Interference of White Light

Task number: 1972

Two coherent rays of white light interfere with a path difference of 2.5 μm. Light components of which wavelenths (colors) exhibit an interference maximum?

Hint 1
What is the condition for two coherent waves to exhibit an interference maximum with respect to the path difference between them?
Solution of Hint 1
Two coherent waves produce an interference maximum when they meet with the same phase — their magnitudes then sum up. This occurs when the path difference \(\Delta l\) between the two waves is equal to an integer multiple of their wavelength \(\lambda\).

We write this condition as:
\[\mathrm{\Delta}l = k \lambda,\]
where k is of the maximum.

An interference minimum occurs when the two waves meet with opposite phases — their magnitudes cancel each other out. This is true when the path difference is an even multiple of the half of the wavelength:
\[ \mathrm{\Delta}l = (2k + 1)\, \frac{\lambda}{2},\]
where k is called the order of the minimum.

We are looking for interference maxima according to the first formula, in this task.
Hint 2
What is the range of wavelengths of the visible light?
Solution of Hint 2
The spectrum of the visible electromagnetic radiation (the light) consist of light of many wavelengths. The range is usually given from some 380 nm to about 740 nm.

Notation

Δl = 2,5 µm = 2,5·10⁻⁶ m	The path difference between the two rays
λ_min = 380 nm = 3,8·10⁻⁷ m	The minimum wavelength of the visible light
λ_max = 740 nm = 7,4·10⁻⁷ m	The maximum wavelength of the visible light
λ = ? (nm)	The wavelength at which occurs an interference maximum

Solution

Our aim is to calculate the wavelengths of the components of the visible light that produce an interference maximum. This occurs when the path difference Δl between the two rays is an integer product of the wavelength λ. We write this as the following condition:

\[\mathrm{\Delta} l = k\lambda,\]

where k is the order of the maximum.

We express the searched wavelength λ:

\[\lambda = \frac{\mathrm{\Delta}l}{k}. \tag{1}\]

To proceed, we need to know the values of k with which the wavelength in Equation 1 falls into the range of visible light.

We express k from the previous formula:

\[k = \frac{\mathrm{\Delta}l}{\lambda}.\]

And we use the limit values of the visible range of wavelengths:

We input the minimal visible wavelength λ_min = 3.80·10⁻⁷ m:

\[k_{min} \,\dot{=} \frac{2.5 {\cdot} 10^{−6}}{3.80 {\cdot} 10^{−7}} = 6.5. \]

And for the maximal visible wavelength λ_max = 7,40·10⁻⁷ m we get:

\[k_{max} \,\dot{=} \frac{2.5 {\cdot} 10^{−6}}{7.40 {\cdot} 10^{−7}} = 3.3. \]

An interference maximum occurs for integer multiples k of the wavelenghts. We take the integers that lay in the interval from 3.3 to 6.5. These are k = 4, k = 5, k = 6.

And we use these values of k in equation (1):

k	λ	“color” of light
4	625 nm	red-orange
5	500 nm	blue-green
6	417 nm	violet-blue

Answer
Interference maxima occur for visible light at wavelengths:
- λ = 625 nm for k = 4
- λ = 500 nm for k = 5,
- λ = 417 nm for k = 6.