Thin film red light green light interference

[arod2812]

Homework Statement

A mixture of red light (wavelength vacuum = 652 nm) and green light (wavelength vacuum = 477 nm) shines perpendicularly on a soap film (n = 1.333) that has air on either side. What is the minimum nonzero thickness of the film, so that destructive interference removes the latter wavelength from the reflected light?

Homework Equations

HOW do I do this problem if this is a mixture of light? Do I simply add the wavelength value and if so at what point in the problem are they added??

The Attempt at a Solution

I used: sin(theta) = m (wavelength/n) and plugged in the value for n=1.333 and but I don't know what to plug in for wavelength.

 

[Doc Al]

Don't be adding any wavelengths, if that's what you're thinking. Read the problem more carefully:

What is the minimum nonzero thickness of the film, so that destructive interference removes the latter wavelength from the reflected light?

That's the wavelength you have to worry about.

 

[Moetasim]

I would approach this problem by first understanding the concept of thin film interference. Thin film interference occurs when light waves reflect off the top and bottom surfaces of a thin film, resulting in the interference of the two waves and the creation of a new wave. This new wave can have a different amplitude and wavelength, depending on the thickness of the film and the refractive index of the medium.

In this problem, we have a mixture of red and green light shining perpendicularly on a soap film with air on either side. The first step would be to determine the minimum nonzero thickness of the film where destructive interference would remove the green light from the reflected light. To do this, we can use the equation for destructive interference:

2nt = (m + 1/2)λ

Where:
n = refractive index of the film
t = thickness of the film
m = order of the interference (0, 1, 2, etc.)
λ = wavelength of the light

In this case, we are looking for the minimum thickness of the film, so we can assume that m = 0. We also know the refractive index of the film (n = 1.333) and the wavelength of the green light (λ = 477 nm). Plugging these values into the equation, we can solve for the thickness of the film (t):

2(1.333)t = (0 + 1/2)(477 nm)
2.666t = 238.5 nm
t = 89.4 nm

Therefore, the minimum nonzero thickness of the film would be 89.4 nm. This means that any thickness greater than this would also result in destructive interference of the green light.

It is important to note that the red light will also experience interference, but since its wavelength is longer (652 nm), the minimum thickness for destructive interference will be larger. This means that the red light will still be present in the reflected light, while the green light will be removed.

In conclusion, to solve this problem, we used the concept of thin film interference and the equation for destructive interference to determine the minimum nonzero thickness of the film where the green light will be removed from the reflected light. This approach can be applied to other similar problems involving thin film interference.