# Homework Help: LLoyd's Mirror Distance Dependence

1. Nov 12, 2012

### alex_b93

Hi, I was wondering if anyone could help with this problem that I am stuck on?

1. The problem statement, all variables and given/known data

I need to work out a formula for the irradiance(intensity) as a function of the distance for a LLoyd's mirror interference arrangement.

The question suggests considering the similarities with the Young Double Slit experiment.

2. Relevant equations

I know that the total irradiance is given by
I = I_1 + I_2 + [2 (I_1 I_2)^0.5] cos(δ)

I know the total irradiance for the double slit experiment is given by
I = 4I_0 cos^2(∏ay/λL)

I have found that the answer for the Lloyd's mirror is
I = 4I_0 sin^2(∏ay/λL)

So it is clearly very similar.

3. The attempt at a solution

I have the working out for the Young's set up, and I can follow that, but I can't get it to work for the LLoyd's set up, mainly because I can't find a decent diagram for the angles.

Any help or if anybody knows of a decent diagram would be much appreciated.

2. Nov 12, 2012

### collinsmark

Hello alex_b93,

The Wikipedia article on Lloyd's mirror, http://en.wikipedia.org/wiki/Lloyd%27s_mirror, provides a basic diagram. Think of the distance between the two "slits" (slit's as in Young's experiment) as the distance between the real source and the virtual (reflection) source. The figure in the wiki gives you an idea on how to extrapolate where the virtual source is (by extrapolating the reflection). You'll have to label the angles yourself, but the skeleton of the diagram is there.

There is an important difference between the two slits in Young's experiment, and the two "slits" in Lloyd's experiment (besides one the slits being virtual). In Lloyd's experiment, the virtual "slit" is out-of-phase by 180o, compared to a hypothetical, real slit, all else being equal, due to the reflection. That has a big impact on what ends up as a maxima as opposed to a minima.

Last edited: Nov 12, 2012