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# Problem Set of Interference and Parallel Plates

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2.7 Suppose a virtual image is located 8 cm to the right of a lens which has a focal length of -12 cm. Find the location of the object. (Answer: 24 cm to the right.)

3.8 Suppose two parallel plates of glass are separated uniformly by a very small distance. Show that constructive interference occurs when the thickness and wavelength are related by
2t = (m + 1/2)(lambda)
Here, t is the thickness of the air gap and m is an integer, 0, 1, 2, etc. You will provide a diagram and a narrative clearly showing how you obtain this expression. (You don't need to worry about the thickness of the glass plates themselves.)

3.9 In the problem above, how would the expression be modified if there were water or oil between the plates instead of air?

3.10 Imagine a coordinate system with marks 1 meter apart, in the x- and y-directions. A speaker at the origin is producing sound of a single frequency, 660 Hz. There is another speaker on the x-axis, at x = 2.0 m, producing an identical sound. Question 1: at the point (0, 3.56 m), is there constructive interference or destructive interference of the sound from the two speakers? Question 2: What about at the point (0.60 m, 2.85 m)? Hint: find the wavelength, and then find the path length difference from the speakers to the point in question. Assume the speed of sound to be 343 m/s. Drawing a diagram will help.

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#### Solution Preview

Please see the attached files.

2.7 Suppose a virtual image is located 8 cm to the right of a lens which has a focal length of -12 cm. Find the location of the object. (Answer: 24 cm to the right.)

For thin lenses the standard equation is , where i is the distance from the center of the lens to the image, o is the distance form the center of the lens to the object, and f is the focal length of the lens. So we have i = 8, f = -12, and we get:

so o = 24 cm. But which way? When f is negative, it means that the lens is a diverging lens, which is thinner in the middle than at the edges. A diverging lens forms an image on the same side as the object. So the object is 24 cm to the right of the lens. Look at the diagram below, which I drew:

What happens is that light leaving the tip of the object (vertical arrow at 24 cm to the right) goes in all directions, but there are two important ones: down through the center of the lens (A), and parallel to the axis (horizontal), at which point it is bent by the lens upward, as if it came from the focal point (B). Light from the bottom of the object, which we have place on the axis, goes everywhere as well, but we care about the light that goes straight through the center of the lens. Any light that goes through the center of the lens at any angle is ...

#### Solution Summary

The solution is attached as a .doc and .pdf file that explains the 4 questions about using speakers and virtual images.

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