# Measurement of (brightness of) double slit experiment

1. Sep 1, 2011

### Lino

Hi,

I think that this is (should be) a straight forward question: when conducting the double slit experiment, is there a relationship between the brightness / measure of the light source and the sum of the "peaks" from the interference pattern? (I have done a number of searches but can't find a specific answer to this - any links / references would be greatly appreciated.)

On a slightly different note, am I right in saying that the quantum equivelant of this experiment (light source, silevered mirrors, half silvered mirror, detectors, etc. arranged in a square) is an idealised / thought experiment of the standard experiment?

Regards,

Noel.

2. Sep 1, 2011

### xts

Pretty simple relation - they are proportional. But that is not so easy to calculate the prortionality factor.
I am not so sure what experiment "in a square" you are thinking about. Some kind of interferometer probably. It is more or less related to double slit, but not strictly equivalent.
I don't understand why you call Young's double slit experiment "standard" and interferometer - "quantum". Both are pre-quantum - were conducted in 19th century.

3. Sep 1, 2011

### JeffKoch

The number of photons hitting the detector is the same regardless of the slit separation, assuming everything else is constant (slit width, distances, source brightness, exposure time) and assuming the angle subtended by the slit separation is very small as viewed from the source. That should give you your answer.

4. Sep 2, 2011

### Lino

Thanks xts and Jeffkoch. Have you ever come across a good article / paper discussing this aspect of the experiment?

xts, I had only ever come across the interferometer version associated with quantum based discussions / articles - hence my misrepresentation - apologies. Especially given the 19th century reference, can this be anything other than a thought experiment? I kind of figured that, even with todays technology, getting the light so harmonised (if thats the right word) and then detecting it (or not) after it has passed through the arrangement would be exceptionall difficult or even (practically) impossible! Am I right?

Regards,

Noel.

5. Sep 2, 2011

### xts

Both experiments: double slit and interferometer may be easily performed at home at cost of few € or none at all. Red laser pointer is a good light source for such play. Just a dark room, some alufoil with holes made with pins, wall as a screen, mirrors, piece of glass, etc.

My son performed Young's experiment when he was 11 years old ...

Google for hints how to make such experiments at home (you'll find lots of cases - from simples to be made in 10 mins, to more precise ones, but still at school level)

6. Sep 2, 2011

Thanks xts.