# PHYS 632 Lecture 14: Images

```Lecture 14 Images Chp. 35
• Opening Demo
• Topics
– Plane mirror, Two parallel mirrors, Two plane mirrors at right
angles
– Spherical mirror/Plane mirror comparison in forming image
– Spherical refracting surfaces
– Thin lenses
– Optical Instruments
Magnifying glass, Microscope, Refracting telescope
• Warm-up problem
Geometrical Optics:Study of reflection and refraction
of light from surfaces using the ray approximation.
The ray approximation states that light travels in straight lines
until it is reflected or refracted and then travels in straight lines again.
The wavelength of light must be small compared to the size of
the objects or else diffractive effects occur.
'
Reflection 1  1
Refraction: Snells Law n1 sin 1  n2 sin 2

 n1
n2
Using Fermat’s Principle you can prove the
above two laws. It states that the path taken
by light when traveling from one point to
another is the path that takes the shortest
time compared to nearby paths.
Plane mirrors
Real side
Angle of
incidence
Normal
Angle of reflection
Virtual side
i=-p
Virtual image
Problem: Two plane mirrors make an angle of 90o. How
many images are there for an object placed between
them?
mirror
eye
object
2
mirror
1
3
Problem: Two plan mirrors make an angle of 60o. Find all
images for a point object on the bisector.
mirror
2
4
object
mirror
5,6
1
3
How tall must be a mirror be for a person to see
his entire reflection in it. Let H be the height of
a man and let l be the height of the mirror?
i=-p
magnification = 1
What happens if we bend the mirror?
Concave mirror.
Image gets magnified.
Field of view is diminished
Convex mirror.
Image is reduced.
Field of view increased.
Thin Lenses: thickness is small compared to object
distance, image distance, and radius of curvature.
Converging lens
Diverging lens

Thin Lens Equation
1 1 1
 
f p i
Lensmaker Equation
1
1 1
 (n 1)(  )
f
r1 r2
What is the sign convention?

Sign Convention
Light
Virtual side - V
Real side - R
r1 r2
p
i
Real object - distance p is pos on V side (Incident rays are diverging)
Radius of curvature is pos on R side.
Real image - distance is pos on R side.
Virtual object - distance is neg on R side Incident rays are converging)
Radius of curvature is neg on the V side.
Virtual image- distance is neg o the V side.
Rules for drawing rays to locate images
•A ray initially parallel to the central axis will pass through the focal point.
•A ray that initially passes through the focal point will emerge from the lens
parallel to the central axis.
•A ray that is directed towards the center of the lens will go straight
through the lens undeflected.
Then go to example
24(b). Given a lens with a focal length f = 5 cm and object distance p
= +10 cm, find the following: i and m. Is the image real or virtual?
Upright or inverted? Draw 3 rays.
Virtual side
Real side
.
.
F1
F2
p
1 1 1
 
i f p
m
1 1 1
1
  
i 5 10
10
m
i  10 cm



y 
i

y
p
10
 1
10
Image is real,
inverted.
24(e). Given a lens with the properties (lengths in cm) r1 = +30, r2 =
-30, p = +10, and n = 1.5, find the following: f, i and m. Is the image
real or virtual? Upright or inverted? Draw 3 rays.
Real side
Virtual side
.
F1
r1
r2
p
.
F2
y 
i

y
p
1 1
1
 n  1  
f
 r1 r2 
1 1 1
 
i f p
1
1  1
 1
 1.5  1 

f
30

30

 30
1 1 1
1

 
i 30 10
15
m
f  30cm
i  15cm
Image is virtual,
upright.
m

15
 1.5
10
27. A converging lens with a focal length of +20 cm is located 10 cm to
the left of a diverging lens having a focal length of -15 cm. If an object is
located 40 cm to the left of the converging lens, locate and describe
completely the final image formed by the diverging lens. Treat each lens
Separately.
Lens 1 Lens 2
+20
-15
f1
f2
f1 f2
40
10
Lens 1 Lens 2
+20
-15
f1
f2
f 1 f2
40
40
10
30
Ignoring the diverging lens (lens 2), the image formed by the
converging lens (lens 1) is located at a distance
1 1 1
1
1
  

. i1  40cm
i1 f1 p1 20cm 40cm
Since m = -i1/p1= - 40/40= - 1 , the image is inverted
This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.
Lens 1 Lens 2
+20
-15
f1
f2
f 1 f2
40
40
10
1 1 1
1
1
  

i2 f 2 p2 15cm 30cm
30
i2  30cm.
Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is
virtual (since i2 &lt; 0).
The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image
has the same size orientation as the object.
Optical Instruments
Magnifying lens
Compound microscope
Refracting telescope
```