mirror equation problems can be
intimidating when you first deal with
them and that's not because the mirror
equation is all that difficult it's kind
of easy it's just a few fractions added
together the place where it gets tricky
is deciding whether these should be
positive or negative so there's a bunch
of positive and negative sign decisions
that you have to make and if you make
even one of those incorrectly you can
get the wrong answer so let's do a few
more equation problems and you can see
how the signs work now I have to warn
you everyone has their own sign
convention there's a lot of different
sign conventions when you deal with
optics the one I'm going to use is the
one that I feel like most textbooks are
using these days and it's the one where
anything on the front side of the mirror
so by front side your eye should be over
here somewhere so let's say your eyes
over here right you're looking at some
object maybe it's an arrow or a crown
right here a blue crayon you're holding
in front of this mirror right here so
this is the mirror right here and the
convention I'm using is that anything on
this side of the mirror is going to be
counted as positive so if it's a focal
point on that side positive if it's an
object distance on this side its
positive and if the image distance comes
out positive you'll know that it's also
on this front side of the mirror
anything that comes out negative so if
we get a negative image distance after
we do our calculation we'll know that
that thing is behind the mirror that
kind of makes sense negative like behind
positive like in front so the side over
here is in front of the mirror positive
and back here would be behind the mirror
and that would be negative so I've got
some numbers in here already let's just
solve this one so what do we do we're
going to use this mirror equation we're
going to say that one over the focal
length and already we have to decide on
a positive or negative sign so this
mirror the way it's shaped right here
based on how we're looking at it is a
concave mirror and with the sign
conventions we just discussed and the
signs I'm using in this formula concave
mirrors always have a positive focal
length so in other words since this
focal point is 4 centimeters from the
center of the mirror I'm gonna have to
plug in the focal length as positive 4
centimeters notice I'm not converting
that's okay if you leave everything in
centimetres you'll just get an answer in
centimeters so it's okay you don't have
to convert as long as everything's in
the same units and we'll set that equal
to one over the object distance just one
other warning sometimes instead of do
you'll see this as so4 object distance
or you might see di as SI for image
distance is the same thing it's just a
different letter so the object distance
again is on this side in front of the
mirror so it's going to be positive 12
centimeters since it's located 12
centimeters from the center of the
mirror so there's also going to be
positive 12 centimeters and now we're
going to add to that one over the image
distance the image distance we don't
know I haven't drawn the image on here
it's going to be a surprise we don't
know what this is going to be but we can
solve for it so we can solve for image
distance I'll subtract 1 over 12 from
both sides which will give us 1 over 4
centimeters minus 1 over 12 centimeters
and that's going to have to equal 1 over
the image distance so 1/4 you could
rewrite as three twelfths so three
twelfths minus 1/12 is just going to be
two twelfths and that's going to equal
one over the image distance but two
twelfths is just 1/6 so 1 over the image
distance is just going to be 1/6 but
that's what 1 over the image distance is
sometimes people forget to flip this at
the end we don't want one over the image
distance we want the image distance so
finally you take one over each side and
we solve and we get that the image
distance is going to be 6 centimeters
and it came out positive that's
important this came out to be positive 6
centimeters so where's our image going
to be so since this image distance came
out to be positive our image is going to
be in front of the mirror so it's going
to be over here it's going to be 6
centimeters from the mirror somewhere
around here so at this point right here
is where our image is going to be but we
don't know how big it's going to be or
whether it's right side up or upside
down to figure that out we have to use a
different equation and that equation is
called the magnification equation says
that the magnification is equal to the
height of the image divided by the
height of the object and that's equal to
negative the image distance divided by
the object distance so it turns out this
race
vo of negative image distance over
object distance is always equal to the
ratio of the height of the image over
the height of the object so what's the
height of our image going to be let's
just solve for it if we solve for the
height of our image we get that the
height of the image is going to be
multiplied both sides by H Oh we'll have
the negative signs already here so
negative height of the object times this
ratio of the image distance over the
object distance and now we can just plug
in numbers the height of the object it
says that's three centimeters tall right
here so it's three centimeters so
negative three centimeters times the
ratio of the image distance was six the
object distance was 12 and so if you
solve this
you'll get 1/2 of negative 3 which is
negative 1.5 centimeters the negative
sign means that this image got inverted
so got flipped over it's going to be
upside down compared to what it was
before and the 1.5 is how tall is going
to be so what we end up with is an image
6 centimeters from the mirror and it's
going to be have a height half as tall
so 1.5 centimeters and it's going to be
upside down because of this negative
