The Thickness of an Air Wedge A beam of light is sent directly down onto a glass

Question

The Thickness of an Air Wedge

A beam of light is sent directly down onto a glass plate (n = 1.5) and a plastic plate (n = 1.2) that form a thin wedge of air (see the drawing). An observer looking down through the glass plate sees the fringe pattern shown in the lower part of the drawing, with dark fringes at the ends A and B.

(a) What, if any, phase change occurs when the light, traveling in the glass, reflects from the interface between the glass and the air wedge? No phase change occurs. A phase change equivalent to %u03BBair occurs, where %u03BBair is the wavelength in the air wedge. A phase change equivalent to %u03BBair occurs, where %u03BBair is the wavelength in the air wedge.


(b) What, if any, phase change occurs when the light, traveling in the air wedge, reflects from the interface between the air and the top surface of the plastic plate? A phase change equivalent to %u03BBair occurs, where %u03BBair is the wavelength in the air wedge. A phase change equivalent to %u03BBair occurs, where %u03BBair is the wavelength in the air wedge. No phase change occurs.


(c) Why is there a dark fringe at A? There is a dark fringe at A because the light reflected from the top surface of the plastic and the light reflected from the bottom surface of the glass are one-half wavelength out of phase and destructive interference between these two light waves occurs. There is a dark fringe at A because the light reflected from the top surface of the plastic and the light reflected from the bottom surface of the glass are one wavelength out of phase and destructive interference between these two light waves occurs. There is a dark fringe at A because the light reflected from the top surface of the plastic and the light reflected from the bottom surface of the glass are one wavelength out of phase and constructive interference between these two light waves occurs. There is a dark fringe at A because the light reflected from the top surface of the plastic and the light reflected from the bottom surface of the glass are one-half wavelength out of phase and constructive interference between these two light waves occurs.


(d) How many wavelengths of light fit into the air between the glass and plastic plates at B? Three wavelengths of light fit. Four wavelengths of light fit. Five wavelengths of light fit. Sixwavelengths of light fit. Seven wavelengths of light fit.

Correct. At the second dark fringe in the bottom part of the drawing (the first is at A), thedown-and-back distance traveled by the light in the air wedge is one wavelength (in air). At the third dark fringe, the down-and-back distance traveled by the light is two wavelengths. This must be the case, so that the one-half wavelength phase difference between the wave reflecting at the glass/air interface and the wave reflecting at the air/plastic interface can lead to destructive interference. Extending this reasoning, we can see that at the seventh dark fringe at B the down-and-back distance must be six wavelengths. The thickness of the air wedge at B is one-half this distance or three wavelengths in air.



(e) The wavelength of the light is 560 nm. What is the thickness t of the air wedge at B?

Number Unit t = Your response differs from the correct answer by more than 10%. Double check your calculations. ---Select---nmPaAKN

Explanation / Answer

i cant understand the question completely as the formating is very bad.... however i will help you how to start.........hope its helpful...... at the second dark fringe (the first is at A), thedown-and-back distance traveled by the light in the air wedge isone wavelength in the air film at the third dark fringe, the down-and-backdistance traveled by the light in the air wedge is twowavelengths therefore, at the seventh dark fringe at B, thedown-and-back distance traveled by the light in the air wedge issix wavelengths since the down-and-back distance is sixwavelengths, the thickness of the air wedge at B is one-half ofthis distance, or three wavelengths the case m = 0 corresponds to the first darkfringe at A therefore, the dark fringe at B corresponds to m= 6 the thickness will be t = (1 / 2) m ?air = ........... nm

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