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Unformatted text preview: Czerny-Turner Monochromator 1 A monochromator selects out a single wavelength from a light beam. It literally
allows a single (mono) color (chrom) to pass through the instrument. 'To accomplish
this, the light must be dispersed into its component colors. The dispersive element could
be a prism or a grating. In most commercial monochromators a metallic reﬂection grat-
ing serves this role and this is the case in a Czerny-Turner monochromator. In this type
of monochromator, a planar grating with a period A is used to diffract a plane wave.
The directions of the various wavelengths are given by the grating equation, A
sin0’ +sin6; = For the common situation of a monochromator in air, n1 equals 1. Similarly, for ﬁrst-
order diffraction, i equals 1. By rotating the grating to an angle p, the freespace wave-
length /\ can be selected to pass through the system. The incident angle is 6’ = p + ¢
and the diffracted angle is 0; = p — d). The quantity 2gb is the mirror-grating-mirror angle
(also known as the Ebert angle) and is ﬁxed for a given monochromator. Derive an expression for the wavelength A that is allowed to pass through the
monochromator. The surrounding medium is air and the diffraction is into the ﬁrst or-
der. Express your result as a function of A, p, and (1) only. Eliminate any sums and dif-
ferences of angles in your expression. Simplify your result to the greatest extent possible.
Put your ﬁnal answer in the space provided below. For a particular Czerny—Turner monochromator, ¢ = 46° and the grating has
1200 grooves/ mm (1 /A). For this monochromator, calculate, showing all work, the
rotation angle of the grating, p, needed to pass 600 nm wavelength light through the
monochromator. Express your answer in degrees accurately to within i0.001°. Put your
answer in the space provided below. p = degrees. CZERNY-TURNER MONOCHROMATOR ENTRANCE EXIT
SLIT ll SLIT EXIT é SLIT é [A
COLLIMATING MIRROR Czerny—Turner Monochromator 1 Grating equation A
sinH’ +sin0; 2 n1 —1
9’ = p+¢
Using sin(p + qﬁ) = sinp cos ¢ + cosp sinqﬁ
sin(p — (b) = sinp 005(1) — cosp sin¢ the grating equation becomes , A
A = 2Asinpcosq§
For ¢ = 46° l/A : 1200mm‘1
A = 600nm then ,0 = sin—1(2Ai‘os¢) = 21.1717deg. ...
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This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Tech.
- Spring '08