sign so it's going to be 1.5 centimeters
tall and upside down that's what you're
going to see if you look into this
mirror it's like a funhouse mirror it's
a weird curved mirror you'd see an
upside-down image right here it might
look like you could reach out and grab
it but it's going to be an optical
illusion there's going to be no object
there it's just going to be the image of
this object here so that's an example
using a concave mirror what would change
what if we did this what if we took our
object so say we don't put it here
anymore we move it inside here so
instead of being at 12 centimeters we
moved it to like 3 centimeters what
would we do differently everything would
be the same we just would plug in
instead of positive 12 down here we'd
plug in positive 3 so the mirror
equation works exactly the same you
still plug in whatever that object
distance is you solve for your image
distance we're of course going to get a
different image distance but whatever
you get that would tell you where the
image is and then you would take that
plug it into the magnification equation
if you wanted to decide how big the
image is going to be
whether it's going to be right side up
or upside down so these numbers going to
be different but you would use this
equation the exact same way now what
would happen if instead of using a
concave mirror
we used a convex mirror let's say we use
the mirror shaped like this so imagine
our eye again is over here looking at
this object inside of the mirror and
we're going to see an image of the
object we're going to see the object
right here but we're also going to see
the image of the object this mirror this
time instead of concave this is a convex
mirror so its focal length is behind the
mirror so what do we do now to figure
out where the images we again use the
mirror equation we're going to use the
same equation we're going to have one
over the focal length and again I
immediately have to make a decision on
the sign with the convention that I'm
using this focal length is behind the
mirror so this focal length for a convex
mirror is going to be negative so this
would be negative four centimeters and
that's going to equal one over the
object distance well again the object is
in front of the mirror twelve
centimeters so this object distance is
going to be positive twelve centimeters
and then we add to that the image
distance which we don't know this is
what we're going to solve for so this
time if we solve we're going to have 1
over negative 4 and again we have to
subtract 1 over 12 and that's all going
to have to be equal to 1 over the image
distance so now on the left hand side we
have negative 1/4 but that's the same as
negative 3 twelfths so negative 3
twelfths minus 1/12 is the same as
negative 4 twelfths and that's got to
equal 1 over the image distance but
negative 4 twelfths is the same as
negative 1/3 so 1 over the image
distance has to be equal to negative 1/3
and if we flip that over we get that the
image distance finally is going to be
negative 3 centimeters so in other words
this here is negative 1/3 so when you
flip that over you get that the image
distance is negative 3 centimeters where
is this image going to be well since it
came out negative that means it's behind
the mirror because that's the sign
convention we're using so it's going to
be 3 centimeters behind the mirror so
it's going to be like over here about at
this point right around here somewhere
three centimeters behind
this mere and again if we want to figure
out how tall it's going to be whether
it's right side up or upside down we're
going to have to use the magnification
equation and that magnification equation
looked like this it said that the height
of the image over the height of the
object had to be negative image distance
over object distance so let's just see
what this right-hand side is going to be
if we just set this equal to it's going
to be negative of negative 3 is the
image distance so I have to plug in the
negative 3 and you you keep that
negative in there this negative out here
comes along always but now we have
another negative inside of this image
distance of negative 3 and the object
distance again was 12 centimeters so
what are we left with we get negative of
negative 3 which is positive 3 over 12
which is positive 1/4 and the
centimeters cancel so what this ratio
tells you which is the magnification is
that the image is not going to be
inverted this positive means it's going
to be right side up and the 1/4 ratio
means it's going to be 1/4 as large the
image I should say is going to be 1/4 as
large as the object is going to be and
the reason is this ratio is equal to
height of the image over the height of
the object so in other words if I can
multiply if I want to multiply both
sides I can say that height of the image
is equal to positive 1/4 times the
height of the object well the height of
our object was 3 centimeters in this
case but whatever your height of the
object is you multiply it by this ratio
of negative DRDO
and you get what the height of the image
is going to be so we're going to get 1 4
3 to get positive 3/4 of a centimeter so
going to be it's going to be tiny it's
gonna be a little little image that's
going to be right side up so it's going
to be right here but it's going to be
teeny it's only going to be like this
3/4 of a centimeter tall that's what our
image is going to look like so recapping
you can use the mirror equation to
figure out where the images are going to
be located the sign convention we're
using is that objects images and focal
lengths in front of the mirror are going
to be positive anything behind the
mirror is going to be negative and you
could use the magnification equation to
figure out how tall the image is going
to be relative to the object by taking
negative the image distance over the
object distance