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Unformatted text preview: 1 University of California, Berkeley Physics H7C Spring 2011 ( Yury Kolomensky ) SOLUTION TO PROBLEM SET 5 Interference, Diffraction Maximum score: 200 points Due: March 4, 2011 Composed and formatted by Yu. Kolomensky, and T.R. Beals 1. (20 points) ( Hecht 9.4 ) Will we get an interference pattern in Youngs experi- ment (Fig. 9.8) if we replace the source S with a single long-filament lightbulb ? What would occur if we re- placed the slits S 1 and S 2 with the same lightbulbs ? 1. Solution If we replace the source S with a long-filament light- bulb, oriented along slits S 1 and S 2 , we will still see the interference pattern, as long as the angular size of the filament , as viewed from each of the slits, is sub- stantially smaller than the angular difference between a minimum and maximum of intensity /d . If the dis- tance from S to the screen with S 1 and S 2 is L , and the diameter of the filament is D , this means that the requirement for spatial coherence of the wavefronts at slits S 1 and S 2 is D L d For a typical distance L 1 m, visible light ( 500 nm), and a relatively large slit width of 500 m, this means D 1 mm. This is reasonable as long as the filament is oriented along the slit direction. If it is perpendicular to the slit direction, then the interference pattern is washed out, unless the filament is very far. In comparison, the angular size of the Sun is about Sun . 5 . 01 (incidentally, the Moons angu- lar size on Earth is about the same), so the interference patterns can only be seen in the visible ( < 780 nm) if d 78 m. Thats a pretty small size, which is why Grimaldis original experiment failed (see Hecht p. 393). If we replace S 1 and light bulbs S 2 , the experiment will fail (no interference pattern), since the sources S 1 and S 2 would be incoherent. 2. (30 points) ( Hecht 9.3 ) Return to Fig. 2.22 and prove that if two electromag- netic plane waves making an angle have the same am- plitude E , the resulting interference pattern on the yx- plane is a cosine-squared irradiance distribution given by I ( y ) = 4 I cos 2 parenleftbigg y sin parenrightbigg Locate the zeros of irradiance. What is the value of fringe separation ? What happens to the separation as increases ? Compare your analysis with that leading to Eq. (9.17). 2. Solution We have two electric fields: E 1 ( y,z,t ) = E cos bracketleftbigg 2 z t bracketrightbigg E 2 ( y,z,t ) = E cos bracketleftbigg 2 ( z cos y sin ) t bracketrightbigg We want to find the spatial distribution of (maximum) intensity, which is given by ( E 1 + E 2 ) 2 . Using a trigono- metric identity, we find the following: ( E 1 + E 2 ) 2 2 = 4 E 2 cos 2 parenleftbigg 2 ( z 1 + cos 2 y sin 2 ) t parenrightbigg cos 2 parenleftbigg y sin z (1 cos ) parenrightbigg The first cos 2 term is the time-dependent part of the ex- pression; it has maximum value of 1 and the average...
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