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### Course: Class 10 Physics (India) > Unit 1

Lesson 7: Mirror formula & magnification# Mirror formula

Let's explore the mirror formula (1/f = 1/v+1/u) and see how to locate images without drawing any ray diagrams. Created by Mahesh Shenoy.

## Want to join the conversation?

- In the
**previous**video,*Mirror formula derivation*, it was said that this formula (1/f=1/v+1/u) was only for one kind of mirror and one kind of image, but in**this video**, it was said that it is used for all the mirrors. I am not sure which one is correct.(4 votes) - If it turns out that the mirror produces a virtual image will the formula tell us that v equals an undefined value?(1 vote)
- No. It means that whatever image is formed due to reflection is formed BEHIND the mirror. Hence, it would be virtual, erect and behind the mirror. V's value would be POSITIVE.

In a convex mirror, in both existing cases, it forms a virtual, erect and diminished image. Only U's value is negative, while V and F are positive.

Even in the last case of a concave mirror, when object is placed between the focal point and pole, it produces a virtual, erect and enlarged image behind the mirror.(3 votes)

- as tolkein said, "one formula to rule them all"(2 votes)
- are all formulas sign sensitive, and how do we know that a formula is sign sensitive(1 vote)
- As far as I know, there are majorly 4 formulae in light: the mirror formula, lens formula and magnification.

All three are sign sensitive as in a mirror and lens, each side has a different denotation.

Even the power of lens formula is sign sensitive, as it is calculated using focal length of a lens.

The magnification also is composed of V and U, or height of image and object, all sign sensitive.

So, any formula having image height, object height, U, V and F, which are all sign sensitive, would automatically become a sign sensitive one.(2 votes)

- n the diagram shown below, an object O\text OO is placed in front of a concave mirror, which forms an image I\text II.(1 vote)
- will we get the answer as infinity for image distance if no real image was formed(i.e when object is kept between focus and pole).(1 vote)

