ASTR 310
Homework 3 Solutions
sin
N when
, for n= 0, 1, 2,... This is the grating term
sin
in the intensity profile for a diffraction grating.
1. Show that
We have to take the limit
sin
lim
. Just plugging in
will give a zero
sin
in the denominator, so we have to use l'Hopital's rule and then use
:
n
lim
n
sin N
sin
lim
n
Ncos
Ncos
N
cos
cos
1
N
2. A single long slit of width d is illuminated by monochromatic, plane-parallel light.
Just behind the slit is a lens which focuses the slit image on a screen 100 cm from the
slit. The wavelength of the light is 6563 Angstroms (the H line). You want the
distance between the two dark fringes on either side of the central peak of the
diffraction pattern to be 10 mm. How wide (d) should you make your slit?
We can find
with the small angle approximation,
tan
5mm
0.005radians
100cm
Then, using the equation
radians, we have
, with n=1,
d=
sin
= 6563 A, and
= 0.005
= 1.31 E-3 m
There is another, slightly more complicated way to do this problem, but it gives
n , and that
the same end result. From your notes, you know that the minima are at
kb
2
sin , where k
and b=d. Plugging in k and b, you find that
2
d
n
sin . This again gives you the result that
, and using n=1,
you'll get the same answer.
3. Take the 14-inch telescope at the UM Observatory. What is the angular radius, in
arcsec, of the first dark ring of the Airy disk at a wavelength of 5000 Angstroms?
Suppose you are observing the star Vega. How far is it in parsec (mention your
reference)? How far from the star in AU is the first dark ring of the Airy disk? (Recall
that at a distance of one parsec, a 1 AU orbit has an angular radius of one arcsec).
The location (angular) of the first dark ring is
1.22
D
D = 14 inches = 0.3556 m. So,
5 10
7
6
0.3556m
5
1.4 10 radians= 8 10 degrees = 0.29 arcsec
The distance to Vega is 7.751 pc (this number will be a little different depending
on your reference). How far away from Vega will the first dark ring be? Because
it’s a small angle, we can use the approximation tan
, so
x
0. 29arcsec=
7. 751pc
,
where x is the physical size of the first Airy ring. From the equation for parallax,
1 AU
d
, we know that if we have distance in parsecs and an angle in
pc
p
arcsec
arcseconds, we get an object size in astronomical units:
x AU
7.751pc 0.29arcsec = 2.25 AU
The first dark Airy ring is located 2.25 AU from Vega on the image.
Some of you did the first part of this problem a different way, where you used
5
A
5 10
D
to get the size of the Airy disk in arcseconds directly, without having to convert
from radians to degrees to arcseconds. This is fine, but remember that A is the
diameter from one side of the Airy ring to the other, not the radius from the star to
the ring (need to divide
A
by 2). You can see this by comparing the value of
A
with the value I found for
above.
4. A diffraction grating has 600 lines per mm. We wish to observe the yellow D lines of
sodium which form a close doublet of wavelengths 5890 and 5896 Angstroms, in the
second order with this grating.
a. What is the resolution, R, needed to separate these lines?
avg
R=
5893A
6A
= 982.2
b. How many lines, N, of the grating must be illuminated to achieve this
resolution?
R= nN , where n=2, so
N=
R 982.2
n
2
= 491 lines
c. How wide must our grating be to meet this requirement?
Width=
Numberoflines,N
491lines
Numberoflinespermm 600lines mm
= 0.8185 mm
Note: See the next page for pictures to help with problems 2 and 3.