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mirror equation problems can be
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intimidating when you first deal with
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them and that's not because the mirror
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equation is all that difficult it's kind
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of easy it's just a few fractions added
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together the place where it gets tricky
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is deciding whether these should be
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positive or negative so there's a bunch
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of positive and negative sign decisions
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that you have to make and if you make
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even one of those incorrectly you can
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get the wrong answer so let's do a few
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more equation problems and you can see
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how the signs work now I have to warn
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you everyone has their own sign
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convention there's a lot of different
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sign conventions when you deal with
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optics the one I'm going to use is the
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one that I feel like most textbooks are
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using these days and it's the one where
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anything on the front side of the mirror
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so by front side your eye should be over
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here somewhere so let's say your eyes
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over here right you're looking at some
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object maybe it's an arrow or a crown
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right here a blue crayon you're holding
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in front of this mirror right here so
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this is the mirror right here and the
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convention I'm using is that anything on
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this side of the mirror is going to be
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counted as positive so if it's a focal
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point on that side positive if it's an
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object distance on this side its
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positive and if the image distance comes
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out positive you'll know that it's also
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on this front side of the mirror
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anything that comes out negative so if
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we get a negative image distance after
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we do our calculation we'll know that
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that thing is behind the mirror that
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kind of makes sense negative like behind
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positive like in front so the side over
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here is in front of the mirror positive
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and back here would be behind the mirror
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and that would be negative so I've got
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some numbers in here already let's just
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solve this one so what do we do we're
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going to use this mirror equation we're
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going to say that one over the focal
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length and already we have to decide on
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a positive or negative sign so this
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mirror the way it's shaped right here
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based on how we're looking at it is a
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concave mirror and with the sign
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conventions we just discussed and the
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signs I'm using in this formula concave
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mirrors always have a positive focal
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length so in other words since this
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focal point is 4 centimeters from the
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center of the mirror I'm gonna have to
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plug in the focal length as positive 4
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centimeters notice I'm not converting
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that's okay if you leave everything in
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centimetres you'll just get an answer in
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centimeters so it's okay you don't have
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to convert as long as everything's in
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the same units and we'll set that equal
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to one over the object distance just one
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other warning sometimes instead of do
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you'll see this as so4 object distance
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or you might see di as SI for image
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distance is the same thing it's just a
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different letter so the object distance
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again is on this side in front of the
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mirror so it's going to be positive 12
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centimeters since it's located 12
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centimeters from the center of the
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mirror so there's also going to be
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positive 12 centimeters and now we're
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going to add to that one over the image
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distance the image distance we don't
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know I haven't drawn the image on here
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it's going to be a surprise we don't
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know what this is going to be but we can
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solve for it so we can solve for image
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distance I'll subtract 1 over 12 from
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both sides which will give us 1 over 4
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centimeters minus 1 over 12 centimeters
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and that's going to have to equal 1 over
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the image distance so 1/4 you could
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rewrite as three twelfths so three
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twelfths minus 1/12 is just going to be
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two twelfths and that's going to equal
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one over the image distance but two
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twelfths is just 1/6 so 1 over the image
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distance is just going to be 1/6 but
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that's what 1 over the image distance is
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sometimes people forget to flip this at
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the end we don't want one over the image
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distance we want the image distance so
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finally you take one over each side and
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we solve and we get that the image
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distance is going to be 6 centimeters
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and it came out positive that's
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important this came out to be positive 6
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centimeters so where's our image going
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to be so since this image distance came
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out to be positive our image is going to
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be in front of the mirror so it's going
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to be over here it's going to be 6
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centimeters from the mirror somewhere
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around here so at this point right here
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is where our image is going to be but we
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don't know how big it's going to be or
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whether it's right side up or upside
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down to figure that out we have to use a
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different equation and that equation is
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called the magnification equation says
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that the magnification is equal to the
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height of the image divided by the
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height of the object and that's equal to
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negative the image distance divided by
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the object distance so it turns out this
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race
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vo of negative image distance over
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object distance is always equal to the
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ratio of the height of the image over
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the height of the object so what's the
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height of our image going to be let's
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just solve for it if we solve for the
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height of our image we get that the
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height of the image is going to be
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multiplied both sides by H Oh we'll have
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the negative signs already here so
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negative height of the object times this
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ratio of the image distance over the
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object distance and now we can just plug
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in numbers the height of the object it
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says that's three centimeters tall right
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here so it's three centimeters so
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negative three centimeters times the
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ratio of the image distance was six the
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object distance was 12 and so if you
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solve this
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you'll get 1/2 of negative 3 which is
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negative 1.5 centimeters the negative
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sign means that this image got inverted
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so got flipped over it's going to be
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upside down compared to what it was
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before and the 1.5 is how tall is going
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to be so what we end up with is an image
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6 centimeters from the mirror and it's
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going to be have a height half as tall
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so 1.5 centimeters and it's going to be
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upside down because of this negative
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sign so it's going to be 1.5 centimeters
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tall and upside down that's what you're
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going to see if you look into this
