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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. - PowerPoint PPT Presentation

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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.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

Interference: Thin Films

Blue travels an extra distance of 2t in the film.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

Interference: Thin Films

And, blue undergoes two refractions and reflects off of a different surface.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

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 wave must 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 fixed end, the reflected wave must 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 wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

Interference: Thin Films

When a wave on a string encounters a free end, the reflected wave does 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 SLOWER medium (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.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

Interference: Thin Films

If na < nf > nw, there WILL be a 180 degree phase difference (/2) due to reflection.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

Interference: Thin Films

There will ALWAYS be a phase difference due to the extra distance of 2t/.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

Interference: Thin Films

When t=/2 the phase difference due to path is 360 degrees (equivalent to no difference)

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

Interference: Thin Films

When t=/4 the phase difference due to path is 180 degrees.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

Interference: Thin Films

Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a/nf.

air

film

water

reflected red interferes withrefracted/reflected/refracted blue.

t

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

Mirror

Half Mirror

Screen

Lightsource

Michelson InterferometerIf 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

Mirror

Half Mirror

Screen

Lightsource

Michelson Interferometer

If the blue beam goes a little extra distance, s, the the screen will show a different interference pattern.

Mirror

Mirror

Half Mirror

Screen

Lightsource

s

Michelson Interferometer

If s = /4, then the interference pattern changes from bright to dark.

Mirror

Mirror

Half Mirror

Screen

Lightsource

s

Michelson Interferometer

If s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift).

Mirror

Mirror

Half Mirror

Screen

Lightsource

s

Michelson Interferometer

By counting the number of fringe shifts, we can determine how far s is!

Mirror

Mirror

Half Mirror

Screen

Lightsource

s

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

Mirror

Half Mirror

Screen

Lightsource

s

Limits on Resolution: further examples: telescope

Light from far away is almost parallel.

objectivelens

eyepiece

fofe

Limits on Resolution: further examples: telescope

The telescope collects and concentrates light.

objectivelens

eyepiece

fofe

Limits on Resolution: further examples: telescope

Light coming in at an angle, in is magnified to out .

objectivelens

eyepiece

fofe

x

Limits on Resolution: further examples: telescope

in = x/fo, out = x/fe; M = out/in = fo/fe

objectivelens

eyepiece

fofe

x

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)

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