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Chapter 27: Problem 16

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00) .\) A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 625 \(\mathrm{nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is 242 \(\mathrm{nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

Short Answer

Expert verified
The refractive index of the film is approximately 1.29.

Step by step solution

01

Understanding the problem

We need to find the refractive index (\( n_2 \)) of a thin film based on given conditions. The conditions given are that the film is on gasoline \( n=1.40 \) and open to the air \( n=1.00 \), the light wave looks bright due to constructive interference for the first nonzero minimum thickness of the film, and the wavelength of light in vacuum is 625 nm.
02

Finding wavelength in the film

To find the wavelength of light within the film, we first need to calculate it from the given wavelength in vacuum. The formula to find the wavelength in the film is:\[ \lambda_2 = \frac{\lambda_0}{n_2} \]where \( \lambda_0 \) is the wavelength in vacuum (625 nm) and \( n_2 \) is the refractive index of the film.
03

Condition for constructive interference

The condition for constructive interference for a thin film is:\[ 2n_2d = m\lambda_2 \]where \( d \) is the thickness of the film (242 nm), \( m \) is the order of the interference (1, in this minimum case), and \( \lambda_2 \) is the wavelength of light in the film.
04

Rearranging for refractive index

Rearrange the constructive interference equation to solve for \( n_2 \):\[ n_2 = \frac{m\lambda_0}{2d} \]Substituting in the values \( m = 1 \), \( \lambda_0 = 625 \text{ nm} \), and \( d = 242 \text{ nm} \):\[ n_2 = \frac{1 \times 625}{2 \times 242} = \frac{625}{484} \approx 1.29 \]
05

Conclusion

By solving the equation for \( n_2 \), we find that the refractive index of the film needs to be approximately 1.29 to fulfill the condition for constructive interference at the given thickness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index, or index of refraction, is a measure of how much light bends when it enters a material. It is denoted by the symbol \( n \).
It's calculated by the ratio of the speed of light in a vacuum to the speed of light in the material:
  • For air, the refractive index is typically around 1.00.
  • For gasoline, it's about 1.40.
A higher refractive index means light travels more slowly in the material, bending more as a result. In this exercise, we need to find the refractive index of a thin film floating on gasoline. To solve this, we'll explore how the light interference happens in between layers.
Thin Film Interference
Thin film interference occurs when light waves reflected by the upper and lower boundaries of a film combine. These reflections can enhance or reduce the intensity of the light due to interference. This effect is commonly observed in oil slicks or soap bubbles.
For constructive interference, the waves must align to strengthen each other. Thin films alter the phase of light waves traveling through them. This phase change affects interference patterns.
Understanding the refractive index of the film helps determine how light bends within it, leading to the different interference effects.
Constructive Interference
Constructive interference happens when two or more light waves meet in phase, amplifying the resultant intensity. This is one of the key reasons why the film looks bright at specific thicknesses.
For thin films, the condition for constructive interference depends on the thickness and the wavelength of the light within the film. The mathematical formula used to determine this specific scenario involves the film's refractive index \( n_2 \), its thickness \( d \), and the wavelength \( \lambda_2 \) of light in the film.The presence of bright colors under certain conditions is a beautiful and practical example of constructive interference in action.
Wavelength in Film
The wavelength of light is influenced by the medium it is traveling through. When light enters a medium, its speed changes, affecting its wavelength.The wavelength within the film can be calculated using:\[\lambda_2 = \frac{\lambda_0}{n_2}\]where \( \lambda_0 \) is the original wavelength in vacuum, and \( n_2 \) is the refractive index of the film. This is crucial in finding out the correct refractive index for the film.
Changes in wavelength within different media are fascinating instances illustrating how the properties of materials significantly impact the behavior of light, due to the refractive index.

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Most popular questions from this chapter

When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is 296 \(\mathrm{nm}\) . What is the vacuum wavelength of the light in \(\mathrm{nm}\) ?

In a Young's double-slit experiment the separation \(y\) between the second- order bright fringe and the central bright fringe on a flat screen is \(0.0180 \mathrm{~m}\) when the light has a wavelength of \(425 \mathrm{nm} .\) Assume that the angles that locate the fringes on the screen are small enough so that \(\sin \theta \approx \tan \theta .\) Find the separation \(y\) when the light has a wavelength of \(585 \mathrm{nm}\).

Light of wavelength 410 nm (in vacuum) is incident on a diffraction grating that has a slit separation of \(1.2 \times 10^{-5} \mathrm{m} .\) The distance between the grating and the viewing screen is 0.15 \(\mathrm{m} .\) A diffraction pattern is produced on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (Hint: The diffraction angles are small enough that the approximation \(\tan \theta \approx \sin \theta\) can be used.) (b) If the entire apparatus is submerged in water \(\left(n_{\text { water }}=1.33\right),\) what is the distance \(y ?\)

Light shines through a single slit whose width is \(5.6 \times 10^{-4} \mathrm{m} .\) A diffraction pattern is formed on a flat screen located 4.0 \(\mathrm{m}\) away. The distance between the middle of the central bright fringe and the first dark fringe is 3.5 \(\mathrm{mm}\) . What is the wavelength of the light?

Two stars are 3.7 \times \(10^{11} \mathrm{m}\) apart and are equally distant from the earth. A telescope has an objective lens with a diameter of 1.02 \(\mathrm{m}\) and just detects these stars as separate objects. Assume that light of wavelength 550 \(\mathrm{nm}\) is being observed. Also assume that diffraction effects, rather than atmospheric turbulence, limit the resolving power of the telescope. Find the maximum distance that these stars could be from the earth.
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