Wave Optics (8hrs) · Engineering Physics

Newton's Ring experiment

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Newton's Ring experiment

1. What It Is

Newton's rings = concentric circular interference fringes formed by division of amplitude in the air film between a plano-convex lens (radius RR large) and a flat glass plate, illuminated from above with monochromatic light.

Observed by reflected light (or transmitted) through a microscope.

2. Geometry and Path Difference

Air film thickness at distance rr from contact point OO (where lens touches plate):

t≈r2/(2R)t ≈ r²/(2R)

derived from circle geometry (R−t)2+r2=R2⇒t=R−√(R2−r2)≈r2/(2R)(R -t)² + r² = R² ⇒ t = R - √(R² - r²)≈ r²/(2R) for r≪Rr≪R.

Light reflects from top of air film (glass→air, no flip) and bottom (air→glass, half-wave π flip). So for reflected system:

Δ=2μtcosr′+λ/2(airμ≈1,nearnormalcosr′≈1)⇒Δ=2t+λ/2Δ = 2μt cos r' + λ/2 (air μ≈1, near normal cos r'≈1) ⇒ Δ = 2t + λ/2

Constructive/destructive at eye:

  • Dark (destructive) reflected:
Δ=(2n+1)λ/2⇒2t=nλΔ = (2n+1)λ/2 ⇒ 2t = nλ
  • Bright reflected:
Δ=nλ⇒2t=(2n−1)λ/2Δ = nλ ⇒ 2t = (2n-1)λ/2

Since t=r2/(2R)t=r²/(2R):

rn(dark)=√(nλR),rn(bright)=√[(2n−1)λR/2]r_n^{(dark)} = √(n λ R) , r_n^{(bright)} = √[(2n-1)λ R/2]

for reflected light. Central spot (t=0t=0) ⇒ $r=0 ⇒ dark in reflected; bright in transmitted (no half-wave difference).

3. Setup (diagram description)

Monochromatic source → 45° glass plate → plano-convex lens (R) on glass plate
                       ↓ reflected beams from top/bottom of air film interfere → microscope sees rings

White light gives coloured rings (few orders). Sodium light λ=589λ=589 nm gives clear.

4. Applications (what examiners ask)

  • Measure λλ: with known RR, measure diameter Dn=2rnD_n =2r_n for nnth dark ring:
λ=Dn2/(4nR)λ = D_n²/(4nR)

. Use two rings to avoid contact error:

λ=(Dn+m2−Dn2)/(4mR)λ = (D_{n+m}² - D_n²)/(4mR)
  • Measure RR (lens curvature):
R=(Dn+m2−Dn2)/(4mλ)R = (D_{n+m}² - D_n²)/(4mλ)
  • Refractive index of liquid: fill air film with liquid μμ: rn2=nλR/μr_n² = nλR/μ ⇒
μ=nλR/rn2μ = nλR/r_n²

5. Transmitted Rings (complement)

Without half-wave both reflections? Transmitted system: Δ=2tΔ=2t (no extra λ/2λ/2) ⇒ bright/dark swapped vs reflected; transmitted rings fainter (less contrast) because intensities differ.

6. Worked Examples

(a) λ=600 nm, R=1.0 m, 10th dark ring: r =√(10·6e-7·1)=√6e-6≈2.45 mm, D≈4.9 mm ✓

(b) Diameters D5=3.0 mm, D15=5.1 mm, λ=589 nm ⇒ R=(D15²-D5²)/(4·10·λ)= (26.01-9)·10^{-6}/(2.356e-5)≈0.72 m ✓

(c) Why central dark? t=0 ⇒ Δ=λ/2 ⇒ destructive ⇒ dark (reflected) ✓

(d) Liquid μ? Air ring D=4 mm becomes 3.5 mm in liquid: μ=(4/3.5)²≈1.31 ✓

7. Exam Points

  • Diagram + formation and path difference Δ=2t+λ/2Δ=2t+λ/2 — 5 marks, almost certain.
  • Derive rn=√(nλR)r_n=√(nλR) (dark reflected) — 7 marks, very frequent.
  • Use λ=(Dn+m2−Dn2)/(4mR)λ = (D_{n+m}²-D_n²)/(4mR) to find λλ or RR or μμ — 5-7 marks, very frequent.
  • Explain central dark (reflected) vs bright (transmitted) — 3-5 marks, frequent.
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