PhysicsWave opticsClass 12

Wave optics: YDSE, diffraction and polarisation: interactive simulation

Light from two narrow slits overlaps on a screen. Where the two waves arrive in step you get a bright fringe, and where they arrive half a wavelength apart they cancel. The spacing of those fringes tells you the wavelength of light.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics 2, A-level Physics, IB Physics · 4 modes in one lab

What you can explore

  1. 1

    Double slit (YDSE)

    Fringe width with wavelength, slit separation and screen distance, plus thin-film fringe shift.
  2. 2

    Single-slit diffraction

    Central maximum width, minima positions and the fall-off of secondary maxima.
  3. 3

    Polaroids & Malus' law

    Two or three polaroids and how much light gets through.
  4. 4

    Brewster angle

    Polarisation by reflection, with reflected and refracted rays at right angles.

Try this

Change the colour, the slit separation and the screen distance and watch the pattern respond. Then cover one slit with a thin film, close down to a single slit, and finish with polaroids and the angle of perfect polarisation.

  1. 1

    Use λ = 600 nm, d = 0.5 mm and D = 1 m.

    What you should see

    β = 600×10⁻⁹ × 1 / 0.5×10⁻³ = 1.2 mm.

  2. 2

    Switch the colour from red to violet.

    What you should see

    The fringes squeeze together because β is proportional to λ.

  3. 3

    Turn on the film with μ = 1.5 and t = 5 μm.

    What you should see

    The whole pattern moves 5 mm towards the covered slit — about 4.2 fringes — and β does not change.

  4. 4

    In Single slit, set a = 0.2 mm and λ = 600 nm.

    What you should see

    The central maximum is 2λD/a = 6 mm wide, twice the width of the other maxima.

  5. 5

    In Polaroids, cross them at 90°, then add a middle polaroid at 45°.

    What you should see

    0 % goes through without it; with it, 100 → 50 → 25 → 12.5 %.

  6. 6

    In Brewster angle, set μ = 1.5 and press 'Jump to Brewster's angle'.

    What you should see

    i ≈ 56.3°, and the reflected and refracted rays are exactly 90° apart.

Formulas behind it

Fringe width

β=λDd\beta = \frac{\lambda D}{d}

Intensity in YDSE

I=4I0cos⁡2(πd yλD)I = 4I_0\cos^2\left(\frac{\pi d\,y}{\lambda D}\right)

Shift due to a thin film

Δy=D(μ−1)td\Delta y = \frac{D(\mu - 1)t}{d}

Single slit: width of the central maximum

2β0=2λDa2\beta_0 = \frac{2\lambda D}{a}

Malus' law

I=I0cos⁡2θI = I_0\cos^2\theta

Brewster's law

tan⁡ip=μ\tan i_p = \mu

Common mistakes

  • Writing β = λd/D. The separation d goes in the denominator.
  • Expecting the fringe width to change when a thin film is added. The pattern only shifts.
  • Taking I = I₀ cos²θ straight after unpolarised light. The first polaroid always halves it first.
  • Thinking the central maximum in single-slit diffraction is the same width as the others. It is twice as wide.
Practise 191 questionsCheck your answer instantly, with a worked solution.Stuck on a wave optics: ydse, diffraction and polarisation question?Snap it and get a whiteboard video solution.

FAQ

What happens to the fringes if the whole YDSE is placed in water?

The wavelength becomes λ/μ, so the fringe width falls by a factor μ (to 3/4 of its value in air for water).

Why do we need coherent sources for interference?

Only sources with a constant phase difference give a fixed pattern. Two separate bulbs change phase randomly, so the fringes wash out.

What is the difference between interference and diffraction?

Interference is the overlap of waves from separate coherent sources; diffraction is the overlap of wavelets from different parts of the same opening. In YDSE both happen at once.

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