Explaining the Mathematical Reasoning of Finding Image in Convex Mirrors

[Ezio3.1415]

I am having a problem understanding the mathematical reasoning of finding the description of image in convex mirror...

We know the relation,
1/v + 1/u =1/f
is true for all mirrors... we can prove this for both concave and convex mirrors... In case of convex mirrors we use the conventional + and - to get to this general form(I guess real positive system)...
But when they used that formula to find out the position of image they again said for concave mirrors f is minus... I don't understand this point... We derived this formula saying r is minus for concave mirrors... Now what is left is the value... But again why are we saying this is minus?
I tried solving without taking it to be minus but that leads to false answer... Why? Please explain...

 

[andrien]

an artificial explanation which I have heard 4 years ago was that distances should be positive.so taking - and then - again makes it positive.

 

[Simon Bridge]

Yeah - it's just a convention so most of the tricky values get to be positive.
I think it also makes imaginary images at a negative distance.

If you like you can try to keep the derivation convention for doing problems. It doesn't matter as long as you are consistent. Personally I could never remember it and just did the geometry off ray diagrams.

 

[Ezio3.1415]

Thank you Andrien... This could be the explanation I guess... :)

Simon, the book contains both ray diagram and the equations... And I like solving using the equation... Who wants to draw all that stuff? :D

 

[Simon Bridge]

Simon, the book contains both ray diagram and the equations... And I like solving using the equation... Who wants to draw all that stuff? :D

Anyone who wants to understand ray optics.

 

[Emilyjoint]

Totally agree with S Bridge. I love drawing the diagrams, they show you what happens to the light in one view.
The equations are good if you need exact answers...usually you cannot draw the diagrams accurate enough (I find) to get exact answers.

 

[Simon Bridge]

You can use the sketched ray-diagrams to get the geometry, from the geometry you get the exact equation you need - it's all just triangles ... then the standard equation doesn't matter.

The only trouble comes when someone specifically wants to test your knowledge of the lens-maker's formula rather than your ability to do optics. Even so - a sketched diagram takes a few seconds and can tell you quickly if you have the sign convention backwards (and other reality checking functions ...) AND gets you more marks in long answers.

 

[Ezio3.1415]

I was talking about the problems from the book... It wants numbers as an answer... I can quickly calculate using the equation... I meant to say who would want draw these for the problems?

 

[Simon Bridge]

I was talking about the problems from the book... It wants numbers as an answer... I can quickly calculate using the equation... I meant to say who would want draw these for the problems?

1. the book is probably testing your knowledge of the equation - which will be why using the equation to answer the book questions is straight forward (except your question here suggests otherwise); 2. you should still sketch the diagrams because; (a) they help you check your working, and (b) the purpose of doing the exercises is to improve your knowledge of ray optics - the ray diagrams will do that better than rote use of the formula - which is an approximation anyway.

Therefore my original answer still stands: "Anyone who wants to understand ray optics."

If you just want to know how to do a subset of ray-optics problems, then you are going the right way about it though :) You don't have to understand every subject you do - it is not mandatory. But you did ask the question.

 

[Ezio3.1415]

Yeah the ray diagrams can help to see whether my solution is correct... Though I only draw it while solving the hardest problems... For the easy ones,I think I can draw it inside my head to check the signs...
Thank u for the answer... :)

 

[Ezio3.1415]

And another thing,for reflection of light I know only 4 special rays... Is it enough? If not,please give a link...