Understanding Focal Length: Solving for Image Position with 1/u + 1/v = 1/f

[RabbitWho]

Homework Statement

u= distance from the object to the mirror
v = distance from the image to the mirror
f = focal length
I don't understand what they are talking about and I would like to

2. Homework Equations
If the image is real 1/u + 1 v = 1/f
If the image is virtual 1/u - 1/v = 1/f

The Attempt at a Solution

I watched some videos to figure out what they meant by focal length. ^{(My book uses the fact that it is designed to be used in class as an excuse not to bother explaining things).} So focal length is the distance between the focus and the mirror. It's how far away from the mirror you have to move an object in order for it to be in focus, whether that be a real in focus image or a virtual one, right?

So right now I have my book and there is a mirror 1m away from me. I hold up the book and it is 2m from its virtual image.
1/u - 1/v = 1/f
1m - 1m = 0
Isn't the answer always 0? but that can't be because the focal length isn't always 0.

What am I doing wrong / misunderstanding?

 

[NascentOxygen]

What am I doing wrong / misunderstanding

These equations are applicable to curved mirrors, specifically, parabolic or spherical mirrors. Actually, not needing the complete sphere, usually, just a piece of a silvered sphere. You will find plenty of web resources with ray diagrams by googling.

 

[RabbitWho]

I did google :( I have watched about 6 videos on the topic and read about it in two separate textbooks, I know it seems really simple to you but "just google it" hasn't helped, so I came here. Thank you for specifying that it only applies to curved mirros, I'll have to buy one.

 

[NascentOxygen]

You may find you have a shiny soup spoon that can serve as a improvised mirror immediately, to demonstrate magnification, inversion, etc. There are shaving mirrors, concave on one side, convex on the other.

Though, I'd expect spherical mirrors for experimenters can be bought on ebay or amazon, too.

 

[RabbitWho]

Could I ask a related question? The book gives me a problem where the image is clearly virtual because the object is placed inside the focal length, but it tells me "let us assume we do not know whether the image is real or virtual and use the formula for a real image" Why is this?
When I use the formula for a virtual image I get the wrong answer, because I know I need it to be negative and it comes out positive.
What is the formula for a virtual image for? Or am I just doing something wrong with my Math

 

[J Hann]

The formulas that you write seem to assume that you know beforehand whether the image will be
real or virtual. It would be better to become familiar with ray tracing diagrams.
Textbooks tend to use different sign conventions for mirrors, but a ray tracing diagram
is superior to trying to memorize the sign conventions for mirrors.
Incidentally, from your comments, the focal point (1/2 the radius of curvature) of a mirror is
the point at which parallel rays incident on the mirror come to a focus.
For a plane mirror, parallel incident rays are parallel on reflection so the focal point is at infinity.

 

[Merlin3189]

u= distance from the object to the mirror
v = distance from the image to the mirror
f = focal length

If the image is real 1/u + 1 v = 1/f
If the image is virtual 1/u - 1/v = 1/f

I don't understand what they are talking about and I would like to

I'm not sure what your problem is here, except that in your example you seem confused about where to take the measurements from.
All measurements are from the centre of the mirror (surface) on the optical axis.
u is the magnitude (*) of the distance from the centre of the mirror surface to the object.
v is the magnitude (*) of the distance from the centre of the mirror surface to the image.
f is the distance from the centre of the mirror surface to the focal point, where rays parallel to the principle axis meet (for concave mirror) and where they appear to come from for a convex mirror. f is given a positive value for a cocave mirror and a negative value for a convex mirror.
(*) your formulae seem to imply that u and v are always positive, hence I say magnitude. But how your formulae handle virtual objects and virtual focal lengths, I'm not sure.

I've never come across this changing the formulae for different cases, I always use 1/u + 1/v = 1/f for all objects, all images and both types of mirror or lens. In that case, u, v and f need to take positive and negative values according to some strict convention. (Just as distances up and down need to be +ve and -ve, for example.) I've always used the convention that anything real is given a positive value and anything virtual is given a negative value, but there are other conventions.

I watched some videos to figure out what they meant by focal length ...So focal length is the distance between the focus and the mirror.

Yes

It's how far away from the mirror you have to move an object in order for it to be in focus, whether that be a real in focus image or a virtual one, right?

No. An object can form an image and be "in focus" at any distance.
The focus refers specifically to where rays parallel to the principal axis meet after reflection or refraction (or in the case of a divergent mirror or lens, the point such rays appear to emanate from.) You can find this using a distant object (Ideally at infinity! So stars or the sun would be a good approximation, but a light 50m away would be reasonable for the sort of experiments you might do.)

