College Physics Quiz: Thin Film Interference
2 questions · exam conditions
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Thin Film InterferenceQuestion 1 of 2

Newton's rings are formed between a convex lens and a flat glass plate. The setup is illuminated with sodium light (λ=589\lambda = 589 nm) and the radius of the 5th bright ring is measured to be 2.22.2 mm. If the same setup is then illuminated with blue light (λ=450\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?

2.42.4 mm; smaller than the 7th sodium ring by 0.40.4 mm
2.82.8 mm; larger than the 7th sodium ring by 0.10.1 mm
2.42.4 mm; larger than the 7th sodium ring by 0.10.1 mm
2.82.8 mm; smaller than the 7th sodium ring by 0.40.4 mm
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College Physics Quiz

College Physics Quiz: Thin Film Interference

Practice Thin Film Interference in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Thin Film Interference, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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Question 1

Newton's rings are formed between a convex lens and a flat glass plate. The setup is illuminated with sodium light (λ=589\lambda = 589 nm) and the radius of the 5th bright ring is measured to be 2.22.2 mm. If the same setup is then illuminated with blue light (λ=450\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?

  1. 2.42.4 mm; smaller than the 7th sodium ring by 0.40.4 mm (correct answer)
  2. 2.82.8 mm; larger than the 7th sodium ring by 0.10.1 mm
  3. 2.42.4 mm; larger than the 7th sodium ring by 0.10.1 mm
  4. 2.82.8 mm; smaller than the 7th sodium ring by 0.40.4 mm
Explanation: For Newton's rings, the radius of the mmth bright ring is rm=(m+1/2)Rλr_m = \sqrt{(m + 1/2)R\lambda}, where RR is the lens radius of curvature. From the 5th bright ring with sodium light: 2.2=5.5R×5892.2 = \sqrt{5.5R \times 589}, solving for RR: R=(2.2)2/(5.5×589)=1.49R = (2.2)^2/(5.5 \times 589) = 1.49 mm. For the 7th bright ring with blue light: r7=7.5×1.49×450=5033=2.4r_7 = \sqrt{7.5 \times 1.49 \times 450} = \sqrt{5033} = 2.4 mm. For the 7th sodium ring: r7=7.5×1.49×589=6590=2.8r_7 = \sqrt{7.5 \times 1.49 \times 589} = \sqrt{6590} = 2.8 mm. The blue ring is smaller by 2.8−2.4=0.42.8 - 2.4 = 0.4 mm.

Question 2

A thin soap film (n=1.33n = 1.33) floating in air shows first-order constructive interference for yellow light (λ=580\lambda = 580 nm) at normal incidence. If this same film is then immersed in water (n=1.33n = 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?

  1. No interference pattern visible; no wavelength shows constructive interference (correct answer)
  2. Pattern shifts; 435435 nm (blue) now shows constructive interference
  3. Pattern reverses; 580580 nm now shows destructive interference, 290290 nm shows constructive interference
  4. Pattern shifts; 773773 nm (infrared) now shows constructive interference
Explanation: Initially in air, the soap film (n=1.33n = 1.33) has phase changes at both surfaces since nair<nsoapn_{air} < n_{soap} and the film is very thin with air on both sides. For first-order constructive interference: 2nt=0.5λ2nt = 0.5\lambda, so t=λ/(4n)=580/(4×1.33)=109t = \lambda/(4n) = 580/(4 \times 1.33) = 109 nm. When immersed in water, the soap film has the same refractive index as water (n=1.33n = 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.