Thin-Film Interference - College Physics

1. Newton's rings are formed between a convex lens and a flat glass plate. The setup is illuminated with sodium light ($\lambda = 589$ nm) and the radius of the 5th bright ring is measured to be $2.2$ mm. If the same setup is then illuminated with blue light ($\lambda = 450$ nm), what will be the radius of the 7th bright ring, and how does this compare to the 7th ring with sodium light?
   Answer: $2.4$ mm; smaller than the 7th sodium ring by $0.4$ mm For Newton's rings, the radius of the $m$th bright ring is $r_m = \sqrt{(m + 1/2)R\lambda}$, where $R$ is the lens radius of curvature. From the 5th bright ring with sodium light: $2.2 = \sqrt{5.5R \times 589}$, solving for $R$: $R = (2.2)^2/(5.5 \times 589) = 1.49$ mm. For the 7th bright ring with blue light: $r_7 = \sqrt{7.5 \times 1.49 \times 450} = \sqrt{5033} = 2.4$ mm. For the 7th sodium ring: $r_7 = \sqrt{7.5 \times 1.49 \times 589} = \sqrt{6590} = 2.8$ mm. The blue ring is smaller by $2.8 - 2.4 = 0.4$ mm.
2. A thin soap film ($n = 1.33$) floating in air shows first-order constructive interference for yellow light ($\lambda = 580$ nm) at normal incidence. If this same film is then immersed in water ($n = 1.33$), what happens to the interference pattern, and what would be the new wavelength in air that would show constructive interference at the same film thickness?
   Answer: No interference pattern visible; no wavelength shows constructive interference Initially in air, the soap film ($n = 1.33$) has phase changes at both surfaces since $n_{air} < n_{soap}$ and the film is very thin with air on both sides. For first-order constructive interference: $2nt = 0.5\lambda$, so $t = \lambda/(4n) = 580/(4 \times 1.33) = 109$ nm. When immersed in water, the soap film has the same refractive index as water ($n = 1.33$), so there are no phase changes at either interface and the optical path difference becomes zero. With no refractive index contrast, there is no reflection and hence no interference pattern. The film essentially becomes invisible.

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