Loading…

Diffraction Pattern Solver

Physics Oscillations and Waves • Superposition and Interference

View all topics

Compute the ideal single-slit diffraction pattern. Minima occur when \[ a\sin\theta_m = m\lambda,\qquad m=\pm1,\pm2,\dots \] so for small angles, \[ \theta_m \approx \frac{m\lambda}{a}. \] The normalized intensity envelope is \[ \frac{I}{I_0} = \operatorname{sinc}^2(\beta), \qquad \beta = \frac{\pi a\sin\theta}{\lambda}, \] with \(\operatorname{sinc}(\beta)=\frac{\sin\beta}{\beta}\). This tool computes minima, central maximum width, and shows both an interactive plot and a contained diffraction animation.

Diffraction setup
Wavelength is converted from nm to meters, slit width from mm to meters, and angular display uses both radians and milliradians.
Visualization
The central maximum extends between the first minima on the two sides, so its angular width is approximately \(2\lambda/a\) in the small-angle limit.
Ready
Contained diffraction animation
A narrow slit produces a broad central maximum and weaker side lobes. The slit, outgoing wavefronts, and screen pattern all stay inside the frame.
Schematic single-slit diffraction animation.
Interactive diffraction plot
Plot normalized intensity versus diffraction angle or see how the first-minimum angle changes as wavelength or slit width changes.
Tip: on narrow screens, scroll horizontally to see the full plot.
Enter values and click “Calculate”.

Rate this calculator

0.0 /5 (0 ratings)
Be the first to rate.
Your rating
You can update your rating any time.

Frequently Asked Questions

What does the diffraction pattern solver calculate?

It calculates single-slit diffraction minima angles, the central maximum width, and the ideal normalized intensity envelope as a function of angle.

How do you find the minima in single-slit diffraction?

The minima occur when a sinθ = mλ for nonzero integers m. For small angles, this becomes θ ≈ mλ/a.

Why does a narrower slit produce a wider diffraction pattern?

Because the minima angle is proportional to λ/a. If the slit width a gets smaller, the diffraction angles become larger, so the pattern spreads out.

What is the role of the sinc-squared function?

It describes the full ideal intensity envelope of the single-slit diffraction pattern. Its zeros correspond to the diffraction minima, and its broad central peak gives the central maximum.