Ray optics
Every point on a luminous or illuminated
surface produces light rays in all
available directions
The image in a plane mirror is called a
virtual image, meaning that no light rays
pass through the image. Rather light rays
only appear to originate at the image.
The paraxial approximation: All rays are
nearly parallel to the optic axis.
I.e. all angles are small
Spherical mirrors
The small angle approximations are valid
whenever the diameter of the mirror is small
compared to its radius of curvature. In this
case, tan(2) is approximately equal to 2, so
that
f  R/2
The Principal of Reciprocity:
If a light ray follows a certain path through an
optical system, an incident ray in the opposite
direction will follow the reversed path.
A ray going through the focal point will
emerge parallel to the optic axis.
hi
si
M
, M
ho
so
h0
tan  
so  R
si hi
R  si
 

so ho
so  R
,
 hi
tan  
R  si
si ( so  R)  so ( R  si )  si so  si R  so R  so si
2 1 1
 2 si so  si R  so R 
 
R si so
1 1 1
 
si so f
Problem: A concave mirror has a radius of
curvature of 24cm. An object of height
8mm is placed 40cm from the mirror. Find
the location and height of the image.
Problem: A concave mirror has a radius of
curvature of 24cm. An object of height
8mm is placed 40cm from the mirror. Find
the location and height of the image.
Solution: f = 12cm,
1/si + 1/so = 1/f => si = f so / (so – f)
=> si = (12cm) (40cm) / [ 40cm – 12cm]
= 480 cm2 / 28 cm = 17.1cm
hi = M ho = - (si / so ) ho
= - (17.1 / 40) (8 mm) = - 3.4 mm
Problem: A concave mirror has a radius of
curvature of 24cm. An object of height
8mm is placed 18cm from the mirror. Find
the location and height of the image.
Solution: f = 12cm,
si = f so / (so – f)
=> si = (12cm) (18cm) / [ 18cm – 12cm]
= 216 cm2 / 6 cm = 36 cm
hi = M ho = - (si / so ) ho
= - (36 / 18) (8 mm) = - 16 mm
Problem: A concave mirror has a radius of
curvature of 24cm. An object of height
8mm is placed 6cm from the mirror. Find
the location and height of the image.
Problem: A concave mirror has a radius of
curvature of 24cm. An object of height
8mm is placed 6cm from the mirror. Find
the location and height of the image.
Solution: f = 12cm,
1/si + 1/so = 1/f => si = f so / (so – f)
=> si = (12) (6cm) / [ 6cm – 12cm]
= 48 cm2 / -6 cm = - 8 cm
hi = M ho = - (si / so ) ho
= - (-8 / 6) (8 mm) = + 10.7 mm
Whenever the paraxial
approximation is valid,
1 1 1
 
si so f
hi
si
M

ho
so
si > 0
si < 0
M> 0
M< 0
=> real image
=> virtual image
=> upright image
=> inverted image
M 1

M 1
 image is diminished
image is enlarged