PHYS7021 · Electricity, Magnetism & Wave Optics

1. Fringes are concentric circles centred at the point of contact of lens and plate.
2. In reflected light the centre is dark (λ/2 phase change at reflection from the denser medium).
3. Diameters: D_n² ∝ n (dark rings) and ∝ (2n−1) (bright rings) ⇒ rings crowd together as n increases (spacing ∝ 1/√n).
4. With white light only a few coloured rings are seen.
5. Used to measure λ and the refractive index of liquids.

∴ Concentric, dark centre, D² ∝ n, crowding with n

1. Formation: a plano-convex lens (large R) on a glass plate gives a wedge-shaped air film of thickness t; light reflected from the top and bottom of the film interferes ⇒ circular equal-thickness fringes.
2. Geometry: t ≈ r²/2R, where r is the ring radius.
3. Reflected light, with the λ/2 phase change: bright when 2t = (2n−1)λ/2 ⇒ r_n² = (2n−1)λR/2 ⇒ D_n² = 2(2n−1)λR.
4. Dark when 2t = nλ ⇒ D_n² = 4nλR.

∴ Bright: D² = 2(2n−1)λR ; Dark: D² = 4nλR

Given: R = 100 cm; D₃ = 0.181 cm; D₁₃ = 0.501 cm (bright rings).

1. Bright rings: D_n² = 2(2n−1)λR.
2. D₁₃² − D₃² = 2λR[(25) − (5)] = 40λR.
3. Numerically: 0.501² − 0.181² = 0.2510 − 0.0328 = 0.2182 cm².
4. λ = 0.2182/(40 × 100) = 5.46×10⁻⁵ cm = 5456 Å.

∴ λ ≈ 5.46×10⁻⁵ cm ≈ 5456 Å

1. Interference: superposition of two coherent light waves producing a stationary pattern of alternate bright (constructive) and dark (destructive) bands.
2. Conditions: (i) coherent sources (constant phase difference); (ii) same frequency/monochromatic; (iii) comparable amplitudes; (iv) same polarization; (v) small source separation; (vi) large screen distance.
3. Otherwise the fringes wash out and only uniform illumination is seen.

∴ Coherent, monochromatic, equal-ish amplitudes, close sources

1. Light is emitted by independent atomic bursts lasting ~10⁻⁸ s.
2. Two separate lamps therefore have a phase difference that changes randomly millions of times per second.
3. The fringe pattern shifts faster than any detector can follow ⇒ the eye/ detector sees only the uniform sum I₁ + I₂.
4. Hence fringes require one source split into two (division of wavefront or amplitude).

∴ Random phase jumps → no sustained pattern → uniform light

1. Coherence in YDSE: a single slit before the double slit makes S₁, S₂ portions of one wavefront ⇒ constant phase relation.
2. Adding y₁ = a sin ωt and y₂ = a sin(ωt+φ): resultant intensity I = 4I₀cos²(φ/2).
3. Maxima: φ = 2nπ ⇔ path difference = nλ; I = 4I₀.
4. Minima: φ = (2n+1)π ⇔ path difference = (2n+1)λ/2; I = 0.
5. Plot: a cos² curve oscillating between 0 and 4I₀ vs φ; average = 2I₀ = I₁+I₂ ⇒ energy conserved (only redistributed).

∴ I = 4I₀cos²(φ/2); max at nλ, min at (2n+1)λ/2

1. Path difference at a point x on the screen: Δ = xd/D.
2. Bright: x_n = nλD/d; dark: x′_n = (2n−1)λD/2d.
3. Bright fringe width: x_n+1 − x_n = λD/d.
4. Dark fringe width: x′_n+1 − x′_n = λD/d — equal to the bright width. Hence β = λD/d.
5. Conditions as in Q4 (coherence, monochromaticity, etc.).

∴ β = λD/d for both bright and dark ⇒ equal widths

Given: d = 0.1 mm = 1×10⁻⁴ m; β = 5 mm = 5×10⁻³ m; D = 1 m. Formula: λ = βd/D.

1. λ = (5×10⁻³ × 1×10⁻⁴)/1 = 5×10⁻⁷ m.
2. = 5000 Å.

∴ λ = 5×10⁻⁷ m = 5000 Å

1. Constructive: φ = 2nπ ⇒ I = I_max = 4I₀ (bright).
2. Destructive: φ = (2n+1)π ⇒ I = 0 (dark).
3. No violation: the energy missing at dark points appears at the bright points; average intensity = 2I₀ = I₁ + I₂.
4. Interference only redistributes energy; it neither creates nor destroys it.

∴ Energy conserved — redistributed, average 2I₀

1. A biprism is a single prism of obtuse angle ≈ 179°; it refracts the wavefront from one slit into two, producing two virtual coherent sources S₁, S₂ (division of wavefront).
2. Their overlapping region shows interference fringes on a screen, with separation d between the virtual sources.
3. Measure d (lens method or geometry), D and the fringe width β with a micrometer eyepiece.
4. Then λ = βd/D.

∴ Two virtual coherent images; λ = βd/D

Given: d = 0.05 cm; D = 75 cm; λ = 5.89×10⁻⁵ cm. Formula: β = λD/d.

1. β = (5.89×10⁻⁵ × 75)/0.05.
2. = 8.84×10⁻² cm ≈ 0.88 mm.

∴ β ≈ 0.088 cm ≈ 0.88 mm

Given: μ_g = 1.5, t_g = 12×10⁻⁵ mm; μ_d = 2.5. Condition: zero shift ⇒ equal extra optical paths.

1. (μ_g − 1)t_g = (μ_d − 1)t_d.
2. (0.5)(12×10⁻⁵) = (1.5)t_d.
3. t_d = 6×10⁻⁵/1.5 = 4×10⁻⁵ mm (= 40 nm).

∴ t_diamond = 4×10⁻⁵ mm