Measure Lloyd's mirror interference a double slit with only one slit Interference with a reflection To achieve interference of light we have to combine two or more beams that remain in step -- we say "in phase" -- for all time. For most light sources this means we must take the same incoming light and provide two different paths for it. As you would see in the physics explanation on this website, one method is to use two slits or pinholes that sample light from a distant source at two places. The light that passes through them will be in phase because the paths from the distant source to the two apertures are the same. in the wave picture the disturbances of the electromagnetic field coming out of them will be in step. in the photon picture of light, each photon has equal probability of going through either aperture. At some distant point where light from either aperture may overlap the amplitudes of the waves add and interference results. Lloyd's mirror was used as early as 1834 to create this situation and demonstrate the wave nature of light, as well as to enable a measurement of its wavelength. Reection Credit: Wikipedia Stigmatella aurantiaca CC by SA73.O In this illustration a small region of light (usually a narrow slit parallel to the mirror's surface) illuminates a flat mirror and nearly grazing incidence and also a screen that is perpendicular to the mirror at its far edge. Light may directly reach the screen (shown here in blue) or may reflect off the mirror (shown in red) to interfere at each point on the screen. it is geometrically equivalent to two slits separated by twice the height of the first above the mirror's surface. Do not confuse the colors shown here with the different colors of light.l They are used so you can tell one path from another, but the two paths are possible for any color or wavelength. The source in this figure is it above the plane of the mirror's surface. It is t' from a vertical plane on which the light will be detected, something usually called a "screen" but it can be in freely open space. Two paths are shown, the direct one to the screen, and a reflected path taking a longer distance and more time to reach the screen. The reflected path appears to come from another "virtual" slit that is below and behind the mirror as seen from the screen. You could imaging removing the mirror and having another source at that virtual point and have the same geometry for the light. This is simply the geometry for the two-slit or double slit interferometer. 1. Explain why the separation of the two slits is 2h . The reflected light is different from the direct path light because of its interaction with the surface. A wave meeting a surface and then developed again leaving it changes its phase by one half a cycle on reflection. Since phase is measured as an angle from O to 360 degrees or from U to Zn radians, we describe this by saying that the phase changes 180 degrees or /r radians upon reflection. Apart from that effect, this experiment is exactly equivalent to Young's double slit. 2. If the two beams have paths that differ by exactly one wavelength A in Young's double slit, do they add constructively to make a bright interference region, or destructively to make a dark one? 3. Asking the same question again, but for Lloyd's mirror, is the interference in this case constructive or destructive? Here is our version of this experiment. The lights source in this photo is an LED flashlight with the reflector removed. lt is a "point" source that illuminates an adjustable slit. A knob on the far side allows us to change the slit opening. The slit is on a slide that has a screw which positions it left or right with respect to the plane of the mirror's surface. lt has been carefully rotated about a horizontal axis to make the line of the slit opening parallel to the plane of the mirror. Every point on the slit is equivalent to every other point when this is done. On our side of the slit, the light that leaves it illuminates a precision flat mirror. There is an aluminum coating on this side of the mirror, it's front surface, from which the light may reflect. From this side looking toward the illuminated slit we see the slit and its mirror image >7 the two slits of the double slit experiment. 7 In the lower View we have added a millimeter scale so that you can see the separation of the slits. While in this image the two slits both appear "real", the one on the right is the direct image and the one on the left is off the mirror. The separation of the real and virtual slits is twice the distance of the slit on the right above the mirrors surface. The dial gauge is a precision tool that uses a a system of gears to alter the reading on a dial when a plunger is depressed. The smallest reading on this scale is 0.001 inches. Since 1 inch is 25.4 mm, the least count of the dial gauge is 0.00254 mm. We frequently use the unit of the micro-millimeter or "micron" um , which is 106 meter or 103 mm. In that unit, one division of the dial scale is 2.54 microns. When the image shown above was taken the dial read 75 thousands of an inch. We may change that separation using the screw adjustment seen in these photos. The fringes disappeared when the dial read 42 thousands. Use the millimeter scale in the photo to make a measurement of the separation of the two slits at this setting. 4. What as the apparent separation of the slits when the dial gauge read 75 based on the millimeter scale in the image above. Estimate the separation to 0.1 mm. Compare that to the dial gauge, which will give the distance of the slit above the mirror plane. Since the light disappears when the gauge reads 42 that means the slit is behind the mirror at that setting. Therefore, = 2 X (D - on; X 0.0254 where D is the dial gauge reading, D\" = 42 is the zero point of the scale, and {I is the separation of the real and virtual slits in millimeters. The factor 0.0254 is the conversion to millimeters from thousands of an inch. How does this compare with the separation we see in the image above? The dial gauge will be more accurate than a millimeter scale when the separations are small. Interference is observed anywhere beyond the mirror in the space where the camera that took these pictures was located. To effectively have a full mirror in front of the image sensor we would have to place the sensor exactly at the back of the mirror. We achieve the optical equivalent by focusing on the near edge of the mirror. In this plane the light from the real and virtual slit has diverged because of diffraction at the slit itself fills the space with overlapping light. Closer inspection reveals interference fringes. They become more distinct when we move the slit closer to the plane of the mirror, effectively reducing the separation of the slit and its mirror reflection. We start with a small separation just large enough to let light through and create a pattern we can record. These images show a close up view of the light spot where interference occurs in Lloyd's mirror. PLEASE I NEED HELP WITH THESE QUESTIONS !!! DExplain why the separation of the two slits is 2h 2)If the two beams have paths that differ by exactly one wavelength it in Young's double slit, do they add constructively to make a bright interference region, or destructively to make a dark one? lConstructively OR De structivelyl 3)lf the two beams have paths that differ by exactly one wavelength A in Lloyd's mirror, do they add constructively to make a bright interference region, or destructively to make a dark one? (Constructively OR De structively) 4)What as the apparent separation of the slits when the dial gauge read 75 based on the millimeter scale in the image? 5)Observe and comment on the patterns you see in the interference fringes as the spacing is changed 6)Using our calibration of Do validated by the direct measurement with a scale in # 4, what is the slit separation h in each case for dial readings -- 43, 45 , 47, 49, 51, 53, and 55 that produced the fringe patterns in these images? Note that the uncertainty in the dial reading is i]. What uncertainty will this introduce into a measurement of wavelength that depends on d? 7 )B ased on that image, what is the ratio of the wavelength or red light to the wavelength of blue light? 8) A. Describe how you made the measurements using the images on the screen and the scale below them. Estimate the errors in the fringe spacing for each case, and then pair up the separation of the slit and its mirror image d with the spacing of the dark fringes Ax and report them here. The wavelength of the light responsible for the patterns would be lambda equals space left parenthesis capital delta x right parenthesis cross times left parenthesis d divided by ell right parenthesis A = (Ax d l l) as describe on the class website. B. For each one, nd a wavelength, then average them to give a mean value with a standard deviation for the set of data. Convert this average and its uncertainty to the commonly uses units of angstroms, microns, and meters. C. In the image for the dial reading 45, the white fringes are big and show obvious color. What is the ordering of the colors and why does that occur