# Diffraction grating angle

Why have they taken the angle to be 90 degrees? And why have they used only one of the wavelengths provided? can you explain in layman terms?

They have taken the angle to be 90° because what is stated in the question:

The light is incident normally on a diffraction grating. Third-order maxima are produced for each of the two wavelengths. No higher orders are produced for either wavelength.

The normal to any surface is always 90°. But as pointed out by Holzner in comments (and waking me up ;-) ), the angle of incidence in optics is also often defined as the angle between the normal to the interface and the incoming ray.

What this means is, the key is that no higher orders are produced.

With diffraction gratings, maxima occur at angles $$θ_m$$ which satisfy the relationship $$\frac{d\sin \theta_m}{\lambda} = |m|$$, where $$\theta_m$$ is the angle between the diffracted ray and the grating's normal vector, $$d$$ is the distance from the center of one slit to the center of the adjacent slit, and $$m$$ is an integer representing the propagation-mode of interest.

When light is normally incident on a diffraction grating, the diffracted light has maxima at angles $$\theta_m$$ given by $$d\sin \theta_{m} = m\lambda$$.

This means that the maximum angle for diffraction maxima is 90°, so the 3rd-order maxima can appear at 90° (or a smaller angle), and there are then no 4th-order maxima.

Working backwards from this position then means that $$d = \frac{m\lambda}{\sin\theta_m}$$ and if $$\theta_m = 90°$$ then $$d = m\lambda$$.
For $$\lambda = 540\text{ nm}$$, $$d = 1.62 \times 10^{-6}\text{ m}$$, and for $$\lambda = 630\text{ nm}$$, $$d = 1.89 \times 10^{-6}\text{ m}$$.

Then, for a diffraction grating with $$d = 1.62 \times 10^{-6}\text{ m}$$, the 3rd-order diffraction maximum for $$\lambda_{540}$$ is $$\arcsin \frac{m\lambda}{d} = \arcsin \frac{3 \times 540 \times 10^{-9}}{1.62 \times 10^{-6}} = 90°$$, and for $$\lambda_{630}$$ would be $$\arcsin \frac{3 \times 630 \times 10^{-9}}{1.62 \times 10^{-6}} = \arcsin (1.167)$$, which is invalid so there can be no 3rd-order maximum for this wavelength.

However, for a diffraction grating with $$d = 1.89 \times 10^{-6}\text{ m}$$, the 3rd-order diffraction maximum for $$\lambda_{540}$$ is $$\arcsin \frac{3 \times 540 \times 10^{-9}}{1.89 \times 10^{-6}} = 59°$$, and for $$\lambda_{630}$$ is $$90°$$. Calculating 4th-order for $$\lambda_{540}$$ to verify that this is not possible shows that this would be $$\arcsin (1.14)$$.

So the longer wavelength is used because the smaller wavelength diffraction angles will be smaller than the result obtained from the longer wavelength.

• However, the angle of incidence in optics is also often defined as the angle between the normal to the interface and the incoming ray, so normal incidence would correspond to an angle of incidence of 0 deg. See e.g. en.wikipedia.org/wiki/Snell%27s_law. So it's always worth checking the conventions used for an equation you want to use. Oct 17, 2019 at 7:49
• Good point *slaps self on forehead* I shall update my answer.
– Mick
Oct 17, 2019 at 8:49