## Video transcript

- [Instructor] Let's say
we have a concave mirror of focal length two centimeters, and in front of it, we're
gonna keep an object, and let's say that the object is kept six centimeters in front of that mirror. Where will its image be? In this video, we're gonna find out exactly where that image would be. All right, now, in some
of our previous videos, we actually discussed about this. We've seen that there are a couple of ways you can solve this. One way would be to actually, you know, draw this on a piece of paper and then use your ruler
to figure out exactly, by drawing real diagrams,
figure out exactly where the image is going to be. But if we don't wanna
do all those drawings, then another method would be
to build a master formula, a master formula which
connects the object distance, the focal length, and the image distance, and the one which will work for all cases, for concave and convex mirrors, and we can just substitute
in that and get our answer. So in this video, we're not going to build the master formula. If you are interested in figuring out how to derive that master
formula, there a bonus video, and you can just go there and watch that. But we will see what
that master formula is, we'll understand it, and then we'll use it to solve the problem, all right? So that one formula,
which is going to help us figure out solve problems like this, that formula looks like this, and it's called the mirror formula, for obvious reason, because
it works for mirrors. And let's see what these things are. There are three things you can see. F stands for the focal length, so that's something we already know. V is the image, image distance, so this is image distance, and that's something we
don't know in this problem. And u is the object distance,
these are the letters that we usually use for
mirrors, object distance. And one super important thing to remember is that this is a sign-sensitive formula, which means when we're
substituting the numbers into it, we need to take care of their signs, they can positive values
and negative values, and to quickly recall why we do that, that's because we didn't differentiate between measurements done
in front of the mirror and measurements done behind the mirror. We assign one of those
measurements as positive, the other measurement is a negative. And do to quickly recall
how we figure that out, which is positive and which is negative, we always choose the incident
direction as positive. So notice that here is the
object and here is the mirror, and the incident ray would be this way. So incident direction,
right side is positive. This means that all the positions to the right side of the mirror
will be positive positions. Anything on the right side of
the mirror will be positive, and anything to the left of the mirror will be negative positions. And if this thing is
not super clear to you, then it would be great to pause this video and go back to previous videos, where we've discussed
this in great detail. Watch that and then come back over here. All right, now that we
have all our signs ready, let's look at what we have. We need to find the image distance, that's what we need to figure out, so this is what we need. What do we know? We know the object distance,
that is six centimeters, object is six centimeters from the mirror, and so u is six centimeters, but remember, we need to
take care of the signs. Notice that the object
lies on the negative side because all the positions
here are negative. So u will be minus six centimeters. Then, what else do we
need, we need focal length. We can see the focal length, the principal focus also
lies on the negative side, and therefore, this focal length will be negative two centimeters. Hope that is clear. Anything that lies on the negative side will get negative distances. All right, okay, so we have two things, f and u, and we need to
figure out what v is. And so now, then, whatever follows is just going to be
substitution and algebra. So, great idea to pause the video and see if you can do
the algebra yourself. Now, before you pause, a quick thing. I want to emphasize on
what these negative signs mean in this particular example. So once we have these sign conventions, when we substitute the
negative six over here, we're telling the formula that the object lies on the negative side, which means in front of the mirror. If we had substituted plus six by mistake, then the formula will assume that the object lies on this side, which is behind the mirror, and I know that sounds silly, but that's what the formula would assume, and then we'll probably
get the wrong answer. Similarly, when we substitute the negative two centimeters over, we're telling the formula that we're dealing with a concave mirror. How, because only if
it is a concave mirror, the focal point will lie
on the negative side, as is in this example. And if we had substituted plus
two centimeters over here, then it would assume that the focal point is on this side, meaning it would assume it's a convex mirror, again, it would give us a wrong answer. Just telling you how important it is to include proper signs in our formula. All right, pause and try to do the algebra yourself now. All right, let's do this. So if we substitute, we'll get one over f,
which is negative two, that will equal one over v plus one over u, which is negative six. Now, I'm not gonna
substitute the centimeters because everything is in centimeters, so my image distance will
also be in centimeters. I like to put the negative side, negative sign on the numerator, so call this minus 1/2 equals one over v, negative times positive is negative, so minus one over six, excellent. Now, I need to isolate one over v because that's what I want to figure out, so I'm gonna try to get rid
of this negative one over six, by adding one over six on both sides. So if we add one over six on both sides, on the left-hand side,
we're adding one over six, we will also add one over
six on the right-hand side, and when we do that,
that'll cancel with this one or negative one over six. And so, we then operate one over v. Okay, so let's solve this. We have to add two fractions. We're gonna make the common denominator, two and six, our cm is six. And so I'm gonna multiply this by three, multiply this by three, it'll get negative three plus one, plus one, that's gonna be negative two, so we can just write that down somewhere. Okay, minus two over six, and that's one over v. And it's getting a little crowded, but you know, it's almost done, minus, we can just cancel this, two goes one times, two goes three times. and so, we have one over v
equals negative one over three, so we want v, so we'll do the reciprocal, v will be negative reciprocal of this one, one over three reciprocal is three, as we get negative three centimeters. And there we have it,
we have found what v is. So what this means is that the image is three centimeters away from the port, from the
center of this mirror. Remember, whenever we
are using this formula, all the distances are from the port. So, three centimeters, but which side? Well, the sign tells me
it's on the negative side, and so, I know it's in
front of the mirror. So it would be somewhere over here because this is two, it's a given, so this would be somewhere
over here, three centimeters. And there we have it, we
have solved our problem. So, to quickly summarize, we learned what the mirror formula is, which connects focal
length, object distance, and image distance, and
to use this formula, we have to use our sign conventions, and the sign convention that we use is we consider the incident
direction is positive and then the other
direction becomes negative. Now, before we wind up,
one thing I need to mention is something that we
should always remember is that a sign convention that we're using is just a standard. If you ever forget that, don't worry, you will still get the right answer, so what I mean is if, by mistake, when you're solving this problem, if you chose the left side as positive, and the right side as negative, but you substituted
your values accordingly, then you would still get the right answer. You would eventually get that the image is three centimeters
in front of the mirror. But usually, we like to choose the incident direction as positive, and we like to stick to
this sign convention.