5. [30 points] Parts A and B of this problem are independent.

    1. A periodic plane wave passes from one medium into another. The speed of the wave in the second medium is less than that in the first. The rays [i.e. wave vectors] shown below right illustrate this situation; however, the boundary between the two regions is not shown.

      1. [7 pts] On the diagram, draw and label a line to represent the location and orientation of a (straight) boundary that is consistent with the information given. Explain your reasoning.

        2. Since the speed of the wave is slower in the second medium, the ray must be bent toward the normal.

         

      2. [5 pts] Is the frequency of the refracted wave greater than, less than, or equal to the frequency of the incident wave? Explain.

      As the wave crosses the boundary, each crest is transmitted as a crest and each trough is transmitted as a trough without delay. Therefore the frequency of the refracted wave is the same as the frequency of the incident wave.

    2. The following experiment is done entirely in a very large pool of calm water. Monochromatic light is incident upon a mask containing a single slit of width w. A diffraction pattern is observed on a screen that is placed far from the mask. (The light source, screen, and mask are all submerged in the water.)

      3.  

      For each of the following possible changes to the setup, determine whether each change would cause the angle to the first minimum to increase, decrease, or remain the same as before. Explain your reasoning in each case.

       

      1. [7 pts] The width of the slit w is decreased.

        2. The location of the first diffraction minimum depends on the width of the slit (a sin q = l ). If the width of the slit is decreased, sin q and therefore q must increase.

         

      2. [5 pts] The screen is moved closer to the mask.

        3. The location of the first minimum (measured as the angle from the center of the screen) does not depend on the distance between the mask and the screen, so the angle will not change.

      3. [6 pts] The pool is drained of water and the experiment is performed again.

Air has a lower index of refraction than water, so the speed of the light in air will be greater than in water but the frequency will remain the same — therefore the wavelength of the light will be longer in air than in water. Since the location of the first minimum is dependent on the wavelength of the light,

(a sin q = l ), if l increases sin q and therefore q must increase.