# If the illuminating source is monochromatic, will result in a series of alternating light and dark bands of gradually changing intensity on the screen

I am currently studying Modern Optical Engineering, fourth edition, by Warren Smith. Chapter 1 presents the following diagram and explanation when discussing interference and diffraction:

If the illuminating source is monochromatic, i.e., emits but a single wavelength of light, the result will be a series of alternating light and dark bands of gradually changing intensity on the screen (assuming that $$s$$, $$A$$, and $$B$$ are slits), and by careful measurement of the geometry of the slits and the separation of the bands, the wavelength of the radiation may be computed. (The distance $$AB$$ should be less than a millimeter and the distance from the slits to the screen should be to the order of a meter to conduct this experiment.)

I'm wondering why, if the illuminating source is monochromatic, the result will be a series of alternating light and dark bands of gradually changing intensity on the screen?

I would greatly appreciate it if people would please take the time to explain this.

• Have you studied about constructive and destructive interference of light waves? Commented Mar 9, 2020 at 3:35
• @GuruVishnu Yes, but I am not experienced with the mathematics of it, so it would be nice if care is taken to explain the mathematics (for instance, when you're presenting wave equations and discussing the phase, etc.) when presenting it. I've been exposed to these concepts, but I am very much a novice, so it may not be clear to me what you're implying unless you take care to explain it explicitly (if that makes sense). Commented Mar 9, 2020 at 3:38
• This should be covered in any optics textbook. Commented Apr 7, 2020 at 7:30
• @RobJeffries Well, not in the one I mentioned. That's why I posted this question. Commented Apr 7, 2020 at 7:32
• From Google books. "general reference book for engineers and assumes a broad knowledge of current optical systems and their design. ". i.e. It isn't an optics textbook. You need to go back a step. Something like Optics by Hecht. Commented Apr 7, 2020 at 7:35

There are two ways you can explain this problem. One using the wavefronts emanating, which I personally find more intuitive, albeit less mathematically rigorous, and one utilizing the path length from each slit. For completeness, I will show you both.

1) Wavefronts. In the above image, each slit is projecting its own wavefront, with a continuous curve indicating a maximum, and a dashed curve a minimum at the recorded time. (NB: the amplitude is maximally positive or negative. It is NOT 0) Because both sources (ie, slits) radiate waves at the same amplitude, any time two maxima or minima overlap, they interfere constructively, and any time a maximum overlaps a minimum, they interfere destructively. Tracing these points gives us the orange curves, which indicate where the bright and dark bands would be found if we placed the screen at any given distance.

2) Path length

We know that waves interfere constructively when their amplitude is pointing in the same direction, and negatively when they are of opposite sign. Because a wave switches sign every half wavelength, we can say we have maximum constructive interference when: $$R_1 - R_2 = \lambda \cdot N$$ With $$\lambda$$ the wavelength, and $$N$$ a whole number. Likewise, for destructive interference, we have: $$R_1 - R_2 = \lambda \cdot ( N + \frac{1}{2})$$ Assuming that $$d$$ is larger than $$\lambda$$, we can use a basic geometric arguments to show that we will have both of these conditions satisfied for some points on the screen.

Finally, for the change in intensity, simply consider that to get to a point further from the center of the screen, the light will have to fan out more, and thus, will decrease in intensity by a factor of $$\frac{1}{\pi \cdot R}$$

• This is immediately too vague to make sense to me. A maximum and minimum of what? The maximum and minimum of the amplitude? Commented Mar 9, 2020 at 15:13
• The maximum and minimum of the wave. If the wave has an amplitude of 10, the maximum is 10, and the minimum is -10. Note that these are momentary values, since we are dealing with a traveling wave. However, these points will consistently interfere as described. Commented Mar 9, 2020 at 15:32
• I see that the maximums coincide with the points where two crests or troughs meet, which must be constructive interference, and the minimums coincide with the points where a crest and a trough meet, which must be destructive interference. So do you mean that the maximum is when the absolute value of the amplitude is $10$, and the minimum is when the absolute value of the amplitude is $0$? Commented Mar 9, 2020 at 15:38
• Commented Mar 9, 2020 at 15:49
• You are correct in stating that destructive interference should result in a value of 0. Perhaps I worded this confusingly. What I meant to say is that both two coinciding phases (positive or negative) result in an absolute amplitude of 2A, wheras two waves that are exactly out of phase interfere destructively, to an amplitude of 0. Wherever the amplitude is 2A, you are at the local maximum of the interference-pattern (A bright band), and wherever it is 0, you are at a local minimum (a dark band). Commented Mar 9, 2020 at 16:41