So right now I have my book and there is a mirror 1m away from me. I hold up the book and it is 2m from its virtual image.
1/u - 1/v = 1/f
1m - 1m = 0?
Isn't the answer always 0? but that can't be because the focal length isn't always 0.

You appear to be using a plane mirror, if the real object is 1m in front of the mirror and the virtual image is 2m away, ie. 1m behind the mirror.
In that case 1/f is indeed 0 and f is infinite (or undefined), because parallel rays striking a plane mirror are still parallel after reflection.

I don't think there is any mirror with focal length 0, but all plane mirrors have 1/f =0
Curved mirrors have different focal lengths, depending on their curvature.

I don't think anything can be gained by trying to guess what you might have meant if you were talking about a curved mirror. If you can explain again the situation you have in mind, giving all distances from the mirror, I'll try to do the right calculation.

 

[RabbitWho]

The formulas that you write seem to assume that you know beforehand whether the image will be
real or virtual. It would be better to become familiar with ray tracing diagrams.

That is what it seems like, right, but the book says that even if you know from the diagram that the image is virtual you should use the formula for real

I've never come across this changing the formulae for different cases, I always use 1/u + 1/v = 1/f for all objects, all images and both types of mirror or lens. In that case, u, v and f need to take positive and negative values according to some strict convention. (Just as distances up and down need to be +ve and -ve, for example.) I've always used the convention that anything real is given a positive value and anything virtual is given a negative value, but there are other conventions.

Here is what it says exactly "An object is placed 12cm in front of a concave mirror of focal length 20cm. Find the position, nature, and magnification of the image.
The object is inside of the focus therefore the image is virtual and the forumla to use is 1/u - 1/v = 1/f. However, let us assume we do not know whether it is real or virtual and use the formula 1/u + 1/v = 1/f"

If you use the other formula you get the wrong answer. (It should be v=-30, I have this sum worked out I know the answer and how to get it) And I don't understand why you have to use the formula for a real image on a virtual image.

 

[Merlin3189]

f = 20cm, u= 12 cm
If you use $\frac{1}{u}-\frac{1}{v}=\frac{1}{f}$, you get
$\frac{1}{12}-\frac{1}{v}=\frac{1}{20}$ , so
$\frac{1}{12}-\frac{1}{20}=\frac{1}{v}$
and v= +30cm so the assumed virtual image must be 30cm behind the mirror

If you use $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$, you get
$\frac{1}{12}+\frac{1}{v}=\frac{1}{20}$ , so
$\frac{1}{v}=\frac{1}{20}-\frac{1}{12}$
and v= -30cm so the image must be virtual and 30cm behind the mirror (because a real image must be a positive distance in front of the mirror)

The advantages of always using $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$, are that ;
you only need one formula and don't need to remember which is which,
you don't need to know in advance whether the image is real or virtual,
it works for both types of mirrors and lenses (converging and diverging)
it even works for virtual objects!

Edit: (additional comment) If you use different formulae for different situations, you end up with 8 formulae;
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ real obj, real im, concave mirror
$-\frac{1}{u}-\frac{1}{v}=-\frac{1}{f}$ virtl obj, virt im, convex mirror
$\frac{1}{u}+\frac{1}{v}=-\frac{1}{f}$ real obj, real im, convex mirror
$-\frac{1}{u}-\frac{1}{v}=\frac{1}{f}$ virt obj, virt im, concave mirror
$\frac{1}{u}-\frac{1}{v}=\frac{1}{f}$ real obj, virt im, concave mirror
$-\frac{1}{u}+\frac{1}{v}=-\frac{1}{f}$ virt obj, real im, convex mirror
$\frac{1}{u}-\frac{1}{v}=-\frac{1}{f}$ real obj, virt im, convex mirror
$-\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ virt obj, real im, concave mirror
and you would need fairly good insight in order to say in advance what sort of image you were expecting.
All in all, it's much easier to have one formula and learn one rule which tells you whether the numbers you put in (or get out) are +ve or -ve. (ie. real objects, images and foci are positive. virtual objects, images and foci are negative. If any answer is a negative number, then it is a virtual object, image or focus.)

Edit2: changed layout of edit 1 and added comments to it.

 

[RabbitWho]

Saw this a week and a half ago, I liked it, it helped me. Now studying lenses rather than mirrors, forgot what you had said but remembered you had answered so came back here and got helped again. Thank you!

 

[Merlin3189]

I appreciate your comment. It's always nice to know that you've been able to help. Good luck with the rest of your course.