Download physics 415/416 supplemental problems

PHYSICS 415/416 SUPPLEMENTAL PROBLEMS
Part 8
8A. (Grating Dispersions)
As we have seen in class, there are a few different quantities to describe the dispersive ability of a grating.
The angular dispersion gives the angular width per wavelength for the cone of diffracted light.
The linear dispersion gives the distance per wavelength of the diffracted light pattern projected onto a screen.
The reciprocal dispersion is the reciprocal of the linear dispersion and tells you how much wavelength range is
contained in a given distance of the projected pattern. This value is sometimes called the “plate length” since
photographic plates used to be used to detect the diffracted light.)
You are to calculate these three dispersions for red light near 650 nm in the third order spectrum of a
transmission grating 6 cm wide with 3500 grooves/cm. A 150 cm focal length lens is used to image the
spectrum.
(a) Find the angular dispersion in /nm.
(b) Find the linear dispersion in mm/nm.
(c) Find the reciprocal dispersion in nm/mm.
(d) Find the theoretical resolving power of the grating under these conditions.
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8B. (Grating)
On the next page you will find the specification sheet for the diffraction grating that we use in the 1/4 meter
monochromator in the Optics laboratory (Oriel Model 77233). The grating is 50 mm wide. The top curve
shows the relative efficiency of the grating (relative amount of reflected light) versus wavelength. Note that the
efficiency is polarization dependent. Light polarized perpendicular to the grooves is called “s-polarization”
while light polarized parallel to the grooves is “p-polarization”. Below the efficiency curve are some other
grating specifications. Note that the grating is blazed so that the efficiency of the grating peaks near 500 nm. It
is blazed to operate in the first order. Refer to the sheet and answer the following questions.
(a) What is the line separation a of this grating?
(b) What is the linear dispersion (in mm/nm) of this grating near the blaze wavelength? Calculate it using
L=1/4 m.
(c) From your answer in (b), calculate the reciprocal dispersion. Is it close to the stated value of 2.86 nm/mm?
(d) What is the theoretical maximum wavelength that can be diffracted? Does this agree with the efficiency
curve? (Look where the efficiency is zero in the shaded portion of the curve.)
(e) Because of the mechanical limits of the monochromator, the grating cannot be rotated so that the incident
and diffracted angles are close to 90º. What is the maximum wavelength due to mechanical limits?
(f) What is the smallest angle of incidence possible for light of this maximum wavelength found in (e)?
(g) What is the theoretical resolving power of this grating? What is the theoretical smallest wavelength
separation (min) measurable near the blaze wavelength?
(h) The true resolving power and min are limited not by the grating, but by the exit slit at the output of the
monochromator. The min, often referred to as the “resolution” of the monochromator, is the product of the
reciprocal dispersion and slit width. Find the resolution and resolving power of the monochromator if an exit
slit width of 25 m is used.
Additional notes:
The efficiency curve for unpolarized light is relatively smooth, while those for polarized light, particularly for
the s-polarization, have several peaks and troughs. These sudden increases and decreases are called “grating
anomalies” (or “Wood’s anomalies”) and are a result of the interaction of the electric fields of the incident and
diffracted light with each other and with the metal coating of the blazed grooves. They are not a result of
imperfect gratings, but rather a consequence of basic optical interactions.
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8C.(Fourier Transforms and Far-Field Diffraction)
We said in class that the far-field intensity distribution for an aperture is proportional to the square of the
Fourier transform of a function that describes the transmission of the aperture (the so-called aperture function).
You are going to calculate and compare the patterns produced by a clear vertical slit and a shaded vertical slit.
Here are some helpful facts about Fourier transforms that you will need.
Fourier Transform
Let g(s) be a function of s. Then the Fourier transform of g(s) is
Relation of the Fourier Transform to the Far-Field Diffraction Integral
Let g(s) describe the transmission properties of the aperture. The function has a value of 1 if the transmission is
100% and 0 if no light is transmitted. The intensity of the light at a point P on the viewing screen is given by
We see that the integral is the Fourier transform of the aperture function where f = y / (λL) is called the spatial
frequency. Recall that y is the position of point P on the screen with y = 0 at the center of the pattern, L is the
distance from the aperture to the screen, and λ is the wavelength of the light.
Similarity Theorem
If
and a is a constant then
Some Functions and Their Fourier Transforms
Name
rectangle function
triangle function
g(s)
1 s  1 / 2
rect(s) = 
0 otherwise
1  s
(s) = 
 0
s 1
otherwise
G(f)
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(a) Consider a clear slit of width b shown below. Write down an aperture function that describes the
transmission of the aperture. Using the Similarity Theorem, write down the Fourier transform G( f ) of the
aperture function. We want the aperture function to look like the function below.
g
1
-b/2
b/2
s
(b) Consider a clear slit of width b but with a piece of plastic in the opening whose opaqueness varies linearly
from the center of the slit to the edges as shown below. Write down an aperture function that describes the
transmission of the aperture. Using the Similarity Theorem, write down the Fourier transform G( f ) of the
aperture function. We want the aperture function to look like the function below.
g
1
-b/2
b/2
s
(c) The intensity of the light in the viewing plane is proportional to the square of the Fourier transform of the
aperture function. If we choose I(0) = 1 for convenience for both intensity patterns, write down the
expressions for the intensity distributions as functions of screen position y. (Recall that sinc(0) = 1 which
you can verify with L'Hospital's rule.)
(d) Let b = 0.5 mm, L = 2 m, and λ = 500 nm. Plot the two intensity patterns (I vs. y ) from y = 0 to 10 mm
Realize that the patterns are symmetric about y = 0. You are only plotting positive positions.
(e) Find the widths of the two central diffraction peaks. You should see that the shaded slit produces a wider
central diffraction peak. Do you?
(f) However, while the central diffraction peak is wider for the shaded lit, you should see that the clear slit
produces higher intensity peaks farther from the pattern center. Do you? (The sharpness of the edges of the
clear slit result in larger intensities for higher spatial frequencies which correspond to positions on the
viewing screen that are farther from the center of the pattern.)
(g) OPTIONAL FOR 5 BONUS POINTS. Prove by hand (no software) that the Fourier transform of the
triangle function, (s), is indeed a sinc2( πf ).