# Visable light

1. Mar 7, 2006

### truckerron1

A tiny layer of oil (n = 1.25) is situated on top of a water puddle (n = 1.33) in a parking lot. If the thickness of the oil is 242 nm, the what color(s) of visible light will give a strong reflection?

2t=m*lambda/n2

2t=242nm/1.25 and then 2t=242nm/1.33
i just dont know what goes where on this problem plz help thanks ron

2. Mar 7, 2006

### Staff: Mentor

242 nm is the thickness of the oil (t), not the wavelength of the light (lambda).

3. Mar 7, 2006

### truckerron1

so how do i set it up if t=242

4. Mar 7, 2006

### truckerron1

so would it be
2(242)=m*1.25/1.33

5. Mar 7, 2006

### lightgrav

No, NONE of the light that gets reflected travels through the water.
The extra path length (which is responsible for the constructive interference)
travels down thru the OIL and back up thru the OIL.
You should expect the OIL thickness to be associated with the OIL index.

You're trying to SOLVE for lambda ...

Last edited: Mar 7, 2006
6. Mar 8, 2006

### Staff: Mentor

In addition to lightgrav's useful comment, I'll add that at this point you may be wondering what to use for m?

Each integer value of m (1, 2, 3, ...) gives a possible wavelength, in principle. However, you specifically need only the wavelengths that are visible, that is, in the range 400nm to 700nm. So start with m = 1 and calculate lambda for each value of m in turn. It will be pretty obvious when you can safely stop (i.e. after you've gone through the entire visible range).