Total energy through a Young experiment

[quartodeciman]

SETUP

A Young experiment is set up using a continuous laser source of monochromatic plane waves with wavelength λ, an opaque barrier with two equal size vertical slits and a sliding light detector that can produce an irradiance graph for all of the light passing through the slits. The width of each slit (b) and the separation distance between the central axes of the two slits (a) are all safely larger than λ. All of these (λ, a and b) are each safely smaller in scale than the distance between the laser and the barrier with slits and also smaller than the distance between the barrier with slits and the track of the sliding light detector. This jointly means that a fraunhofer analysis of the setup should yield a valid forecast of the irradiance graph produced.

A derivation of the irradiance distribution yields this formula:

I(θ) = 4I₀(cos²α)(sin²β/β²)

where
θ is the angle that a position of the sliding detector makes with the center point between the slits,
I(θ) is the irradiance at a position of the sliding detector represented by angle θ,
I₀ is what one would expect for the irradiance at the position directly opposite one slit in a setup with just one of the two slits, if that one slit were positioned directly between the laser and the detector,
α is the difference in waves between light from one slit and light from the other slit, given by πasinθ/λ,
β is the difference in waves between light from one edge of a slit and light from the other edge of that slit, given by πbsinθ/λ,
(and to repeat,) a is the separation distance between the centers of the two slits and b is the width of each slit.

This irradiance distribution formula for a fraunhofer analysis is given in two references.
Eugene Hecht,Optics 4th ed.,Addison Wesley(2002)
Grant Fowles,Introduction to Modern Optics wnd ed.,Dover(1975)
(slight but insignificant difference in notation due to differences in defining I₀)

QUESTION

What would be the computed prediction of the integrated and averaged irradiance over the range of the detector, based on this setup and the irradiance distribution formula?

I assume that a computer and software capable of producing a Young irradiance graph would also be able to sum it and compute a mean value. The authors of the two references don't do this and I don't know how to figure the answer.

(thx to hydr0matic for the idea!)

 

[Hydr0matic]

Can anyone help us figure the answer ?

I was discussing this with quarto and I came to the following conclusion..

originally posted by Hydr0matic

I₀ = intensity from a single slit where 100% of the light goes through. (in slit 1 & 2 (double-slit) only 50% goes through each)
then
I₀ = 4 I₁ = 4 I₂

and as you said - "Across the screen, this should average out to twice the intensity from just one component"... i.e. when A₁ and A₂ are interfering,

I_tot ~ 2 I₁ = 2 I₂

so

I₀ > I_tot

i.e. the intensity of the light is a lot lower with the double slit.

 

[quartodeciman]

Hydr0matic's naming is different from mine. I follow Hecht and define I₀ as the intensity projected to the center position of the detector track or screen by one slit centrally placed, with slit width b. The formula for one slit is

I(θ) = I₀(sin²β/β²)

. θ is the angle from the normal placed at the slit barrier center, β is πbsinθ/λ (that's a PI in there!),λ is light wavelength.

Compared to Hydr0matic's quote, my I₀ ~ I₁ and I₀ ~ I₂.

The intensity at the center of the screen, when θ is zero, is I₀, then it decreases to the left and right, with shorter fringes beyond that.

The formula given for two slits of width b centered on the slit barrier, with distance a between slit centers is

I(θ) = 4I₀(sin²β/β²)(cos²α)

. α is πasinθ/λ (that's a PI in there!).

The intensity at the center of the screen, when θ is zero, is 4I₀, then it decreases to the left and right, with shorter fringes beyond that. There are also interference plunges in intensity under the diffraction envelope of intensities, left and right.

The maximum intensity at the center of the detector path or screen is four times that of the one slit case. That is explained by the coherence of the light and conditioning. Vector amplitudes from both slits add at the center and are aligned parallel, so their total magnitude is double. Squaring total amplitude gives a central intensity (properly called "irradiance") of four times that of a single slit.

Check this! Take the special case where a = b. Then the slit separation is equal to the slit width and the two slits are joined exactly. The formula becomes

I(θ) = 4I₀(sin²β/β²)(cos²β)

, since a = b means α = β. But sin²β(cos²β) = ½(sin(2β)), so

I(θ) = 4I₀(sin²(2β)/(2β)²)

, and that is just the formula for one slit at the center, with double width (2β), just as it should be. Note the forefold increase of the irradiance at the center of the detector path or screen.

My curiosity is whether the larger maximum of irradiance at the center is sufficient to offset any interference plunges, so that the total luminosity produced for the detector (or screen) preserves what is provided separately by the two slits. In short, would the sum of overall luminosities for two different single slit experiments with slit width b yield the same overall luminosity as the two slit experiment does, in spite of interference?