Supplementary Material

Supplementary Material This set of slides contains material dealing with thin films and with the Michelson Interferometer. Both of these phenomena can be explained nicely by the interference of waves. Also included are slides about the telescope.

Interference: Thin Films Before, we had several different parts of a wide beam interfering with one another. Can we find other ways of having parts of a beam interfere with other parts?

Interference: Thin Films We can also use reflection and refraction to get different parts of a beam to interfere with one another by using a thin film. reflected red interferes with refracted/reflected/refracted blue. air film water

Interference: Thin Films Blue travels an extra distance of 2t in the film. reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films And, blue undergoes two refractions and reflects off of a different surface. reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films When a wave encounters a new medium: • the phase of the refracted wave is NOT affected. • the phase of the reflected wave MAY BE affected.

Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wavemust interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

Interference: Thin Films When a wave on a string encounters a free end, the reflected wavedoes NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

Interference: Thin Films • When light is incident on a SLOWERmedium (one of index of refraction higher than the one it is in), the reflected wave is 180 degrees out of phase with the incident wave. • When light is incident on a FASTER medium, the reflected wave does NOT undergo a 180 degree phase shift.

Interference: Thin Films If na < nf< nw, BOTH red and blue reflected rays will be going from fast to slow, and no difference in phase will be due to reflection. reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films If na < nf> nw, there WILL be a 180 degree phase difference (/2) due to reflection. reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films There will ALWAYS be a phase difference due to the extra distance of 2t/. reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films When t=/2 the phase difference due to path is 360 degrees (equivalent to no difference) reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films When t=/4 the phase difference due to path is 180 degrees. reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a/nf. reflected red interferes with refracted/reflected/refracted blue. air film t water

Interference: Thin Films • reflection: no difference if nf < nw; 180 degree difference if nf > nw. • distance: no difference if t = a/2nf 180 degree difference if t = a/4nf Total phase difference is sum of the above two effects.

Interference: Thin Films Total phase difference is sum of the two effects of distance and reflection • For minimum reflection, need total to be 180 degrees. • anti-reflective coating on lens • For maximum reflection, need total to be 0 degrees. • colors on oil slick

Thin Films: an example An oil slick preferentially reflects green light. The index of refraction of the oil is 1.65, that of water is 1.33, and of course that of air is 1.00 . What is the thickness of the oil slick?

Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).

Thin Films: an example • Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). • Since we have nf > nw, we have 180 degrees due to reflection.

Thin Films: an example • Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). • Since we have nf > nw, we have 180 degrees due to reflection. • Therefore, we need 180 degrees due to extra distance, so need t = a/4nf where a = 500 nm, nf = 1.65, and so: t = 500 nm / 4(1.65) = 76 nm.

Michelson Interferometer Split a beam with a Half Mirror, the use mirrors to recombine the two beams. Mirror Half Mirror Light source Mirror Screen

Michelson Interferometer If the red beam goes the same length as the blue beam, then the two beams will constructively interfere and a bright spot will appear on screen. Mirror Half Mirror Light source Mirror Screen

Michelson Interferometer If the blue beamgoes a little extra distance, s, the the screen will show a different interference pattern. Mirror Half Mirror Light source Mirror s Screen

Michelson Interferometer If s = /4, then the interference pattern changes from bright to dark. Mirror Half Mirror Light source Mirror s Screen

Michelson Interferometer If s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift). Mirror Half Mirror Light source Mirror s Screen

Michelson Interferometer By counting the number of fringe shifts, we can determine how far s is! Mirror Half Mirror Light source Mirror s Screen

Michelson Interferometer If we use the red laser (=632 nm), then each fringe shift corresponds to a distance the mirror moves of 316 nm (about 1/3 of a micron)! Mirror Half Mirror Light source Mirror s Screen

Limits on Resolution: further examples: telescope Light from far away is almost parallel. eyepiece objective lens fo fe

Limits on Resolution: further examples: telescope The telescope collects and concentrates light. eyepiece objective lens fo fe

Limits on Resolution: further examples: telescope Light coming in at an angle, in is magnified to out . eyepiece objective lens x fo fe

Limits on Resolution: further examples: telescope in = x/fo, out = x/fe; M = out/in = fo/fe eyepiece objective lens x fo fe

Limits on Resolution: further examples: telescopes telescopes • magnification: M =out/in = fo /fe • light gathering: Amt D2 • resolution: 1.22  = D sin(limit) so in = limit and out = 5 arc minutes so limit 1/D implies Museful = 60 * D where D is in inches • surface must be smooth on order of 

Limits on Resolution: Telescope Mmax useful = out/in = eye/limit = 5 arc min / (1.22 *  / D) radians = (5/60)*(/180) / (1.22 * 5.5 x 10-7 m / D) = (2167 / m) * D * (1 m / 100 cm) * (2.54 cm / 1 in) = (55 / in) * D

Limits on Resolution: Telescope What diameter telescope would you need to read letters the size of license plate numbers from a spy satellite?

Limits on Resolution: Telescope • need to resolve an “x” size of about 1 cm • “s” is on order of 100 miles or 150 km • limit then must be (in radians) = 1 cm / 150 km = 7 x 10-8 • limit = 1.22 x 5.5 x 10-7 m / D = 7 x 10-8 so D = 10 m (Hubble has a 2.4 m diameter)

Supplementary Material