Question:How to Draw Ray Diagrams for Diverging Lenses

[suf7]

Hi i have a question about a lens,im not sure if this is the right place to ask but il ask anyways...please help I am really stuck!

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??

 

[zanazzi78]

[tex]
\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 didn`t look quite right did it ... (latex is harder that i 1st thought!)

1/f = 1/u + 1/v

 

[krab]

Zanazzi: Your slash before the ending tex tag should be forward not backwards

 

[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} [/tex]

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).

 

[suf7]

[tex]
\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 didn`t 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??

 

[suf7]

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} [/tex]

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 don't know if the concave lens is positive or negative,it doesn't say in the question??

 

[zanazzi78]

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, what`s happened? The image has changed scale, so it`s a question about proportions! have a think and if you can work it out ...

 

[zanazzi78]

is that correct?? i don't know if the concave lens is positive or negative,it doesn't say in the question??

I`ve 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!

 

[suf7]

Your question stated the distance in cm so the asnwer you worked out is also in cm!

When it comes to magnification, what`s happened? The image has changed scale, so it`s 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??

 

[suf7]

I`ve 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??

 

[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,
[tex] - \frac {72}{24} = -3 [/tex]

you can also do it if you know the heights of the image and object

basically [tex] Mag = \frac{image height}{object height}= - \frac {v}{u} [/tex]

 

[zanazzi78]

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

 

[suf7]

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,
[tex] - \frac {72}{24} = -3 [/tex]

you can also do it if you know the heights of the image and object

basically [tex] Mag = \frac{image height}{object height}= - \frac {v}{u} [/tex]

thats cool but i don't get where you got the 72 from in the equation -72/24??..isnt the number ontop of the fraction meant to be 1/72??

 

[suf7]

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.

 

[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
[tex] - \frac {1}{18} = \frac{1}{24} + \frac {1}{v} [/tex]

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;
[tex] Mag = - \frac {-72}{24} = +3 [/tex]

therefore you have an image 3 times bigger than the object

i`ve attached a ray diagram to help

keep practising it will become natural soon as you`ve allready got a good understanding of optics!

A my idol once said " Optics are either very easy or very hard" - Richard Feynman

 

[suf7]

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
[tex] - \frac {1}{18} = \frac{1}{24} + \frac {1}{v} [/tex]

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;
[tex] Mag = - \frac {-72}{24} = +3 [/tex]

therefore you have an image 3 times bigger than the object

i`ve attached a ray diagram to help

keep practising it will become natural soon as you`ve 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 can't 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??

 

[suf7]

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

 

[zanazzi78]

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??

[tex] \frac{1}{72} = \frac{1}{v} [/tex]
multiply both sides by v
[tex] 1 = \frac {v}{72}[/tex]
multiply both sides by 72
[tex] 72=v [/tex]

ok!

don`t worry too much about what letters to use as long as you remember

[tex] \frac{1}{focal length} = \frac {1}{object distance} + \frac {1}{image distance} [/tex]

Maybe using the notation crosson gave may be a little less confusing, it doesn`t matter what the letters are as long as the equation is applied correctly!

 

[zanazzi78]

Thanks,that all great..the only problem i have is when i try to use that equation i can't 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??

as long as the drawing is in proportion it will be fine. The one I've done is to scale, 1:10, the focus I've placed on the opposite side to the object to remind me it`s 'negative' (from the sign convention) and the image is on the same side showing that it is virtual

 

[suf7]

cheerz

as long as the drawing is in proportion it will be fine. The one I've done is to scale, 1:10, the focus I've placed on the opposite side to the object to remind me it`s 'negative' (from the sign convention) and the image is on the same side showing that it is virtual

thanks,i think I've got the understanding of that now and should be able to do a similar question if it was on a convex lens...can i ask you another question??its a pretty long question and i don't know how to tackle it??

 

[zanazzi78]

thanks,i think I've got the understanding of that now and should be able to do a similar question if it was on a convex lens...can i ask you another question??its a pretty long question and i don't know how to tackle it??

Of course you can, fire away!

 

[suf7]

New Question

Of course you can, fire away!

Design a converging achromatic lens of focal length 100cm. Refractive indices and dispersive powers for the two glasses available are 1.51 and 0.016 for crown glass and 1.61 and 0.026 for flint glass. The converging element of the achromat is to be bi-convex and the diverging element is to be plano-concave. Which glass is to be used for each of the two elements of the lens, and determine the radii of curvature of the lens surfaces?

Thats the question,ist relly long and i don't understand nearly all of it,can you help?

 

[zanazzi78]

Thats the question,ist relly long and i don't understand nearly all of it,can you help?

Well lets's start by you telling me what you do understand from the question and then I'll be able to help!

Do you know what a bi-convex and plano-concave lense are?
Do you understand the term "converging achromatic lens"?
What do you know about dispersive powers?

 

[Gamma]

Post #15.

[tex] - \frac {1}{18} = \frac{1}{24} + \frac {1}{v} [/tex]

v = - 7/72 cm = - 10.3 cm

According to the ray diagram, image is in the same side as the object, image is upright and shrunken. I have attached a diagram.

 

[suf7]

Well lets's start by you telling me what you do understand from the question and then I'll be able to help!

Do you know what a bi-convex and plano-concave lense are?
Do you understand the term "converging achromatic lens"?
What do you know about dispersive powers?

No i don't understand any of those three questions,i only understand that i have to design something..lol..sorry I am not very good at this..how do i even start a question like this?

 

[suf7]

Post #15.

[tex] - \frac {1}{18} = \frac{1}{24} + \frac {1}{v} [/tex]

v = - 7/72 cm = - 10.3 cm

According to the ray diagram, image is in the same side as the object, image is upright and shrunken. I have attached a diagram.

you have given me a different answer??..and your ray diagram is different?...im confused,which is right?

 

[Gamma]

If you go back to post #15 and read it, you will see that there is a calculation error.


[tex] - \frac {1}{18} = \frac{1}{24} + \frac {1}{v} [/tex]

does not give 72 cm. Rather it gives v= -10.3 cm.

Also did you draw a ray diagram? do you agree with the ray diagram that I posted?

 

[suf7]

If you go back to post #15 and read it, you will see that there is a calculation error.


[tex] - \frac {1}{18} = \frac{1}{24} + \frac {1}{v} [/tex]

does not give 72 cm. Rather it gives v= -10.3 cm.

Also did you draw a ray diagram? do you agree with the ray diagram that I posted?

ok,thanks for knowticing the calculation error...i don't kno how to draw a ray diagram but thers one in post #15??...its different to yours tho??

 

[Gamma]

I did a google search and found this one which explains how to draw ray diagrams for the case of diverging lenses.

http://www.glenbrook.k12.il.us/gbssci/phys/Class/refrn/u14l5ea.html

If you are learning about 'light-lenses' you may have to go back to your notes or book to learn more about image formation by converging and diverging lenses and how to draw ray diagrams.

 

[suf7]

I did a google search and found this one which explains how to draw ray diagrams for the case of diverging lenses.

http://www.glenbrook.k12.il.us/gbssci/phys/Class/refrn/u14l5ea.html

If you are learning about 'light-lenses' you may have to go back to your notes or book to learn more about image formation by converging and diverging lenses and how to draw ray diagrams.

thanks,il try and read through that site.