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\(\bullet\) A thin film of polystyrene of refractive index 1.49 is used as a nonreflecting coating for Fabulite (strontium titanate) of refractive index \(2.409 .\) What is the minimum thickness of the film required? Assume that the wavelength of the light in air is 480 \(\mathrm{nm}\) .

Short Answer

Expert verified
The minimum thickness is approximately 80.54 nm.

Step by step solution

01

Understanding Thin Film Interference

To find the minimum thickness of the polystyrene film, we need to apply the concept of thin film interference. For the film to be non-reflective, the path difference between light reflecting from the top and the bottom of the film should be half a wavelength, causing destructive interference.
02

Wavelength in the Film

First, determine the wavelength of light in the polystyrene film. Use the formula \( \lambda_f = \frac{\lambda_0}{n} \), where \( \lambda_0 \) is the wavelength in air (480 nm) and \( n \) is the refractive index of polystyrene (1.49). Thus, \( \lambda_f = \frac{480}{1.49} \approx 322.15 \text{ nm} \).
03

Condition for Minimum Thickness

For destructive interference, the thickness \( t \) of the film must be \( \frac{\lambda_f}{4} \). This ensures that the reflected waves are out of phase. Thus, \( t = \frac{322.15 \text{ nm}}{4} \approx 80.54 \text{ nm} \).
04

Conclusion

The minimum thickness required for the polystyrene film to act as a nonreflecting coating is approximately 80.54 nm.

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Key Concepts

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

Non-reflective Coatings
Non-reflective coatings are fascinating solutions to minimize unwanted reflections from surfaces like glass or lenses. Such coatings are designed to reduce light reflection. This is achieved by using thin films with specific optical properties. When light reaches the surface, it passes through the thin film coating before hitting the substrate beneath. The goal is for the reflected light from the top surface and the interface of the coating and the substrate to interfere destructively, cancelling each other out. This makes the surface appear less reflective. Non-reflective coatings are widely used in applications such as camera lenses, eyeglasses, and display screens, enhancing the quality of the visual experience by increasing transparency and reducing glare.
By choosing the right thickness and material for the coating, engineers skillfully control how light is reflected and transmitted, achieving the desired optical effect.
Refractive Index
Refractive index is a measure of how much light slows down when it enters a material. It's a key concept in optics as it helps us understand light behavior across different media. This value is expressed as the ratio of the speed of light in a vacuum to the speed of light in the material. The higher the refractive index, the slower light travels through the material. In our example, the polystyrene coating has a refractive index of 1.49, meaning light travels slower than it does in air (which has a refractive index of approximately 1).
The concept of refractive index is crucial in designing non-reflective coatings, as it enables the calculation of the needed film thickness for effective interference.
  • Polystyrene: Refractive index = 1.49
  • Fabulite (Strontium titanate): Refractive index = 2.409
Understanding these differences helps in creating optimal conditions for minimizing reflections.
Destructive Interference
Destructive interference occurs when two waves combine to form a smaller amplitude wave. In the context of thin film coatings, when two reflected light waves are out of phase, they cancel each other out. This happens when the path difference between them is half a wavelength (\( rac{ ext{a}}{2}\) phase shift). For destructive interference to be used effectively in non-reflective coatings, the film thickness must be calculated precisely. This ensures that the reflected rays from the top and bottom surfaces of the film interfere destructively.
For example, in a polystyrene film, the wavelength of light within the film changes due to its refractive index. After calculating, the minimum thickness for destruction interference is found using the formula: \[ t = \frac{\lambda_{f}}{4} \] where \( \lambda_{f} \) is the wavelength of the light in the film. Here, for light of 480 nm originally, \( \lambda_{f} \) becomes 322.15 nm, resulting in a required thickness of about 80.54 nm for effective interference.
This neat trick effectively reduces the reflection of certain wavelengths, giving rise to their name—non-reflective surfaces.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illumi- nated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), includ- ing the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ?\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

\(\bullet\) Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wave- length. It then falls on two slits separated by 0.460 \(\mathrm{mm} .\) In the resulting interference pattern on a screen 2.20 \(\mathrm{m}\) away, adjacent bright fringes are separated by 2.82 \(\mathrm{mm}\) . What is the wave- length of the light that falls on the slits?

\(\bullet\) If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 \(\mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 \(\mathrm{nm} ) ?\)

\(\bullet$$\bullet\) A glass sheet measuring 10.0 \(\mathrm{cm} \times 25.0 \mathrm{cm}\) is covered by a very thin opaque coat- ing. In the middle of this sheet is a thin, straight scratch 0.00125 \(\mathrm{mm}\) thick, as shown in Figure \(26.50 .\) The sheet is totally immersed beneath the surface of a liquid having an index of refraction of \(1.45 .\) Monochromatic light strikes the sheet perpendicular to its surface and passes through the scratch. A screen is placed under water a distance 30.0 \(\mathrm{cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on this screen are 22.4 \(\mathrm{cm}\) apart. What is the wavelength of the light in air?

\(\bullet\) Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured care- fully on a screen 1.20 \(\mathrm{m}\) away from the double slit, and the cen- ter of the 20 th fringe (not counting the central bright fringe) is found to be 10.6 \(\mathrm{mm}\) from the center of the central bright fringe. What is the separation of the two slits?

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