## Lens

Question:
An object is placed 24cm infront of a concave mirror of focal length 18cm. Where is the image formed and what is its magnification?Include an accurately dranw ray diagram??

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 $$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$ where f is the focal lenth of the lense,v is the object distace and u the image distance. [note: allways good to say what you have allready tried!] that didnt look quite right did it .... (latex is harder that i 1st thought!!) 1/f = 1/u + 1/v
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## Lens

This is an exercise that belongs in some homework section of the forums. From a mental ray diagram, I think the image is upright, shrunken and is a virtual image on the same side of the lens as the object.

It is really quite simple. Use the lens equation:

$$\frac{1}{s_0} + \frac{1}{s_i} = \frac{1}{f}$$

The only things you must be careful about are the signs (+,-) of your quantities. Do we count a concave lens as negative or positive? Look that up in your notes (and read about the lens equation).

 Quote by zanazzi78 $$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} [\tex] where f is the focal lenth of the lense,v is the object distace and u the image distance. [note: allways good to say what you have allready tried!] that didnt look quite right did it .... (latex is harder that i 1st thought!!) 1/f = 1/u + 1/v oh right,thanks,that was really helpfull...so would the equation turn out like this: 1/f = 1/u +1/V 1/18 = 1/u + 1/24 u = 1/72 so is this the image distance?? or should the lengths have been in Metres?? and also how do i work out the magnification??  Quote by Crosson This is an exercise that belongs in some homework section of the forums. From a mental ray diagram, I think the image is upright, shrunken and is a virtual image on the same side of the lens as the object. It is really quite simple. Use the lens equation: [tex] \frac{1}{s_0} + \frac{1}{s_i} = \frac{1}{f}$$ The only things you must be careful about are the signs (+,-) of your quantities. Do we count a concave lens as negative or positive? Look that up in your notes (and read about the lens equation).
thanks for your help,the equation you have given is the same as this one:
1/f = 1/u + 1/v

is that correct?? i dont know if the concave lens is positive or negative,it doesnt say in the question??

 Quote by suf7 so is this the image distance?? or should the lengths have been in Metres?? and also how do i work out the magnification??
Your question stated the distance in cm so the asnwer you worked out is also in cm!

When it comes to magnification, whats happened? The image has changed scale, so its a question about proportions! have a think and if you can work it out ...

 Quote by suf7 is that correct?? i dont know if the concave lens is positive or negative,it doesnt say in the question??
Ive always used some thing calle da sign convention (there are diffrent ones but they all give the same answers!)

For the "real is positive" convention:

both f and R are positive for concave mirrors/lenses (real focus)
both f and R are negative for convex mirrors/lenses (virtual focus)
u is positive for real objects and negative for virtual objects
v is positive for real images and negative for virtual images

this becomes very important when you start putting more than one lense in a system!

 Quote by zanazzi78 Your question stated the distance in cm so the asnwer you worked out is also in cm! When it comes to magnification, whats happened? The image has changed scale, so its a question about proportions! have a think and if you can work it out ...
ok so i think if the object distance is 24cm and the image distance is 0.01cm then its magnification is 2400???..im nots ure i quite understand that??

 Quote by zanazzi78 Ive always used some thing calle da sign convention (there are diffrent ones but they all give the same answers!) For the "real is positive" convention: both f and R are positive for concave mirrors/lenses (real focus) both f and R are negative for convex mirrors/lenses (virtual focus) u is positive for real objects and negative for virtual objects v is positive for real images and negative for virtual images this becomes very important when you start putting more than one lense in a system!
Whats "R"???...oh so if i had the same question but the lens was convex then i would use the same equation as before but the focal length would be negative??

 Maybe the language I used confused you, sorry If the object is 24 cm away from the lense and the image formed is 72cam away then, the magnification is the relationship between the two distances, $$- \frac {72}{24} = -3$$ you can also do it if you know the heights of the image and object basically $$Mag = \frac{image height}{object height}= - \frac {v}{u}$$

 Quote by suf7 Whats "R"???...oh so if i had the same question but the lens was convex then i would use the same equation as before but the focal length would be negative??
You got it!

R is the radius of a mirror, sorry (again) I should have explained all the terms

 Quote by zanazzi78 Maybe the language I used confused you, sorry If the object is 24 cm away from the lense and the image formed is 72cam away then, the magnification is the relationship between the two distances, $$- \frac {72}{24} = -3$$ you can also do it if you know the heights of the image and object basically $$Mag = \frac{image height}{object height}= - \frac {v}{u}$$
thats cool but i dont get where you got the 72 from in the equation -72/24??..isnt the number ontop of the fraction meant to be 1/72??

 Quote by zanazzi78 You got it! R is the radius of a mirror, sorry (again) I should have explained all the terms
so when the lens is convex does the magnification formula change or does it stay as "image distance/object distance???"...thanks, what your telling me really is making sense to me.

 Ok first I must readdress an errror I made! I gave you the wrong sign convention! Ooops So ... for the Real is positive convention ; f is positive for a convex lense (real focus) f is negative for a concave lense (virtual focus) u is positive for real objects and negative for virtual objects v is positive for real images and negative for virtual images so your lense equation should look like this $$- \frac {1}{18} = \frac{1}{24} + \frac {1}{v}$$ giving you an answer of v = -72 cm (not 1/72) the minus sign indicates that the image formed is virtual (as it would be for ALL concave lense images!) the magnification therefore goes ike this; $$Mag = - \frac {-72}{24} = +3$$ therefore you have an image 3 times bigger than the object ive attached a ray diagram to help keep practising it will become natural soon as youve allready got a good understanding of optics! A my idol once said " Optics are either very easy or very hard" - Richard Feynman

 Quote by zanazzi78 Ok first I must readdress an errror I made! I gave you the wrong sign convention! Ooops So ... for the Real is positive convention ; f is positive for a convex lense (real focus) f is negative for a concave lense (virtual focus) u is positive for real objects and negative for virtual objects v is positive for real images and negative for virtual images so your lense equation should look like this $$- \frac {1}{18} = \frac{1}{24} + \frac {1}{v}$$ giving you an answer of v = -72 cm (not 1/72) the minus sign indicates that the image formed is virtual (as it would be for ALL concave lense images!) the magnification therefore goes ike this; $$Mag = - \frac {-72}{24} = +3$$ therefore you have an image 3 times bigger than the object ive attached a ray diagram to help keep practising it will become natural soon as youve allready got a good understanding of optics! A my idol once said " Optics are either very easy or very hard" - Richard Feynman
Thanks,that all great..the only problem i have is when i try to use that equation i cant get -72??..i keep gettin 1/72 or -7/72???..how do you end up with just -72??...and finally when i do my ray diagram it should look like the one you have attatched except that my lengths will be different,is that right??

 Also you have written v = -72......but i thought we were working out u?? sorry,ive confused myself again.