RMPh 2023, Теория — Задача 3

Problem 3: Diffraction

Consider a diffraction grating made up of a sequence of identical equidistant slits.

The gratings with a better brightness, i.e. $I_{dif} > I_{inc}/4,$ are the blazed (reflection) gratings, in which the plane of the individual grooves makes an angle $\alpha$ with the grating plane P$_1$P$_2$. (see figure).

The grating from (d) is used in a monochromator in which the diffraction spectrum is projected onto the exit slit with a concave mirror having the radius of curvature $R = 1$ m.

Note. The normal incidence on the slits and gratings is considered only for a., b., c. Diffraction in parallel light is considered.

Problem proposed by
Dr. Marian Băzăvan, Faculty of Physics, University of Bucharest, Romania

Solution:

a. The distribution of the electric field intensity, depending on the diffraction angle θ, is given by the relation (Fraunhofer type diffraction):

(0.2pt)
where k is the modulus of the wave vector. The light intensity distribution is in this case:

(0.4pt)
which results in:

(0.4pt)

b. The diffraction patterns, given as the eletric field, for the two grattings are:

and

(0.2pt)
respectively.
Adding the two field distributions, we get the field distribution as if there was no aperture:

(0.2pt)
The main maxima of the 1st order are formed in the geometric shadow and therefore:

(0.3pt)
and

(0.3pt)

The two gratings with slits of width $a$ and $l-a$ will have $I_{dif\,a}= I_{dif\,l-a}$ and therefore the main maxima of the 1st order will have equal intensity (This follows from the Babinet principle. i.e. According to the Babinet principle for complementary apertures, the light intensity distribution in the geometric shadow is the same.)

c. We also consider the complementary grating (see point b,) illuminated with the same incident beam. In this situation we have the following relationships:

1. $I_{dif\,a} = I_{dif\,l-a}$ given the Babinet principle. (0.4pt)
2. $I_{0a}/a^2 = I_{0\,l-a}/(l-a)^2$ for the in the intensity in the 0th order. (0.4pt)

. (0.6pt)

From 1., 2., 3., we get :

(0.6pt)

The maximum occurs for $a=l/2$ which leads to $I_{dif\,max}=1/4*I_{incident}$. (0.2+0.3pt)

This grating is a pure amplitude grating, which causes the diffraction maximum corresponding to a slit to overlap with the main maximum of order 0.

d. To avoid this problem, the grating must cause a periodic variation of the phase of the incident wave (phase gratings). In practice, reflection gratings are mainly used, the surface of their features is inclined to the plane of the grating, i.e. "blazed gratings".

(0.4pt)
For m=1 and $i=\alpha$ (maximum intensity) we obtain:

(0.4pt)

$\alpha = 30°$. (0.2pt) .

e. The width of the spectral range Δλ is:

(0.4pt)
$\Delta\lambda=0.17$ Å. (0.2pt)
Coherence length=optical path difference for which the interference fringes are no longer observed:
Intensity distribution in a two-beam interferometer illuminated with radiation from the spectral range $[\lambda_o-\Delta\lambda/2; \lambda_o+\Delta\lambda/2]$ and $[\nu_o-\Delta\nu/2; \nu_o+\Delta\nu/2]$ respectively is:

(1pt)
Finge visbility:

(0.6pt)
Where L is the optical path difference corresponding to the interfering waves. The first disappearance of the fringes is obtained given the following condition:

(0.4pt)

(0.2pt)
$L_c$=1.5 cm (0.2pt)

f. The resolution power for the 1st order:

(0.4pt)
The smallest range that could be resolved by the grating is then δλ=0.025 Å. (0.2pt)
From the point of view of the matrix detector, the minimum spectral range is limited by the distance between two consecutive pixels 1 mm/ 200 pixels =5 μm

(0.4pt)
$\delta\lambda' = 0.043$ Å (0.2pt)
Therefore, the minimum spectral range resolved by the spectrometer is 0.043 Å. (0.3pt)