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Unformatted text preview: 1017 Chapter 33 Interference and Diffraction Conceptual Problems *1 Determine the Concept The energy is distributed nonuniformly in space; in some regions the energy is below average (destructive interference), in others it is higher than average (constructive interference). 2 Determine the Concept Coherent sources have a constant phase difference. The pairs of light sources that satisfy this criterion are ( b ), ( c ), and ( e ). 3 Determine the Concept The thickness of the air space between the flat glass and the lens is approximately proportional to the square of d , the diameter of the ring. Consequently, the separation between adjacent rings is proportional to 1/ d . 4 Determine the Concept The distance between adjacent fringes is so small that the fringes are not resolved by the eye. 5 Determine the Concept If the film is thick, the various colors (i.e., different wavelengths) will give constructive and destructive interference at that thickness. Consequently, what one observes is the reflected intensity of white light. *6 ( a ) The phase change on reflection from the front surface of the film is 180 ; the phase change on reflection from the back surface of the film is 0 . As the film thins toward the top, the phase change associated with the films thickness becomes negligible and the two reflected waves interfere destructively. ( b ) The first constructive interference will arise when t = /4. Therefore, the first band will be violet (shortest visible wavelength). ( c ) When viewed in transmitted light, the top of the film is white, since no light is reflected. The colors of the bands are those complementary to the colors seen in reflected light; i.e., the top band will be red. Chapter 33 1018 7 Determine the Concept The first zeroes in the intensity occur at angles given by . sin a = Hence, decreasing a increases and the diffraction pattern becomes wider. 8 Determine the Concept Equation 332 expresses the condition for an intensity maximum i slit interference. Here d is the slit separation, the wavelength of the light, m an integer, a the angle at which the interference maximum appears. Equation 3311 expresses the condition for the first minimum in singleslit diffraction. Here a is the width of the slit, the wavelength of the light, and the angle at which the first minimum appears, assuming m = 1. 9 Picture the Problem We can solve m d = sin for with m = 1 to express the location of the firstorder maximum as a function of the wavelength of the light. The interference maxima in a diffraction pattern are at angles given by: m d = sin where d is the separation of the slits and m = 0, 1, 2, Solve for the angular location 1 of the firstorder maximum : = d 1 1 sin Because green light < red light : light red light green < and correct. is ) ( a *10 Determine the Concept The distance on the screen to...
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This homework help was uploaded on 02/26/2008 for the course PHYSICS 11 taught by Professor Licini during the Spring '07 term at Lehigh University .
 Spring '07
 Licini
 Energy, Diffraction

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