"Physical optics" on the AP Physics B exam refers to phenomena arising from light acting as a wave. The two most common situations to deal with are (1) slits -- single, double, and multiple -- and (2) thin films.

I'm not including all of the prerequisite facts about waves, sound, and light. (The "Just the facts" post on waves is found here.)

When two waves travel different distances and then interfere, the positions of constructive interference are

[So if a problem doesn't seem to fall in the categories of slits or thin films, use this general condition as a guide.]

*

When the angle

For a

A

A

When light passes through a

The wavelength

Light changes phase anytime it reflects off of a material with a higher index of refraction. If the light changes phase zero or two times, then who cares. If the light changes phase only once, though, reverse conditions for constructive and destructive interference.

When two waves travel different distances and then interfere, the positions of constructive interference are

*generally*given by setting the path difference equal to a whole number of wavelengths. The positions of destructive interference are generally given by setting the path difference equal to a half number of wavelengths.*[So if a problem doesn't seem to fall in the categories of slits or thin films, use this general condition as a guide.]

*

*Sillily, most textbooks try to be mathematically correct and concise for the Ph.D.'s by writing something like "(m + 1/2) where m is a nonnegative integer." I'm not even going to try to write the textbook expression for an odd nonnegative integer for sound-in-closed-pipes.*

*I just say "half number" and the students know exactly what I mean. Often in introductory physics clarity must be prized above precision. Those who take a second-year physics class will transition to mathematical notation just fine.**For light or sound passing through slits, the path difference is equal to*

*d*sin*θ,*where*d*is the distance between slits and*θ*can determine the position of a spot on the screen. This leads to the equation*d***sin***θ***=***mλ.*Here,*m=*0 represents the central maximum. When*m*is a whole number, you get constructive interference; when*m*is a half-number you get destructive interference.When the angle

*θ*is small and the spots on the screen are close together, the equation*x =**mλL/d*can be used instead. Here*L*represents the slit-screen distance and*x*is the distance along the screen from the central maximum. Usually this equation is fine to use.For a

**double slit**, the*central maximum is the brightest spot. The intensity on the screen fades gradually from maximum to minimum and back to maximum, producing "fringes".*A

**diffraction grating**is a set of many many slits. You see bright dots at the position of constructive interference; everywhere in between the dots is dark.A

**single slit**produces a wide, bright central maximum. Outside the central maximum, the brightness fades in and out as in a double slit, but the conditions for constructive and destructive interference are reversed: whole numbers give DEstructive interference positions, half-numbers give CONstructive positions.When light passes through a

**thin film**, the path difference is twice the thickness*t*of the film, leading to the equation*2t = m**λ*_{n}.The wavelength

*λ*represents the wavelength inside the thin film, which is equal to the wavelength in air divided by the index of refraction of the film itself._{n}Light changes phase anytime it reflects off of a material with a higher index of refraction. If the light changes phase zero or two times, then who cares. If the light changes phase only once, though, reverse conditions for constructive and destructive interference.

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