Image Formed by a Flat Mirror
Consider rays from an object, O, a
distance p in front of the mirror.
They reflect, with angle of reflection
= angle of incidence. To an
observer, the look like they came
from a distance q behind the mirror.
I is called the image of O.
Note that q = p.
Proof that q = p
ϕ = ϕ’ (Law of reflection))
ϕ’ = ϕ’’ (interior angles)
Therefore tan ϕ = tan ϕ’’.
p/s = q/s
Therefore p = q.
Note that this is true for all
rays from the object (point
source) that hit the mirror, i.e.
all values of s and ϕ. → All
reflected rays seem to come
from the same point: unique
image at q.
s
s
ϕ
ϕ’’
Because rays do not actually pass through the image (i.e. cannot put a
screen there to observe the image), it is called a “virtual image”.
ϕ’
This is true even when the point object is not directly in front
of the mirror
Point source
p
Plane of mirror
q
image
The image is the same distance behind
the plane of the mirror as the source is in
front of it, directly behind the source
(i.e. the line between the source and
image is perpendicular to the plane of
the mirror).
Magnification ≡
image height/object height:
M = h’ / h = +1
+ sign: image is upright
Extensive object
Where is the image and
what does it look like?
D
Extensive object
C
A
B
Choose some “extreme
points” outlining the
object.
D
Extensive object
C
A
B
B
A
C
D
Locate the point image
of these extreme points.
D
Extensive object
C
A
B
B
A
Image (note that it has
the same size as the
object).
C
D
D
Extensive object
C
A
B
z
y
x
B
A
C
D
Note that the mirror image
switches the front and
back (z) (not top/bottom
or sides (x,y))
Problem: Determine the minimum height of a vertical flat mirror in which a
person 178 cm (5’10”) can see his whole image.
Light from top of head travels path 5-4-3
to eyes. Light from feet travels path 1-2-3
to eyes. Therefore, the minimum length
of the mirror = L = a+b.
But the height of the man H = 2b + 2a.
Therefore, L = H/2 = 178 cm/2 = 89 cm.
Note the answer does not depend on how
far the man is standing from the mirror.
Convex Spherical Mirror
1/p + 1/q = 2/R = 1/f
If p = ∞, 1/q = 1/f – 1/p = 1/f – 0
q=f
As p decreases, 1/q = 1/f – 1/p,
q increases.
When p = f, 1/q = 0, q = ∞.
(i.e. reverse the arrows in the top picture).
If p = f, 1/q = 1/f – 1/p = 0
q=∞
If p < f, 1/q = 1/f – 1/ p < 0
→ q < 0:
A virtual image is formed
behind the mirror.
Mirrors
Object distance = p, image distance = q, radius of curvature = R, focal length = f
1/p + 1/q = 2/R = 1/f
If p,q,R,f in “front” of mirror, they are positive (e.g. q > 0 → real image).
If p,q,R,f in “back” of mirror, they are negative (e.g. q < 0 → virtual image)
R, f > 0 for concave mirror
R,f < 0 for convex mirror
R,f = ∞ for plane mirror.
1/p + 1/ q = 1/∞ = 0
q = -p
Problem:
Where is the position of the image if an object is p = 12 cm in front of a
a) Convex mirror with |R| = 10 cm
b) Concave mirror with R = 10 cm
c) Concave mirror with R = 30 cm
Problem:
Where is the position of the image if an object is p = 12 cm in front of a
a) Convex mirror with |R| = 10 cm
R = -10 cm, f = R/2 = -5 cm
1/q = 1/f – 1/p = -1/5cm – 1/12cm = -0.283/cm
q = -3.53 cm (virtual, behind mirror)
b) Concave mirror with R = 10 cm
f = R/2 = 5 cm
1/q = 1/f – 1/p = 1/5cm – 1/12cm = +0.117/cm
q = + 8.57 cm (real, in front of mirror)
c) Concave mirror with R = 30 cm
f = R/2 = 15 cm
1/q = 1/f – 1/p = 1/15cm – 1/12cm = -0.0167/cm
q = - 60 cm (virtual, behind mirror)