|
ID/Type |
Web Link or WA Question Code |
Local download |
Launch from browser |
Description |
|
L19A |
L19-01 |
point-interference-02.iwp |
point-interference-02.iwp |
Waves are incident from the left on a barrier. At t = 0, two apertures open in the barrier. Waves emerging from the two apertures interfere. At positions where the two waves reach the screen in phase, the waves will interfere constructively. Note that when Point P is at the midpoint of a fringe, the ratio of the Path Difference to the Wavelength (see Outputs) is an integer. This is the condition for constructive interference. Determine the wavelength. Note that the distance between grid markings is 10 cm.
Refer to the Enlarged View to see the geometrical relationships between the various parameters. |
|
L19A |
L19-02 |
point-interference-03.iwp |
point-interference-03.iwp |
Waves are incident from the left on a barrier. At t = 0, two apertures open in the barrier. Waves emerging from the two apertures interfere. The pattern of bright interference fringes is shown on the screen. Determine the separation of the slits. |
|
L19A |
L19-03 |
point-interference-04.iwp |
point-interference-04.iwp |
Waves are incident from the left on a barrier. At t = 0, two apertures open in the barrier. Waves emerging from the two apertures interfere. Determine the vertical position on the screen of the center of the bright fringe for m = 2 and the dark fringe for m = 1/2. |
|
E.28.02 |
APB-28-02-05b |
thin-film-jc-01.iwp |
thin-film-jc-01.iwp |
A light wave of the given frequency is incident from air on a thin film of the given thickness and index of refraction. The wavelength of the light decreases in the film due to the fact that the latter's index of refraction is greater than that of air. The wave that reflects from the upper boundary undergoes a phase inversion because the light is reflecting off a medium with a greater index of refraction than the incident medium. However, there is no phase inversion when the wave reflects from the lower boundary.
Note that the colors of the waves are simply intended to help distinguish the waves and are not indicative of the actual color of the light. Note also that the applet does not correctly depict the relative amplitudes of the incident, refracted, and reflected waves. |
|
E.28.02 |
APB-28-02-06 |
thin-film-jc-01.iwp |
thin-film-jc-01.iwp |
A light wave of the given frequency is incident from air on a thin film of the given thickness and index of refraction. The wavelength of the light decreases in the film due to the fact that the latter's index of refraction is greater than that of air. The wave that reflects from the upper boundary undergoes a phase inversion because the light is reflecting off a medium with a greater index of refraction than the incident medium. However, there is no phase inversion when the wave reflects from the lower boundary.
Note that the colors of the waves are simply intended to help distinguish the waves and are not indicative of the actual color of the light. Note also that the applet does not correctly depict the relative amplitudes of the incident, refracted, and reflected waves. |
|
E.28.02 |
APB-28-02-01t |
air-wedge-1a.iwp |
air-wedge-1a.iwp |
Monochromatic light is incident on an air wedge composed of two glass slides. The angle of the wedge may be changed by changing the height of the triangular post. Playing the animation advances the position of the incident ray. The effective path length is given as an output in multiples of wavelengths in air. This is the total distance (in wavelengths) traveled by the ray between the glass plates plus 0.5 wavelength for the phase inversion upon reflection from the lower plate.
The rays reflected from the lower surface of the upper plate and the upper surface of the lower plate interfere. The interference is constructive if the effective path length is an integral number of wavelengths and is destructive if the effective path length is an odd half-integral number of wavelengths.
All distance units other than the effective path length are micrometers (10^-6 m).
Refraction of the rays in the glass plates is not shown. |
|
E.28.02 |
APB-28-02-02a |
air-wedge-3a.iwp |
air-wedge-3a.iwp |
Light of the given wavelength is incident on an air wedge. Determine the height of the post. Greatest accuracy is achieved by changing the angle of incidence to 0 deg and positioning the incident ray at the position of the post (in order to maximize the effective path length). |
|
E.28.02 |
APB-28-02-03t |
thin-film-1b.iwp |
thin-film-1b.iwp |
A ray of light is incident from air (blue/1) on a thin film (yellow/2). The film is deposited on a transparent medium (gray/3). Light is reflected from the upper and lower surfaces of the film. The reflected rays interfere. The difference in phase changes (phase difference) between the reflected rays is given in units of wavelengths. Playing the animation will advance the wavelengths through the range of visible light.
1. Rank the media in order of increasing index of refraction. (What evidence do you have for your choice?)
2. Based on your rankings, which of the reflected rays, if either, undergo phase reversal on reflection?
3. Determine to the nearest nanometer the wavelength of visible light, if any, that undergoes complete constructive interference for the given film thickness and an angle of incidence of 0 deg.
4. Determine to the nearest nanometer the wavelength of visible light, if any, that undergoes complete destructive interference for the given film thickness and an angle of incidence of 0 deg. |
|
E.28.02 |
APB-28-02-04b |
thin-film-2b.iwp |
thin-film-2b.iwp |
A ray of light is incident from air on a thin soap film (yellow). The medium below the film is also air. Light is reflected from the upper and lower surfaces of the film. The reflected rays interfere. Playing the animation will increase the film thickness by the indicated increment.
a. Set the angle of incidence to 0 deg. For a given wavelength and film thickness, determine the phase difference.
b. Determine to the nearest nanometer the smallest film thickness for which the reflected rays undergo complete destructive interference at an angle of incidence of 0 deg.
c. Determine to the nearest nanometer the next higher film thickness for which the reflected rays undergo complete destructive interference at an angle of incidence of 0 deg. |