# What does the differential of $d_s\sin(\theta) = m\lambda$ help us see, with respect to waves through diffraction gratings?

With respect to waves traveling through a diffraction grating, we have an equation like this one: $$d_s\sin(\theta) = m\lambda.$$

Where $d_s$ is the distance between slits in the grating, $\theta$ is an approximate angle at which the waves bend through each slit of the grating, $\lambda$ is the wavelength of the waves passing through the gradient, and $m$ is the number of wavelengths by which distances traveled by one wave from one slit differ from an adjacent slit. $d_s$ and $m$ are usually given a remain constant in the scenarios I'm working with.

My physics book says that the differential of the above mentioned equation is $$d_s \cos(\theta)d\theta = md\lambda$$ (without confusing the single $d_s$ (distance) with the ones in $d\theta$ and $d\lambda$).

What does it mean to call the second equation the "differential" of the first? I am trying to understand the concept behind the differentials more so that I may later make sense of the physics.

EDIT: In user6786's question, user6786 states that "according to the formula $dy=f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential)". I'm trying to see how that works.

-
""In Electromagnetic Wave Theory as it pertains..."" What is "electromagnetic" in this relation? 1st this was known well before of electromagnetic waves, and 2nd it fits to sound waves diffracted by a lattice fence. –  Georg Nov 28 '11 at 10:53
Aaah true, so "diffraction of waves through small slits." –  trusktr Nov 29 '11 at 1:00

There's nothing fancy going on here. Taking the differential of an expression is "physics code" for computing a first order Taylor expansion of the expression around its initial value. In other words, say that you have some physical quantity $Y$ which is a function of a variable $X$. Let $X_0$ represent the initial value of $X$. Then if $X$ changes to $X_0 + \delta X$, the value of $Y$ becomes

$$Y(X_0 + \delta X) \approx Y(X_0) + (\delta X) Y'(X_0)$$

and the change in $Y$ is

$$\delta Y = Y(X_0 + \delta X) - Y(X_0) \approx (\delta X)Y'(X_0)$$

The expression on the right is the differential of $Y$. This is really just a simple way of expressing how much $Y$ changes when you change $X$ by a certain (small) amount.

In your example, you can identify $\theta$ as $Y$ and $\lambda$ as $X$ (simply because we think of wavelength as the independent variable, and the angle of a diffraction fringe as a dependent variable). So the equation

$$d_s\cos\theta\;\mathrm{d}\theta = m\;\mathrm{d}\lambda$$

tells you how much you expect the angular position of the fringe to change if you adjust the wavelength by a small amount, or how much angular separation you would get between two nearby wavelengths (for resolving closely spaced spectral lines), etc. Using this equation to calculate, say, "these fringes are .04 radians apart" is a lot more convenient than calculating the absolute position of one fringe using the original equation $d_s\sin\theta = m\lambda$, calculating the absolute position of the other one using the other wavelength, and subtracting to see what the difference is.

-
Thanks. Nice info. This is what I realized when I replaced the word "differential" with "derivative" which makes SO much more sense! I couldn't understand what they meant by the second equation being the differential of the first. It makes more sense to say that the second equation is the derivative of the first and can be used to find small differentials of the values in the first equation. –  trusktr Nov 29 '11 at 1:06
In this new light (better verbage), I'd rather just use the original equation (not its derivative) since I already know what the equation is and don't have to worry about finding any sort of anti-derivative. –  trusktr Nov 29 '11 at 1:32

This differential tells you how the peak angles move around when you change the wavelength. It's useful if you have a spread in frequency, so that the peaks are not sharp, and you want to know how much spread is in the peaks because your light is not monochromatic, for instance.

-

I am assuming by now you have moved past this point, however for new viewers be careful not to confuse differential equations with derivatives. DE's are similar but slightly more complex and very powerful tools. You could always think of a derivative as a DE if you differentiate each side individually. There are further implications as to why you need to understand the differential relationship in this case. However if you want to see why the theory leading up to this equation is not valid check out an experiment I have performed. search chinstrap effect in you tube to see the video. If you wish to see the math behind it email me at john.blaszynski@mohawkcollege.ca and i will send u a copy of the math.

-