Recent content by milleniumbug

getting an exact reading of light intensity, is too complex for this level. Substitute light intensity for current, as they are both affected by resistance in the same way. As resistance increases, it means that light intensity will be decreasing, and so will the current.

after speaking to my lecturer about it, he says that getting a figure for the light intensity is far too complex, and even he wouldn't know where to begin. He told me to use the diffraction grating, and put the LDR in a circuit, using a power supply and a ammeter. Use the currrent to represent...

thank you for the help so far, and yes, resistance goes down as light intensity increases. I do not have to explain in great detail about the workings of the LDR, as the assignment is on a 500 word limit. My main problem is actually working out what the light intensity is, based on the...

I know that as the light intensity decreases, the resistance increases, but do not know a formula for working out an exact intensity. As an error I have put ambient light, and limited display on meter.

You can add me to the list of people working on this as well :rolleyes: For equipment, I have put: Power Supply Ray Box Prism Paper LDR Ammeter I've chosen to use the paper method with prism, as I was unsure about using the diffraction grating. In my method, I have got as far as...