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Q & A: Measuring light intensity

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Most recent answer: 01/08/2012
Q:
Im doing a science fair project on making a lightbulb light i want to compare the difference in brightness how do i demonstrate that
- ray (age 12)
si ny us
A:
Hi Ray,
Nice project.  You'll need some sort of light measuring device.  There are many varieties.  A photographic light meter will work but may cost you $100 or so.   A cheap and straightforward way is to go down to Radio Shack and buy a CdS (Cadmium Sulfide) photoresistor pack.  A 5-pack of them ,stock number 276-1657, costs about 3 bucks.
You'll also need a multimeter to measure the resistance of the device before and after the light is shone on it.   If you don't have one you should buy one. It's a must for  every young experimenter.   Radio Shack has several models in the $20 region. 

Take the largest one of the lot and use the ohm-meter to measure the resistance both pointed toward the light and then covered up.  You should get about a factor of 10 or so difference.   The reciprocal of the resistance (after you subtract the reciprocal of the dark resistance) is proportional to the light intensity.  In other words  I= k((1/R)-(1/RD)),
where I is the light intensity, R is the resistance, and RD is the resistance in the dark. If you vary the distance from the light source to the photoresistor the intensity should fall off as the square of the distance.

LeeH

(published on 05/16/2013)

Follow-Up #1: Measuring light intensity

Q:
In your answer regarding the measurement of light intensity with photoresistors you use the equation I=k((1/R)-(1/Rd)). The variable k is not defined in your response. Could you please supply a definition? Thanks, Jon
- Jon Green (age 49)
Austin, TX, USA
A:
Hello Jon,

The value of k in the equation is a calibration constant.   It will vary from device to device so you have to determine it by illuminating the device with a "standard candle".   So the best you can do with a device of this sort is to measure relative intensities.  Making an absolute measurement can be done but it's tricky.

LeeH

(published on 01/08/2012)

Follow-up on this answer.