lambda D/d$; white light gives overlapping coloured fringes |
| Comparable amplitudes | Makes dark fringes nearly zero, maximising contrast |
| Narrow, closely spaced slits | Keeps $\theta$ small so $\Delta \approx dx/D$ holds and fringes are resolvable |
Worked example
In a Young's double-slit experiment the slits are separated by $0.28\ \text{mm}$ and the screen is placed $1.4\ \text{m}$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2\ \text{cm}$. Determine the wavelength of light used.
The fourth bright fringe sits at $x_4 = \dfrac{4\lambda D}{d}$, so $\lambda = \dfrac{x_4\,d}{4D}$.
Substituting $x_4 = 1.2\times10^{-2}\ \text{m}$, $d = 0.28\times10^{-3}\ \text{m}$, $D = 1.4\ \text{m}$:
$$\lambda = \frac{(1.2\times10^{-2})(0.28\times10^{-3})}{4 \times 1.4} = 6.0\times10^{-7}\ \text{m} = 600\ \text{nm}.$$
The wavelength of the light used is about $600\ \text{nm}$ — in the orange-red region of the visible spectrum.
Young's double slit — the must-knows
- Two slits $S_1$, $S_2$ derived from one source are coherent; two independent lamps are incoherent and give no fringes.
- Path difference $\Delta = d\sin\theta \approx dx/D$ for small angles.
- Bright fringes at $x = n\lambda D/d$; dark fringes at $x = (n+\tfrac{1}{2})\lambda D/d$.
- Fringe width $\beta = \lambda D/d$; fringes are equally spaced; central fringe is bright.
- Intensity $I = 4I_0\cos^2(\phi/2)$, varying between $4I_0$ and $0$ with energy conserved.
- In a medium of index $n$, $\beta$ becomes $\beta/n$; angular fringe width $\lambda/d$ is independent of $D$.