Recent content by sorrel

current is directly proportional to intensity..so the graph of current against wavelength would be the same for graph of intensity against wavelength which means u don't actually need to work out the intensity.

oh and btw thanks megegg!

united fan- i think measuring the current in the circuit is right but i don't think the lamp should be part of the circuit. it should be separate. hope that helps

hey I am confused on part b...how the wavelength of the light falling on the LDR is determined...as if u r using a diffraction grating or prism how do u know?? sorrel