I am having some problems calculating wavelengths from some given information about a grating spectrum.

A diffraction grating with a spacing of 3μm is used in a spectrometer to investigate the emission spectrum of a mercury vapour discharge lamp.

The spectrum is examined over the range of angles from 30° to 50°, and maxima of intensity are observed at the angles and with the colours shown in the table. No other maxima are observed in this range of angles.

Violet: 32.7°, 42.4°  
Blue:   35.5°, 46.6°  
Green:  33.1°, 46.7°  
Yellow: 35.2°, 35.4°

To calculate wavelengths, I know that: $n\lambda = dsin \theta$.
So for the first angle: $n\lambda = 3 \times 10^{-6} \times sin 32.7$
However, I'm not sure where to go from there as I don't know what order of maxima they are as 0° to 30° is not examined. I have tried guessing numbers to make sure the wavelengths are in the range 400nm to 700nm. For example, with values of n=4 (35.5°) and n=5 (46.6°) for the blue light I get wavelength values that both round to 436nm, but this is apparently incorrect.

Also I don't understand how the two yellow maxima are so close together because this would normally mean there is a tiny wavelength, much smaller than that of visible light.

Any pointers would be much appreciated.


closed as off-topic by user10851, John Rennie, ACuriousMind, CuriousOne, user36790 May 22 '16 at 10:36

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  • $\begingroup$ If there are no maxima in violet between 32.7 and 42.4, what does that tell you about the relationship between the value of $n$ for each? $\endgroup$ – DilithiumMatrix May 21 '16 at 23:30
  • $\begingroup$ While Physics Stack Exchange isn't a homework help site, if you do want that kind of help you can take a look at this thread for a list of free online homework help resources. $\endgroup$ – Ulad Kasach May 25 '16 at 22:51

For each wavelength you have been given (exactly) two values $\theta$, so

$$\begin{align}n\lambda &= d\sin\theta_1\\ (n+1)\lambda &= d\sin\theta_2\end{align}$$

subtracting these two equations, we get

$$\lambda = d\left(\sin\theta_2-\sin\theta_1\right)$$

You should be able to figure it out from there...

  • $\begingroup$ Thanks, I didn't realise that the value for n does not have to be an integer, and so I didn't think about forming this equation. $\endgroup$ – Gone Phishing May 22 '16 at 15:01

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