A monochromatic source $S$ emitting unpolarised light illuminates a two-slit arrangement with slits $S_1$ and $S_2$. A polariser $P$, whose transmission-axis direction is unspecified, is placed so that it covers only the beam travelling toward slit $S_2$; the light reaching $S_1$ passes through no polariser. Let $I_0$ be the intensity of the principal (central) maximum when no polariser at all is present. For the present arrangement (polariser in front of $S_2$ only), calculate the intensity of the principal maximum and the intensity of the first minimum.

A monochromatic source S emitting unpolarised light illuminates a two-slit arrangement with slits S₁ and S₂. A

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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.

Note

For constructive interference (bright band): path difference = $n\lambda$ (whole number of wavelengths)

For destructive interference (dark band): path difference = $(n + \frac12)\lambda$ (half-integer number of wavelengths)

Here $\lambda$ is the wavelength of the light, and $n = 0, 1, 2, \dots$

The Geometry

Let the slits be separated by distance $d$ . The screen is far away at distance $D$ ( $D \gg d$ ). For a point on the screen at angle $\theta$ from the centre:

The path difference $\Delta x = d \sin\theta$
Bright bands occur when $d \sin\theta = n\lambda$
Dark bands occur when $d \sin\theta = (n + \frac12)\lambda$

The position $y$ of the $n$ -th bright band on the screen (measured from the centre) is:

y_n = \frac{n\lambda D}{d}

The spacing between consecutive bright bands (fringe width) is:

\beta = \frac{\lambda D}{d}

\text{Fringe width } \beta = \frac{\lambda D}{d}

What This Tells You

Larger $\lambda$ → 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 $S_1$ and $S_2$ , 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 $D \gg d$ ).

The two waves start in phase at the slits (same source).
They travel different distances to reach $P$ .
This path difference $\Delta x$ causes a phase difference $\Delta \phi$ .

Key relation:

\Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta x

Why? Because one full wavelength $\lambda$ corresponds to a phase change of $2\pi$ radians.

2. Finding the Path Difference

From the geometry (see diagram in any textbook):

For a point $P$ at angle $\theta$ from the central axis, the extra distance travelled by the wave from the farther slit is approximately:

\Delta x = d \sin\theta

Why approximate? Because we assume $D \gg d$ , 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:

\Delta \phi = 2\pi m \quad (m = 0, \pm1, \pm2, \dots)

Using $\Delta \phi = \frac{2\pi}{\lambda} \cdot d\sin\theta$ , we get:

\frac{2\pi}{\lambda} \cdot d\sin\theta = 2\pi m

Cancel $2\pi$ to obtain the bright fringe condition:

\boxed{d \sin\theta = m\lambda}

$m$ 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 $\pi$ (half a cycle):

\Delta \phi = (2m+1)\pi

Substitute again:

\frac{2\pi}{\lambda} \cdot d\sin\theta = (2m+1)\pi

Cancel $\pi$ to get the dark fringe condition:

\boxed{d \sin\theta = \left(m + \frac12\right)\lambda}

5. From Angle to Position on Screen

For small angles (typical in exam problems), $\sin\theta \approx \tan\theta = \frac{y}{D}$ , where $y$ is the distance from the central maximum on the screen.

Bright fringe position:

y_m = \frac{m\lambda D}{d}

Dark fringe position:

$y_m = \frac{(m + \frac12)\lambda D}{d}$ …

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The polariser in front of $S_2$ halves that slit's intensity and fixes its polarisation. Only the component of $S_1$ 's unpolarised light parallel to the polariser axis can interfere with it; the perpendicular component contributes a steady, non-interfering background. Working this out gives a principal maximum of $\tfrac{5}{8}I_0$ and a first minimum of $\tfrac{1}{8}I_0$ .

Reference (no polariser)

Let the intensity from each slit alone be $I_s$ . With both slits open and fully coherent, the principal maximum is

I_0=(\sqrt{I_s}+\sqrt{I_s})^2=4I_s\quad\Rightarrow\quad I_s=\frac{I_0}{4}.

With the polariser in front of $S_2$

Slit $S_2$ : unpolarised light through a polariser is fully polarised along the axis, its intensity halved:

I_{S_2}=\frac{I_s}{2}.

Slit $S_1$ : unpolarised, intensity $I_s$ . Resolve it into two independent components, one parallel and one perpendicular to the polariser axis, each of intensity $I_s/2$ .

The parallel component ( $I_s/2$ ) shares $S_2$ 's polarisation, so it interferes with $S_2$ 's light.
The perpendicular component ( $I_s/2$ ) has no partner at $S_2$ ; it cannot interfere and simply adds as a constant background at every point.

Interfering pair

Both interfering beams have intensity $I_s/2$ , i.e. amplitude $\sqrt{I_s/2}$ each.

In phase (principal maximum): …

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Concept understanding — Double Slit Interference

Double Slit Interference: From Ripples to Light

The Core Idea

The Geometry

What This Tells You

Why It Matters

Why this formula?

Double Slit Interference: Why the Formula Holds

1. The Core Idea: Path Difference Creates Phase Difference

2. Finding the Path Difference

3. Condition for Constructive Interference (Bright Fringes)

4. Condition for Destructive Interference (Dark Fringes)

5. From Angle to Position on Screen

Reference (no polariser)

With the polariser in front of $S_2$

Interfering pair

Method: Interference When One Beam Is Polarised and the Other Is Not

Steps

Reference (no polariser)

With the polariser in front of S2S_2S2​

Interfering pair

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With the polariser in front of $S_2$

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