Announcements
•Exam 2 is scheduled for two weeks from
today (March 28). Will cover telescopes and
light from today and whatever we cover
next week.
•Homework: Chapter 6 # 44 & 45 plus
Supplemental Problems.
Telescope basics
(r)
Telescopes either use refraction or reflection to focus
light to a point. For refraction, the basic law is Snell’s Law
The law of reflection is a much simpler: i  r
If the surfaces of a piece of glass are
curved, they will focus light to a point
 1 1  n  noutside  d 
1
  n  noutside    

f
 R1
R2
nR1 R2

Lens Makers Equation: R1 and
R2 are the radii of curvature of
the two faces, n is the index of
refraction of the glass and d is
the center thickness of the
lens. If the outside medium is
air noutside = 1
The focal length, f, is the distance from the lens
axis to the focal point or focal plane. The focal
length of a lens depends on the material of the
lens and the curvature of the surfaces and is
found using the Lens Makers Equation
Sign Convention for Lenses
If the center of curvature is on the opposite
side of the surface as the incoming light, R is
positive.
If the center of curvature is on the same side
of the surface as the incoming light, R is
negative.
A positive focal length is a converging lens
and a negative focal length is a diverging lens
Examples
Determine the focal length of a biconvex lens made
of crown glass (n = 1.52) whose surfaces both have a
radius of curvature of 25 cm. The lens is 1.50 cm
thick at its center.
Determine the focal length of a plano-convex lens
made of flint glass (n = 1.65). Plano-convex means
one side is flat (R = ∞) and the other side has a
convex shape. The convex side has a radius of
curvature of 20 cm and the lens is 1.0 cm thick at the
center. Does it make a difference which side of the
lens the light is incident on?
Example 1 Solution
For a biconvex lens R1>0 and R2<0. For this problem
R1 = +25 cm and R2 = -25 cm with d = 1.50 cm and n = 1.52
1
1
1  n  1 d 
  n  1  


f
nR1 R2 
 R1 R2
 1
1.52  11.50cm  

1
 1.52  1 



 25cm 25cm 1.52  25cm  25cm  
 0.04117cm 1
f  24.3cm
Example 2 Solution
For a plano-convex lens R1 = ∞ and R2<0. For this problem
R1 = ∞
R2 = -20 cm
d = 1.0 cm and n = 1.65
1
1
1  n  1 d 
  n  1  


f
nR1 R2 
 R1 R2
 1
1.65  11.0cm  

1
 1.65  1 



 cm 20cm 1.52  cm  20cm  
 0.0325cm 1
f  30.8cm
A concave mirror will also
focus light to a point
Sign Convention
For a spherical mirror the focal
length is just half the radius of
curvature of the mirror. For
other shapes the formula is
somewhat more complicated
f and r are positive for a concave mirror
f and r are negative for a convex mirror
For any telescope, the most
important property is the
Light Gathering Power (LGP)
do is the diameter
of the objective in
mm. This
compares the
light gathering
power of the
telescope to that
of the human eye
(diameter = 7 mm)
2
0
d
LGP 
49
Examples
Compare the light gathering power of the Hubble
Space Telescope (2.4 m diameter) to a 10” Newtonian
reflector.
APSU has a 20” telescope at the APSU Observatory.
Compare the light gathering power of the telescope
to the human eye (d = 7mm).
Example Solutions
To compare the light gathering power of two telescopes, take
the ratio of the square of their diameters.
d10” = 10” x 2.54 cm/in = 25.4 cm = 0.254 m
2
The Hubble Space Telescope has
2.4m 

d02
LGP  2 
 89.3 almost 90 times the light gathering
2
d1  0.254m 
power of a 10” diameter telescope
1. dHubble = 2.4 m
2. Dtele = 20” x 25.4 mm/in = 508 mm
LGP 
 508mm   5266.6  5270
d

2
d
 7.0m 
2
0
2
1
2
d1 = 7.0 mm
The ability of a telescope to
resolve fine detail is given by
the Rayleigh Criterion
sin     1.22

d
 is the wavelength of the light
being used and d is the
diameter of the aperture.  is
the smallest resolvable angle
of the telescope. Note that this
angle is in radians, not degrees
Example
Using a 10” diameter Newtonian telescope in visible
light ( = 500 nm) what is the smallest angular detail
that can be resolved? At the distance of the Moon
(384,400 km), how large is this in kilometers. The
general formula relating arc length (s), angle () and
radius (r) is: s = r
How large a telescope would be needed to resolve
one of the Apollo lunar landers (~10 meters across)
from Earth using visible light ( = 500 nm)?
Example Solution
1. d = 10” x 0.0254 m/in = 0.254 m
rMoon = 384,400 km
 = 500 x 10-9 m

500 109 m
sin     1.22  1.22
 2.402 106 radians
d
0.254m
using s = r where r = rMoon


s  r   384, 400km  2.402 106 radians  0.923km
To see the lunar lander first find the angle  then find d
s
10m
s  r    
 2.601108 radians
r 384, 400km 1000 m km


500 109 m
sin     1.22  d  1.22  1.22
 23.4m
8
d

2.60110 radians