Download Using math withmirrors and ray diagrams

Mirror equation
 How can we use ray diagrams to determine where to
place mirrors in a telescope/
 We need to know where the image will be formed, so
that it can later be magnified.
We need two equations
Mirror equation
 F = focal length of the
mirror
 Do = distance to object
 Di = distance to the
image
 We are using reciprocals
here!
Magnification equation
 M = magnification
 Hi = height of the image
 Ho = height of the object
 Di = distance to image
 Do = distance to object
 NOTICE it is NEGATIVE
Di/Do
Let’s try one – from your book!
 A 4.00-cm tall light bulb
is placed a distance of
45.7 cm from a concave
mirror having a focal
length of 15.2 cm.
Determine the image
distance and the image
size.
 What do we know
 F = 15.2 cm
 Do = 45.7 cm
 Di = ?
M=?
 Hi = ?
 Ho = 4cm
Lets do it
1/f = 1/do + 1/di
So we have : 1/(15.2 cm) = 1/(45.7 cm) + 1/di
 Use the equation :

 Using the inverse key on our calculator gives you :
0.0658 = 0.0219 + 1/di
 0.0439 = 1/di, Then Di = ?
 di = 22.8 cm
Let’s continue
 What do we know
 F = 15.2 cm
 Do = 45.7 cm
 Di = 22.8
M=?
 Ho = 4cm
 Hi = ?
Step 2 Solve for magnification
 M = hi/ho OR - di/do
 So, M =hi /(4.0 cm) OR
 -Di/Do = - (22.8 cm)/(45.7 cm)
 This gives us M = -.498
 Remember h0= (4.0 cm)
Let’s try one
 What do we know
 F = 15.2 cm
 Do = 45.7 cm
 Di = 22.8
 M = -.498
 Ho = 4cm
 Hi =
Now use the other M to find Hi
 M =-.498
 Object height was 4 cm
 Final image would be
 4 x -.498 = -1.99 cm
Remember:
 What do we know
 F = 15.2 cm
 Do = 45.7 cm
 Di = 22.8
 M = -.498
 Ho = 4cm
 Hi = -1.99
What do the answers tell us
F+
Concave mirror ALL OF OUR MIRRORS
F-
Convex mirror ( we don’t use convex mirrors)
Di +
Real image on object side
Di -
Virtual image behind the mirror ( we don’t use these
either)
Hi +
Upright
Hi -
inverted
M+
larger
M-
smaller
f is + if the mirror is a concave mirror
f is - if the mirror is a convex mirror
di is + if the image is a real image and located on the object's
side of the mirror.
di is - if the image is a virtual image and located behind the
mirror.
hi is + if the image is an upright image (and therefore, also
virtual)
hi is - if the image an inverted image (and therefore, also
real)
Final analysis
 From our answers we






find:
F = 15.2 cm concave
mirror
Do = 45.7 cm
Di = 22.8 real image
M = -.5 smaller
Ho = 4cm
Hi = -1.99 inverted
Note:
 What we have done so
far applies only to
mirrors
 When we add lenses we
will use the same q
equations but the
 LOST part will be much
different