Question

Part I

Physics 8B Lab 8-Diffraction rev 4.0 Part I: Single-Slit Diffraction Here light shines on a single narrow slit of width a. The name may suggest only one wave passes through the slit, but that is misleading. The slit will be treated not as a single point source, but rather as having some finite size. So what you are studying is the interaction between all the waves that pass through this narrow opening. They will result in a diffraction pattern on a distant screen. Position On Screen Dist. To Screen D screen Slit Width a If you work through the superposition of all the waves passing through the slit for Intensity Minima: [email protected]??2?3A this particular geometry, you find that for Single-Slit Diffraction (slit width - a) (Although not exact, for small angles, the intensity maxima occur roughly halfway between the minima, i.e. Intensity Maxima: asi 8-..., ,0, (slit width- a) , NOTE: These are NOT just the previous equations reversed. The best way to think of it is that you applied the same principles of superposition, and these equations are what the mathematical manipulations resulted in. Set up the optical bench as follows Position the laser at one end of the optical bench. Position the screen at the other end of the optical bench. Put the slide with the 3 slits (widths of 0.4 mm, 0.2 mm, 0.1 mm) in the slide holder halfway between the laser and the screen. Orient the slide so the slits are VERTICAL 246

Answer 1

a] One shall see a diffraction pattern on the screen.

The light waves reaching the slit has a constant phase difference between them and so when their path is partially obstructed by the slit, the deviated beam now develops a path difference with respect to the other waves. If the path difference for the waves reaching at the same position on the screen is odd multiple of half wavelengths, destructive interference occurs and one observes a dark band there. If the path difference is of even multiples of wavelength, constructive interference occurs and one observes a bright band there.

b] For a slit aligned vertically, the diffraction pattern should be obtained as a horizontal intensity distribution on the screen.

The highest theoretical bright band obtained on the screen should be the m^th order where:

$\small m = \frac{2d}{\lambda}$

in the experiment, however, only 3 or 4 orders from the central maxima will be obtained due to absorption from the screen, diffusive reflection from the screen and scattering from the air between the slit and the screen.

c] If the slit is now rotated to a horizontal orientation, the diffraction pattern intensity distribution should be vertical.

d] The fringes will be aligned vertically with intensity dropping for every bright band above and below the central maxima. Since the slit width is not changed here, the spacing between the fringes should be the same as when the slit was vertical.