Math Problem Statement

Das an der Oberfläche einer Seifenblase reflektierte Tageslicht erscheint bei Betrachtung unter dem Winkel α = 50° gegenüber dem Einfallslot grün; es hat eine Wellenlänge λ1 = 540 nm. Es wird die Beugungsordnung m = 1 betrachtet, die Flüssigkeit habe die Brechzahl n = 1.33. a) Welche Dicke d hat die Flüssigkeitshaut? b) Welche Wellenlänge λ2 und Farbe hat das reflektierte Licht bei senkrechter Betrachtung (in Richtung des Einfallslotes)?

Solution

To solve the given problem, let's break it down step-by-step.


Given Data:

  1. Incident angle: α=50\alpha = 50^\circ
  2. Wavelength λ1\lambda_1: 540nm540 \, \text{nm} (green light)
  3. Order of interference: m=1m = 1
  4. Refractive index nn: 1.331.33

Part (a): Thickness dd of the liquid film

When light reflects from a thin film, interference conditions apply. The reflected light undergoes a phase change, and constructive interference occurs when:

2dncos(α)=mλ12d n \cos(\alpha) = m \lambda_1

Plugging in the given values:

  • α=50\alpha = 50^\circ
  • m=1m = 1
  • λ1=540nm=540×109m\lambda_1 = 540 \, \text{nm} = 540 \times 10^{-9} \, \text{m}
  • n=1.33n = 1.33

Calculating dd:

2d1.33cos(50)=540×1092d \cdot 1.33 \cdot \cos(50^\circ) = 540 \times 10^{-9}

We solve for dd:

d=540×10921.33cos(50)d = \frac{540 \times 10^{-9}}{2 \cdot 1.33 \cdot \cos(50^\circ)}

Now, let's compute it:

cos(50)0.6428\cos(50^\circ) \approx 0.6428

d=540×10921.330.6428d = \frac{540 \times 10^{-9}}{2 \cdot 1.33 \cdot 0.6428}

d540×1091.709d \approx \frac{540 \times 10^{-9}}{1.709}

d316nmd \approx 316 \, \text{nm}

Thus, the thickness of the liquid film is approximately 316 nm.


Part (b): Wavelength λ2\lambda_2 and color for normal incidence

For normal incidence (α=0\alpha = 0^\circ), the interference condition simplifies to:

2dn=mλ22d n = m \lambda_2

Using the thickness d=316nmd = 316 \, \text{nm} and m=1m = 1:

2316×1091.33=λ22 \cdot 316 \times 10^{-9} \cdot 1.33 = \lambda_2

Calculating λ2\lambda_2:

λ2=2316×1091.33\lambda_2 = 2 \cdot 316 \times 10^{-9} \cdot 1.33

λ2=841nm\lambda_2 = 841 \, \text{nm}

This wavelength (λ2=841nm\lambda_2 = 841 \, \text{nm}) corresponds to infrared light, which is not visible to the human eye.


Summary of Answers

  1. (a) The thickness dd of the liquid film is 316 nm.
  2. (b) The wavelength of the reflected light at normal incidence is 841 nm (infrared light).

Do you need further details or clarifications?

Related Questions:

  1. How does the interference condition change if the refractive index changes?
  2. What happens if the angle α\alpha increases beyond 5050^\circ?
  3. How does the order mm affect the thickness calculation?
  4. Can destructive interference be used to determine the thickness?
  5. What other colors can appear due to thin-film interference?

Tip:

In thin-film interference, constructive and destructive interference depend on both the film's thickness and the incident angle. Adjusting either can produce different colors.

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Math Problem Analysis

Mathematical Concepts

Optics
Interference of Light
Thin Film Interference

Formulas

2d n cos(α) = m λ (constructive interference)
2d n = m λ (for normal incidence)

Theorems

Interference Conditions in Thin Films

Suitable Grade Level

Grades 10-12 or early university level (Physics/Optics)