Brightness of a bulb -- quantitative demonstration

Pop quiz, hotshot:  The brightness of a light bulb depends on the bulb's

(A) voltage

(B) current

(C) resistance

(D) power

Answer:  POWER. 

Sure, for *identical* bulbs, voltage and/or current will kind of relate to brightness.  For a fixed resistance, the equations I²R and V²/R show that with the same resistance, power, and therefore brightness, will increase as current or voltage increases.  However, you'll eventually get into trouble assuming that a bulb carrying more current is necessarily brighter than one carrying less current.

It's pretty straightforward to show experimentally that voltage does not necessarily correlate with a bulb's brightness.  Just get a bunch of miniature flashlight bulbs and holders from Radio Shack or Harbor Freight or somewhere.  Be sure to get bulbs with different voltage and current ratings, so that their resistances are different.  Connect them in parallel to a battery -- they will take the same voltage, but will NOT be just as bright as one another.

Michael Gray, a veteran of my 2010 AP Summer Institute and a frequent contributor to this column, came up with a much cleverer and more subtle demonstration.  He showed with a single light bulb that the bulb's brightness depends on the power, not the voltage or current.  How?  He measured the bulb's brightness directly with a Vernier light sensor.  Of course!  Brilliant.

He connected a bulb to a battery.  He showed that, by the equation V²/R, doubling the bulb's voltage should not just double the bulb's brightness, but quadruple the brightness.  He darkened the room, and placed a light sensor a fixed distance from the bulb.  He zeroed the sensor for the ambient light, and turned the bulb on.  When he doubled the bulb's voltage, the sensor reading quadrupled.  Physics works.

I will certainly use this demonstration next year.  Thanks, Michael!

Mail Time: Is Color Determined by Wavelength or Frequency?

Visible spectrum from betesoft.com

Darren Tarshis, a physics teacher in Hayward, CA, has some physical optics questions:

Imagine that red light with a wavelength of 600 nm passes from air to a chunk of, say, diamond. In the diamond, I know the speed slows, which causes the wavelength to shorten (because the frequency remains constant). In the diamond, would the light have a different color because of its new wavelength?

I always teach my students that for a sound wave, pitch is determined by the wavelength/frequency, and for a light wave, color is determined by the wavelength/frequency. But I'm starting to think this may be incorrect, and the pitch is actually determined by frequency only, not wavelength, and color is determined by wavelength only, not frequency.

 

Yup. Frequency determines color and pitch.  The red light stays red even in diamond.

As a quick example: My voice is baritone. Imagine that you are in the pool with your ears just under water, and I am standing on deck talking to you. When the sound waves from my voice enter the water, they start moving about 4 times faster. The frequency doesn't change -- frequency of a wave NEVER changes when the wave changes materials -- so the wavelength increases by a factor of four as well. If pitch were determined by wavelength, then my voice would sound not only soprano, but squeaky soprano. 

Similarly, have you ever stood underwater and looked up at the trees overhanging the pool?  The leaves of the trees still look green, even though the light speed (and thus the wavelength) has decreased by 25%. 

I also propose a fanciful biological rationale for pitch being related to frequency only. The eardrum vibrates in response to incoming sound waves. It is the rate of vibration -- the frequency -- that can be measured by the ear and converted to a frequency. But how would an ear measure wavelength? With a meterstick? With a teeny weeny tape measure that an invisible goblin sticks out of the ear to measure the peak-to-peak distance of the incoming sound wave?

As long as the sound wave is in room temperature air, or as long as the light wave is in a vacuum (or air), then wavelength and frequency can be used to desribe color and pitch interchangably. That's why it's perfectly okay to say that red light is about 700 nm, and violet is about 400 nm. Those wavelength values must change when the light enters diamond, but the frequency of a given color will never change.

GCJ