# how does the mirror equation works and what lead to using of sign convention?

EVERYTHING HERE IS FOR CONCAVE MIRROR

Everywhere I see the derivation of the mirror equation is given by placing an object before the focus and then proving similarity of the triangles to get to the form

$\frac 1u + \frac 1v = \frac 1f$ .......(1)

( $v$ is toward the reflecting side of mirror)

It absolutely true and i completely agree with it , it should be valid for all the cases where you keep an object before the focus , my question is that how is it valid for the cases where object is kept between focus and pole ?? The triangles whose similarity we had used to get to the equations are not even formed here .....how then are we using the equation here ??

Moreover the '$v$' that we used here is actually the side length of a triangle , it cannot be negative this clearly shows that the equation is invalid for object kept between the focus and centre ... Because we get a negative '$v$'

After thinking for some time and calculations i have come to an answer for the requirement Of a sign convention

If we keep an object between the focus and the pole and extend the rays backward to form a virtual image , we can prove (again by similarity of different triangles ) that for object placed between the focus and pole

$\frac 1u - \frac 1v = \frac 1f$ ........(2)

( $u$ and $v$ are the object and image DISTANCE respectively from pole) ($v$ is towards the opposite of mirror)

Now we can see that for an object kept before the focus we can use eq(1) and for object between focus and pole we can use eq(2) to get image distance respectively

Where is the sign convention now ??

What if we want to generalise the formula i.e use a single formula for both cases .. Can we do that ?!

Lets see First case For cases of objects before focus we can clearly use equation 1

Second case For objects between focus and pole we can use equation 1 to get v

What if we use equation 1 for the second case It us logically absurd but lets see what happens ..

Acc. To equation 1

$v=\frac {uf}{(u-f)}$ ( v is towards reflecting side)

And acc to equation 2 $v=\frac{uf}{(f-u)} = -\frac{uv}{(u-f)}$ ($v$ is opposite to reflecting side)

So we see even though it is absurd to use equation 1 for the second case logically but due to some type of similarity between the two equations the result we get from using equation (1) for the second case is negative of the result from equation (2) (which is correct equation for the second case) for the second case

So if we ise equation (1) in second case we get a negative answer (I)Negative answer means that we are using wrong equation and we have to use eq(2) (II)By using eq(2) we get the same answer of $|v|$ but this time opposite side

So we can generalise equation (1) by using a method that negative answer of '$v$' means image is at distance $|v|$ opposite to the mirror

This method is sign convention

This is the reason for involvement if a sign convention system in mirrors ....

Clearly it is a method to generalise all equations to a single mirror equation ..

I just want reviews and want to know that isnt it right ?

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Example: An object is on the left, and the concave mirror is on the right. They are relatively far away $(d_o>f)$. Light rays travel to the right from the object to the mirror. So, light traveling from the object to the mirror moves a positive amount, and $d_o$ is positive. After the light reflects it is moving back to the left, so left is the positive direction for the reflected light. For this example, the image will be formed to the left of the mirror, and since left is positive, $d_i$ will be positive.
Another example: An object on the left, and concave mirror on the right, just as before, but now they are very close $(d_o<f)$. The light traveling from the object to the mirror moves right, so right is positive, and $d_o$ is positive, as before. The reflected light travels back to the left, making left the positive direction for reflected light, also as before. However, in this example, the reflected rays diverge. If you use geometric ray tracing on this example, you will trace the rays backwards behind (to the right of) the mirror, until they converge. But since left is positive for the reflected light, the image to the right will have a negative $d_i$.