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mirror it's like a funhouse mirror it's
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a weird curved mirror you'd see an
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upside-down image right here it might
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look like you could reach out and grab
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it but it's going to be an optical
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illusion there's going to be no object
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there it's just going to be the image of
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this object here so that's an example
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using a concave mirror what would change
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what if we did this what if we took our
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object so say we don't put it here
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anymore we move it inside here so
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instead of being at 12 centimeters we
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moved it to like 3 centimeters what
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would we do differently everything would
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be the same we just would plug in
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instead of positive 12 down here we'd
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plug in positive 3 so the mirror
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equation works exactly the same you
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still plug in whatever that object
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distance is you solve for your image
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distance we're of course going to get a
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different image distance but whatever
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you get that would tell you where the
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image is and then you would take that
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plug it into the magnification equation
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if you wanted to decide how big the
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image is going to be
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whether it's going to be right side up
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or upside down so these numbers going to
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be different but you would use this
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equation the exact same way now what
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would happen if instead of using a
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concave mirror
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we used a convex mirror let's say we use
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the mirror shaped like this so imagine
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our eye again is over here looking at
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this object inside of the mirror and
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we're going to see an image of the
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object we're going to see the object
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right here but we're also going to see
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the image of the object this mirror this
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time instead of concave this is a convex
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mirror so its focal length is behind the
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mirror so what do we do now to figure
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out where the images we again use the
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mirror equation we're going to use the
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same equation we're going to have one
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over the focal length and again I
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immediately have to make a decision on
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the sign with the convention that I'm
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using this focal length is behind the
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mirror so this focal length for a convex
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mirror is going to be negative so this
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would be negative four centimeters and
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that's going to equal one over the
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object distance well again the object is
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in front of the mirror twelve
217
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centimeters so this object distance is
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going to be positive twelve centimeters
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and then we add to that the image
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distance which we don't know this is
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what we're going to solve for so this
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time if we solve we're going to have 1
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over negative 4 and again we have to
224
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subtract 1 over 12 and that's all going
225
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to have to be equal to 1 over the image
226
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distance so now on the left hand side we
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have negative 1/4 but that's the same as
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negative 3 twelfths so negative 3
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twelfths minus 1/12 is the same as
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negative 4 twelfths and that's got to
231
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equal 1 over the image distance but
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negative 4 twelfths is the same as
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negative 1/3 so 1 over the image
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distance has to be equal to negative 1/3
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and if we flip that over we get that the
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image distance finally is going to be
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negative 3 centimeters so in other words
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this here is negative 1/3 so when you
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flip that over you get that the image
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distance is negative 3 centimeters where
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is this image going to be well since it
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came out negative that means it's behind
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the mirror because that's the sign
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convention we're using so it's going to
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be 3 centimeters behind the mirror so
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it's going to be like over here about at
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this point right around here somewhere
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three centimeters behind
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this mere and again if we want to figure
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out how tall it's going to be whether
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it's right side up or upside down we're
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going to have to use the magnification
253
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equation and that magnification equation
254
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looked like this it said that the height
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of the image over the height of the
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object had to be negative image distance
257
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over object distance so let's just see
258
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what this right-hand side is going to be
259
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if we just set this equal to it's going
260
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to be negative of negative 3 is the
261
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image distance so I have to plug in the
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negative 3 and you you keep that
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negative in there this negative out here
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comes along always but now we have
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another negative inside of this image
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distance of negative 3 and the object
267
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distance again was 12 centimeters so
268
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what are we left with we get negative of
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negative 3 which is positive 3 over 12
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which is positive 1/4 and the
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centimeters cancel so what this ratio
272
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tells you which is the magnification is
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that the image is not going to be
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inverted this positive means it's going
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to be right side up and the 1/4 ratio
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means it's going to be 1/4 as large the
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image I should say is going to be 1/4 as
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large as the object is going to be and
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the reason is this ratio is equal to
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height of the image over the height of
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the object so in other words if I can
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multiply if I want to multiply both
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sides I can say that height of the image
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is equal to positive 1/4 times the
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height of the object well the height of
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our object was 3 centimeters in this
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00:10:07,139 --> 00:10:08,970
case but whatever your height of the
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00:10:08,970 --> 00:10:11,220
object is you multiply it by this ratio
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00:10:11,220 --> 00:10:12,870
of negative DRDO
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00:10:12,870 --> 00:10:14,190
and you get what the height of the image
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00:10:14,190 --> 00:10:15,630
is going to be so we're going to get 1 4
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00:10:15,630 --> 00:10:20,550
3 to get positive 3/4 of a centimeter so
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00:10:20,550 --> 00:10:21,899
going to be it's going to be tiny it's
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00:10:21,899 --> 00:10:23,670
gonna be a little little image that's
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00:10:23,670 --> 00:10:25,350
going to be right side up so it's going
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to be right here but it's going to be
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00:10:27,180 --> 00:10:29,870
teeny it's only going to be like this
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00:10:29,870 --> 00:10:32,610
3/4 of a centimeter tall that's what our
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00:10:32,610 --> 00:10:34,740
image is going to look like so recapping
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00:10:34,740 --> 00:10:37,199
you can use the mirror equation to
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00:10:37,199 --> 00:10:39,180
figure out where the images are going to
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be located the sign convention we're
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using is that objects images and focal
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00:10:43,470 --> 00:10:44,939
lengths in front of the mirror are going
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to be positive anything behind the
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00:10:47,100 --> 00:10:48,509
mirror is going to be negative and you
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could use the magnification equation to
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00:10:50,160 --> 00:10:51,600
figure out how tall the image is going
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to be relative to the object by taking
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negative the image distance over the
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object distance