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Chapter 35: Problem 34

35.34. Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

Short Answer

Expert verified
(a) 36.5 waves; (b) phase difference is \(74\pi\).

Step by step solution

01

Convert Wavelength from nm to m

The given wavelength of the light in air is 648 nm. To perform calculations, we first convert this to meters: \(648 \text{ nm} = 648 \times 10^{-9} \text{ m}\).
02

Calculate Wavelength in the Film

The wavelength of light changes as it enters a different medium. We can find the wavelength in the film using the formula: \(\lambda_n = \frac{\lambda}{n}\), where \(n\) is the refractive index of the film. \(\lambda_n = \frac{648 \times 10^{-9}}{1.35} = 480 \times 10^{-9} \text{ m}\).
03

Calculate the Number of Waves in a Round Trip

The number of waves in the round trip within the film is given by: \(N = \frac{2L}{\lambda_n}\), where \(L\) is the thickness of the film. Substituting the values, \(N = \frac{2 \times 8.76 \times 10^{-6}}{480 \times 10^{-9}} = 36.5\). Hence, there are 36.5 waves contained along the path in the round trip.
04

Determine Phase Difference for Reflected Parts

The phase difference between the two parts of the light as they leave the film can be calculated by considering both the half-wavelength phase change upon reflection from the higher-index surface, and the additional phase change due to the path inside the film: \(2\pi N + \pi\), where \(N = 36.5\) is the round trip number of waves. The phase difference is thus: \(2\pi \times 36.5 + \pi = 73\pi + \pi = 74\pi\).

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Key Concepts

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

Wavelength Conversion
When light travels from one medium to another, its speed changes, and so does its wavelength. The wavelength conversion is especially important in studying thin films. Here鈥檚 how it works:

If light enters a medium with a refractive index different from where it came, its speed will change according to that medium's properties. However, while the speed and wavelength change, the frequency remains constant. To find the new wavelength of light in a different medium, use the formula \(\lambda_n = \frac{\lambda}{n}\), where \(\lambda\) is the original wavelength (in air), and \(n\) is the refractive index of the new medium.

For light with a wavelength of 648 nm in air entering a film with a refractive index of 1.35, the new wavelength becomes 480 nm. This shift illustrates how the wavelength conversion process allows us to understand the light's behavior in different environments.
Refractive Index
The refractive index is a crucial property of materials that describes how they affect the speed of light passing through them. It鈥檚 a measure of the bending of a ray of light when it passes from one medium into another.

Here are a few essentials about refractive index:
  • Defined as the ratio of the speed of light in a vacuum to the speed of light in the medium.
  • Indicates how much the path of light is bent, or refracted, when moving into the material.

If air has a refractive index close to 1 and the film in our exercise has a refractive index of 1.35, the light will slow down as it enters the film. This is why the wavelength changes, although the frequency stays consistent. A higher refractive index means a greater reduction in speed and shortening of the wavelength inside the material.
Phase Difference Calculation
When multiple reflections occur in thin films, the light waves emerging can interfere with each other, depending on their phase difference. Calculating phase difference is essential for understanding interference patterns which can result in phenomena such as vibrant colors across soap bubbles or oil puddles.

Here鈥檚 how phase difference calculation works in thin films:

For light reflecting off the film's surface, a half-wavelength phase shift occurs upon reflection (from higher refractive index). Besides this shift, the light traveling through the film contributes additional phase changes based on its path length, calculated using the formula \(2\pi N\), where \(N\) is the number of waves in the round trip. This exercise calculated \(N = 36.5\), resulting in a phase difference of \(2\pi \times 36.5 + \pi = 74\pi\).

This calculation helps in predicting whether the waves will constructively or destructively interfere when they reunite, resulting in the light intensities observed.

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Most popular questions from this chapter

35.22. GPSTransmission. The GPS (Global Positioning System) satellites are approximately 5.18 \(\mathrm{m}\) across and transmit two low-power signals, one of which is at 1575.42 \(\mathrm{MHz}\) (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42-MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (a) At how many other angles in the range \(0^{\circ}<\theta<90^{\circ}\) is the intensity also 2.00 \(\mathrm{W} / \mathrm{m}^{2} ?\) (b) Find the four smallest angles in the range \(0^{\circ}<\theta<90^{\circ}\) for which the intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (c) What is the intensity at a point on the circle at an angle of \(4.65^{\circ}\) from the centerline?

35.39. The radius of curvature of the convex surface of a planoconvex lens is 95.2 \(\mathrm{cm}\) . The lens is placed convex side down on a perfectly flat glass plate that is illuminated from above with red light having a wavelength of 580 \(\mathrm{nm}\) . Find the diameter of the second bright ring in the interference pattern.

35.54. White light reflects at normal incidence from the top and bottom surfaces of a glass plate \((n=1.52) .\) There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 \(\mathrm{nm}\) . What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 \(\mathrm{nm}\) ?

35.11. Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except light of a single wavelength. 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 wavelength?

35.29. Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed The plates are illuminated at normal incidence by 546 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.
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