Diffraction grating

Figure 12.12: Diffraction grating

We consider the transmission diffraction grading shown in Figure 12.12. The transmission grating is essentially a periodic arrangement of $N$ slits, each slit of width $D$ and slit spacing $d$. The spacing between successive slits $d$ is referred to as the ``grating element'' or as the ``period of the grating". Each slit acts like a source, and the diffraction grating is equivalent to the chain of sources shown in Figure 12.10.

The intensity pattern of a diffraction grating is the product of the intensity pattern of a single slit and the intensity pattern of a periodic arrangement of emmiters

$$I = I_0 \frac{\sin^2 (N \alpha)}{\sin^2(\alpha) } \mbox{sinc}^2(\beta)$$ (12.28)

where

$$\alpha = \frac{\pi d \sin \theta}{\lambda} \hspace{1cm} \mbox{ and} \hspace{1cm} \beta = \frac{\pi D \sin \theta}{\lambda}$$

Typically the slit spacing $d$ is larger than the slit width ie. . Figure 12.13 shows the intensity pattern for a diffraction grating. The finite slit width causes the higher order primary maximas to be considerably fainter than the low order ones.

Figure 12.13: Intensity pattern of a diffraction grating

The transmission grading is a very useful device in spectroscopy. The grating is very effective in dispersing the light into different wavelength components. For each wavelength the $m$th order primary maximum occurs at a different angle determined by

$$\sin \theta_m=m \frac{\lambda}{d}$$ (12.29)

The diffraction pattern, when two different wavelengths are incident on a grating, is shown in the Figure 12.14.

Figure 12.14: Diffraction pattern for two wavelengths (Intensity vs $\beta$)

The dispersive power of a grating is defined as

$${\cal D}= \left( \frac{d \theta_m}{d\lambda} \right) = \frac{m}{d \cos \theta_m}$$ (12.30)

We see that it increases with the order $m$ and is inversely proportional to $d$. The finer the grating (small $d$) the more its dispersive power. Also, the higher orders have a greater dispersive power, but the intensity of these maxima is also fainter.

Figure 12.15: Chromatic resolution

The Chromatic Resolving Power(CRP) quantifies the ability of a grating to resolve two spectral lines of wavelengths $\lambda$ and $\lambda + \Delta \lambda$ . Applying Rayleigh's criterion, it will be possible to resolve the lines if the maximum of one coincides with the minimum of the other.

The minimum corresponding to $\lambda$ (Figure 12.15) is at

$$\Delta \theta = \frac{\lambda}{N d \cos \theta_m},$$ (12.31)

from the maximum and the maximum corresponding to $\lambda + \Delta \lambda$ is at

$$\Delta \theta = \frac{m \Delta \lambda}{d \cos \theta_m}.$$ (12.32)

Equating these gives the CRP to be

$${\cal R} \equiv \frac{\lambda}{\Delta \lambda } = N m$$ (12.33)

The chromatic resolving power increases with the number of surfs or rulings in the grating. This makes the grating a very powerful dispersive element in spectrometers.

Problems

1. For a slit of dimensions $1 \, {\rm mm} \times 1 \, {\rm cm}$, what are the positions of the first three minima's on either side of the central maxima? Use $\lambda=550 \, {\rm nm}$ and $0.1 \, {\rm mm}$.
2. For a rectangular slit whose whose smaller dimension is $D$, what are the positions of the maxima for light of wavelength $\lambda$? (Ans. $\beta$ $\sim\pm 1.43\pi$, $\pm 2.46\pi$, $\pm 3.47\pi$, etc.)
3. Calculate the ratio of intensities of the first intensity maximum and central maximum for the previous problem. (Ans. $\sim$21)
4. Compare the angular resolutions of two circular apertures a. $D=1 \, {\rm mm}$ and $\lambda=550 \, {\rm nm}$and b. $D=45 \, {\rm m}$ and $\lambda=1 \, {\rm m}$.
5. A plane wave of light with wavelength $\lambda =0.5 \, \mu {\rm m}$ falls on a slit of width at an angle $30^{circ}$ to the normal. Find the angular position, with respect to the normal, of the first minima on both sides of the central maxima.

6. The collimator of a spectrometer has a diameter of 2cm. What would be the largest grating element for a grating, which would just resolve the Sodium doublet at the second order, using this spectrometer. (Sodium doublet: nm and $D2=589.6$nm, Ans. $d\sim$0.04mm)

7. Obtain the expression of intensity for a double slit with separation $d$ between the slits and individual slit width $D$, as a special case of N=2.

8. Plot intensity profile as a function of $\theta$ for a double slit with $d=0.1$mm and $D=0.025$mm. Assume wavelength of the incident monochromatic light to be equal to nm. Keep $\vert\theta\vert< 2.5^{\circ}$. Notice that there are no 4th order and the 8th order intensity maxima in the above. These are called the missing orders in the pattern.

9. Missing orders: Find the ratio of $d$ and $D$ for the following double slit diffraction, Figure 12.16.

   Figure 12.16: A double slit diffraction pattern

   

10. Find an expression for the intensity of a double slit diffraction when one of the slits is having a width C and the other is having a width D and the separation between them is $d$.