2. Single-, Double- and N-Slit Diffraction¶

Module

Wave Optics · Module 2

Try it live

Explore this on the interactive simulator with real-time sliders and graphs: 🔬 Launch in Simulator

Aim¶

To observe how the diffraction pattern evolves from a single slit through a double slit to an N-slit grating, and to verify the grating equation and resolving power relation.

Theoretical Background¶

A single slit of width a produces a diffraction pattern with a broad central maximum flanked by weaker secondary maxima; the first minima occur where a*sinθ = λ. When a second identical slit is added at spacing (a+b), the two-slit interference term multiplies the single-slit diffraction envelope, giving the familiar Young's-fringe-inside-an-envelope pattern.

Extending this to N equally spaced slits (a diffraction grating) produces sharp principal maxima at angles satisfying (a+b) sinθ = nλ, with N-2 much weaker secondary maxima appearing between consecutive principal maxima. As N increases, the principal maxima become progressively narrower and more intense — the basis of the diffraction-grating spectrometer.

The ability of a grating to separate two closely spaced wavelengths is quantified by its resolving power R = λ/Δλ = nN, where n is the diffraction order used. A larger number of illuminated slits (or grooves) therefore gives finer spectral resolution for a given order.

Governing Formulas¶

Single-slit minima:

\[ a\sin\theta = m\lambda,\quad m=\pm1,\pm2,\ldots \]

N-slit (grating) principal maxima:

\[ (a+b)\sin\theta = n\lambda \]

Resolving power:

\[ R = \dfrac{\lambda}{\Delta\lambda} = nN \]

Apparatus layout diagram for this experiment

Labelled apparatus layout — original illustration prepared for this guide (not a photograph). See the drag-to-rotate 3-D schematic below for a spatial view of the same setup.

3-D apparatus schematic · drag to rotate, scroll to zoom · Laser → N-slit grating → screen

Simulator Controls¶

Control	Symbol	Range	Unit
Number of slits	N	1 – 12	—
Slit width	a	4 – 40	μm
Slit spacing	a+b	20 – 150	μm
Wavelength	λ	400 – 700	nm

Procedure¶

Set N = 1 and record the single-slit diffraction pattern; note the angular position of the first minimum for at least three slit widths a.
Set N = 2 and note how the double-slit interference fringes appear inside the same diffraction envelope.
Increase N progressively (2, 4, 6, 8, 12) at fixed a and (a+b); for each N record the apparent sharpness/width of the principal maxima and the computed resolving power R = nN.
For a fixed N ≥ 4, vary the wavelength λ and tabulate the angular shift of the first-order principal maximum.
Plot R against N and verify the expected linear relationship.

Observation Table¶

Table: Resolving power as a function of number of illuminated slits N

S. No.	N	a (μm)	(a+b) (μm)	λ (nm)	R = nN (n=1)
1
2
3
4
5
6

Graph¶

Plot Resolving power R (y-axis) against Number of slits N (x-axis).

Expected trend: a straight line through the origin (R = nN).

R vs N at fixed diffraction order n = 1

15×15 grid — choose your own scale per square; plot from your observation table

Calculation¶

For each row compute the theoretical first-order angle from (a+b) sinθ = λ and compare with the simulator's rendered maximum position.
Compute R = nN for each N and plot against N; confirm a straight line of slope n through the origin.

Result¶

Table: Resolving power verification

N	R (calculated)	R (from slope of graph)	% deviation
1
2
3

Maximum Permissible Error¶

From (a+b) sinθ = nλ, differentiating gives a cotθ term for the angle uncertainty in addition to the wavelength's fractional error.

Precautions¶

Keep the diffraction order under consideration (usually n = 1) fixed when comparing resolving powers across different N.
Distinguish clearly between the diffraction envelope minima (single slit) and the interference principal maxima (multi-slit).

Self-Check Quiz¶

1. For a single slit of width a, the first diffraction minimum occurs when:

A) a sinθ = λ/2
B) a sinθ = λ
C) a sinθ = 2λ
D) sinθ = a/λ

Answer: B — The single-slit minima condition is a sinθ = mλ, so the first minimum (m = 1) is at a sinθ = λ.

2. As the number of slits N in a grating increases (spacing fixed), the principal maxima become:

A) Broader and dimmer
B) Narrower and more intense
C) Unchanged
D) They disappear

Answer: B — More illuminated slits sharpen the principal maxima and increase their peak intensity — this is exactly why grating spectrometers use many slits/grooves.

3. The resolving power R = λ/Δλ = nN tells you that, for a fixed diffraction order n, resolving power improves by:

A) Increasing the number of illuminated slits N
B) Increasing the slit width a
C) Decreasing the wavelength only
D) Reducing N

Answer: A — R is directly proportional to N at fixed order n, so illuminating more slits/grooves of the grating improves its ability to separate close wavelengths.

4. Which pattern do you get by multiplying the single-slit diffraction envelope by the two-slit interference term?

A) A pure single-slit pattern
B) A pure N-slit grating pattern
C) Young's fringes modulated inside a single-slit diffraction envelope
D) A completely uniform intensity

Answer: C — The classic double-slit pattern is the fine Young's fringes (from two-slit interference) riding inside the broader single-slit diffraction envelope.

Viva-Voce Questions¶

Why do secondary maxima become negligible once N is large?
How does the resolving power of a grating relate to the total number of illuminated grooves rather than the groove spacing alone?
How does the width of the central diffraction maximum change if the slit width a is reduced?
Why are the secondary maxima in single-slit diffraction much weaker than the central maximum?
What is the physical difference between the diffraction envelope and the interference fringes seen in a double-slit pattern?
How does increasing the number of slits N change the sharpness of the principal maxima in grating diffraction?
Define the resolving power of a diffraction grating and state the factors it depends on.