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Assuming the incident beam is perpendicular, the grating equation is as follows:

$$n\lambda = d \sin \theta$$

where $n$ is the order, lambda is the wavelength, $d$ is the spacing, and $\theta$ is the outgoing angle.

Spectral resolution can be described as

$$\Delta \lambda = \dfrac{\lambda}{n N}$$

in which $N$ is the number of lines/grooves in the grating.

I'm having trouble getting here.

I try to differentiate the grating equations as follows, but I don't know what to do next. Maybe I'm not on the right track. I'm also thinking you need to assume the outgoing angle is small but I'm not sure. I've always taken the spectral resolution equation for granted. I haven't tried to derive it before.

$$\dfrac{\Delta \lambda}{\Delta \theta} = d \cos \theta/n$$

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    $\begingroup$ Have a look at this answer. It starts almost exactly where you stopped, except that they've rearranged your last equation into $n \, \Delta \lambda = d \, \Delta \theta \cos \theta$. $\endgroup$
    – A. P.
    Commented Jan 17, 2021 at 9:19

1 Answer 1

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Suppose our spectrum consists of two wavelengths $\lambda_1 = \lambda$ and $\lambda_2 = \lambda + \Delta \lambda$. According to the Rayleigh criterion the two wavelengths are resolved by the grating if the first diffraction minimum for wavelength $\lambda_1$ coincides with a maximum for wavelength $\lambda_2$

enter image description here

In order to apply this criterion is we need to relate the transversal distance on the screen to the wavelength. Hence, the answer needs the following parts:

  1. Obtain a relationship between the transversal distance of a maximum $y$ on the screen and the wavelength $\lambda$.
  2. Find the distance between the maximum and it's first minimum for $\lambda = \lambda_1$.
  3. Apply Rayleigh's resolution criterion.

Step 1: The simplest way to obtain this relationship is to use the small (diffraction) angle approximation, $\theta \approx \sin{\theta} \approx \tan\theta = y/D$, where $D$ is the distance from the grating to the screen and $y$ is the distance measured on the screen. The distance between to neighbouring maxima is $y_{max}$.

enter image description here

Entering this into the grating equation yields $ n\lambda = d \sin{\theta} \approx d \frac{y}{D} $ or, after rearrangement $$ \tag{1} y_n(\lambda) = y_n \approx \frac{nD}{d} \cdot \lambda $$ If we have a spectrum consisting of two wavelengths $\lambda_1, \lambda_2$ the two $n^{th}$ maxima are shifted by $$ \tag{2} y_{n}(\lambda_2) - y_{n}(\lambda_1) = \frac{nD}{d} \cdot \Delta \lambda $$

Step 2: We have to calculate the distance between the maximum and its first minimum. Although this calculation can be done, I'd like to use a conceptional approach instead. We know that between the $n$ and $n'=n+1$ maxima the diffraction pattern of a grating with $N$ grooves contains several local maxima. In fact the number of maxima is approx. $N$ and they are approx. equally spaced. Hence, we dividing the length $y_{max}$ (see picture above) into $N$ equal parts, each part has a length $y_{max}/N$. Thus, the distance between the $n^{th}$ maximum and it's first minimum is approx. $y_{max}/N$. Using eq. (1) we find $$ \tag{3} y_{max}(\lambda) - y_{min}(\lambda) = \frac{y_{max}}{N} = \frac{y_{n+1} - y_n}{N} = \frac{D}{N d} \cdot\lambda $$

Step 3: Rayleigh's resolution criterion states that eq. (2) is equal to eq. (3). Hence, it reads $$ \frac{D}{N d} \cdot\lambda == \frac{nD}{d} \cdot \Delta \lambda $$ or after rearranging $ \lambda / \Delta \lambda = n N $

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