Quick questions on Diffraction gratings and spectra: WACE Year 12 Physics Unit 4

For a grating with slit spacing $d$ (the distance between adjacent slits), bright maxima occur where

Because the diffraction angle depends on wavelength, each colour in white light emerges at a slightly different angle in each order. The result is a spectrum spread out by wavelength, with violet (short wavelength) deviated least and red (long wavelength) most. Unlike a prism, a grating spreads the colours in a predictable, calculable way, which is why gratings are used in spectrometers to measure the wavelengths emitted by light sources.

Since $\sin\theta$ cannot exceed $1$, only a finite number of orders exist for a given grating and wavelength. The highest visible order is the largest integer $m$ for which $m\lambda/d\le 1$. Long wavelengths and finely ruled gratings (small $d$) produce fewer orders.

The grating is most often used in reverse: measure the angle of a known order and solve for the wavelength. Rearranging the grating equation gives $\lambda=\dfrac{d\sin\theta}{m}$, so a measured first-order angle and a known slit spacing give the wavelength directly. Using a higher order improves precision, because the same wavelength is spread to a larger angle where an angular measurement error matters less. This is exactly how a spectrometer determines the wavelengths of the bright lines emitted by an excited gas, linking this topic to atomic spectra: the grating disperses the light, the angles are measured, and each line's wavelength (and hence the energy-level transition that produced it) is calculated.

Convert lines per millimetre to a spacing $d$ in metres before substituting, and remember $m$ must be a whole number. If a calculation gives $\sin\theta>1$ for some order, that order does not exist for that wavelength.