# How to explain interference pattern in our eye?

Suppose we got a Lamp L that emits some light. The light afterwards hits a diffraction Grating G at a distance a. Now if you were to look through the grating with your Eye E, you were to see the interference Maxima M that are a distance x away from the origin of the Lamp (for the arrangement look below).

The formula for calculating the wavelength $$\lambda$$ of the light is said to be

$$\mathrm{\lambda = g\cdot\dfrac{x}{\sqrt{a^2 + x^2}}}$$

where only the lattice constant $$\textsf{g}$$, the distance $$\textsf{a}$$ and the spacing $$\textsf{x}$$ have to be known (assuming first order).

$$\underline{\text{What is totally unclear to me is why this is the case.}\:}$$ Of course I know the simple derivation of the formula above, but that assumes that the $$\textsf{grid}$$ and the $$\textsf{screen}$$ are at a distance $$\textsf{a}$$ apart.

I don't understand why the distance between $$\textsf{object}$$ and $$\textsf{grid}$$ matters because the diffraction only happens after the light passes the grating. According to my understanding the distance between light and grating doesn't matter if you consider the interference pattern at a screen.

(But of course here we don't have screen but our prying eye)

Maybe I'm misunderstanding your question, but the distance $$a$$ doesn't matter in the sense that if you move the light source farther away, you would still look in the exact same direction to see the diffraction peak. In other words, if you increase $$a$$ by some factor, $$x$$ also increases by the same factor. Consequently, the direction you must look in to see the diffraction peak—making an angle $$\theta$$ relative to the normal—remains the same, with $$\sin\theta=\frac{x}{\sqrt{a^2 + x^2}}=\frac \lambda g.$$
To put it yet another way, the diffraction grating doesn't care how far away the light came from. It essentially "bends" the outgoing light by a particular angle $$\theta$$ (or sometimes several). This angle determines the direction you must look in to see the deflected beam.
• Good answer. For even more clarity I'll point out that $a$ appearing in the formula is an artefact of how we would usually measure this in a lab. You'd put up a ruler with length markings in the same plane as your light source and then read off at what distance you see the bright spot. But nothing stops you from placing your ruler closer or further away, which then determines your $a$ to use instead of the distance of the light source.