Optics: Find & Confirm Image Positions w/ Ray Diagrams

[questions_uk]

Hi. "A biconvеx lеns has a focal length of 100mm. an object is placеd at distancеs of 150 mm and 75 mm from teh lens. Sketch a ray diagram to find the position of thе respective images. confirm these positiones via calculatinons.

This is the diagram I have drawn:
http://www.freewebs.com/questionsuk/diagram.JPG

Does this seem ok?

However, what equations would be used to confirm the position?

Any help is appreciated.

 

[dlgoff]

Give this wikipedia page a read and see if it helps. In paticular the part on the http://en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation".

 

[questions_uk]

Thanks for your help, although how does that equation consider the position of the object from the lens?

 

[dlgoff]

I should have said the http://en.wikipedia.org/wiki/Lens_(optics)#Imaging_properties". Use the thin lens formula.

 

[questions_uk]

Test

 

[questions_uk]

Test

 

[Doc Al]

This is the diagram I have drawn:
http://www.freewebs.com/questionsuk/diagram.JPG

Does this seem ok?

Not really. To learn how to draw a ray diagram to calculate the image distance, read this: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/raydiag.html#c1".

So if am using 1 / f = 1 / s 1 + 1 / s 2 that would be (according to the diagram I've done) 1 / 75 + 1 / 75 which approximately equals 1 / 100 but what about 1 / 150 + 1 / 150 where that is 0.0067 which does not equal 1/100? Unless the diagram needs to be re drawn?

Think of the thin lens equation as 1/f = 1/d_o+ 1/d_i. Here d_o is the object distance, the distance between the object and the lens, and d_i is the image distance, the distance between the image and the lens. Use the equation to solve for d_i.

Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html#c1"

Also, it says "an object" so perhaps the 75 mm means 75 mm from the lens in one direction and 150 mm in the other direction? I've assumed the object has been placed at the two distances at one side of the lens.

You are supposed to figure out the location of the image for two cases: one where d_o = 75 mm and one where d_o = 150 mm. (The big difference is that in one case the object is inside the focal length, while in the other it is outside that length.)

 

[questions_uk]

Not really. To learn how to draw a ray diagram to calculate the image distance, read this: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/raydiag.html#c1".Think of the thin lens equation as 1/f = 1/d_o+ 1/d_i. Here d_o is the object distance, the distance between the object and the lens, and d_i is the image distance, the distance between the image and the lens. Use the equation to solve for d_i.

Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html#c1"You are supposed to figure out the location of the image for two cases: one where d_o = 75 mm and one where d_o = 150 mm. (The big difference is that in one case the object is inside the focal length, while in the other it is outside that length.)

Thanks for the reply.

Just a question about the diagram, does this seem more correct?

http://www.freewebs.com/questionsuk/diagram.JPG (you might need to refresh)

Test

 

[Doc Al]

Just a question about the diagram, does this seem more correct?

http://www.freewebs.com/questionsuk/diagram.JPG (you might need to refresh)

Yes, much better. But to get a numerical result you must draw it carefully and to scale.

With the 75 mm object, would the respective virtual image be behind the 150 mm object? This is what am not sure about.

What does the lens equation tell you?

 

[questions_uk]

Yes, much better. But to get a numerical result you must draw it carefully and to scale.


What does the lens equation tell you?

Thanks for the reply. For the 150 mm object I worked the image out to be 250 mm from the lens.

For the 75 mm however I've done

1 / 100 = 1 / 75 + 1 / v

= 0.003

= 333 mm (I think it's a negative as it is behind the focal length)

 

[Doc Al]

Thanks for the reply. For the 150 mm object I worked the image out to be 250 mm from the lens.

Redo that calculation.

For the 75 mm however I've done

1 / 100 = 1 / 75 + 1 / v

= 0.003

= 333 mm (I think it's a negative as it is behind the focal length)

Redo that more carefully. Your equation is correct, just solve for the unknown.

 

[questions_uk]

Redo that calculation.Redo that more carefully. Your equation is correct, just solve for the unknown.

How comes? I did 1 / f = 1 / u + 1 / v

1 / 100 = 1 / 150 + 1 / v2

0.01 = 0.006 + 1 / v2

0.01 - 0.006 = 0.004 which is equivalent to 1 / v 2 so

v2 = 0.004

= 250 mm ?

v2 by the way is not v x 2

Thanks.

 

[Doc Al]

How comes? I did 1 / f = 1 / u + 1 / v

1 / 100 = 1 / 150 + 1 / v2

0.01 = 0.006 + 1 / v2

Don't round off. That should be 1/150 = 0.006666...

Even better, do this:

1/100 - 1/150 = 3/300 - 2/300 = 1/300 = 1/v2.

 

[questions_uk]

Don't round off. That should be 1/150 = 0.006666...

Even better, do this:

1/100 - 1/150 = 3/300 - 2/300 = 1/300 = 1/v2.

Thanks! For both I got 300 mm although the object at 75 mm is a virtual image.

 

[Doc Al]

Good!

 

[questions_uk]

Also, it has given the index as 1.6 and has asked to calculate the radii of the curvature of the lens.

I've used this equation:

Radii of curvature R = 1 / u + 1 / v = 1 / f (n1 – 1) (1 / R1 – 1 / R2)

Now, I assume it's asking for one radius so I've not included the 1 / R2 in that equation.

To make it simpler the equation is: 1 / f (n1 – 1) (1 / R1 – 1 / R2)

I did: 0.01 = 0.6 [1 / R1] although there may also need to be a -1 / R2 there as well.

I then made R1 to be 6mm. Any replies are appreciated!

 

[Doc Al]

A more reasonable assumption would be to assume that the biconvex lens is symmetrical--both sides have the same radius. (And be more careful with your arithmetic.)

 

[questions_uk]

Ah yes thanks very much!

One last thing, it says "calculate the longitudinal and transverse magnification of the image if the object is 10 mm high and 5 mm long".

For the longitudinal it has given fo x fi / xo^2 fo is the focal length of the object
fi is the focal length of the image
xo is the distance from the object to the focal length

For the 150 mm object I got -4 and the 75 mm object 16

However, where does the objects dimensions (10 mm high and 5 mm long) come into this? Perhaps it just comes into the transverse equation which is

fo / xo = xi / fi = -yi / yo = xo xi = fo fi

??

 

[questions_uk]

Actually, the yo and yi represent the lengths of the object and its respective image. Not sure about the 5mm long though.

 

[Doc Al]

Only the longitudinal magnification will depend on the object size.

 

[questions_uk]

Hi. Thanks for the reply. However, the longitudinal equation that has been provided is fo x fi / xo^2

where

fo is the focal length of the object
fi is the focal length of the image
xo is the distance from the object to the focal length

The dimensions do however seem to be needed for the transverse.

 

[Doc Al]

Only the longitudinal magnification will depend on the object size.

Correction: As long as the object is small, neither transverse or longitudinal magnification will depend on the size of the object.

 

[questions_uk]

Correction: As long as the object is small, neither transverse or longitudinal magnification will depend on the size of the object.

Then why did they give the dimensions of it? Apparently the "long" bit is part of the transverse equation.

 

[Doc Al]

Then why did they give the dimensions of it?

Perhaps they want you to calculate the apparent size of the image, not just the magnification.

Apparently the "long" bit is part of the transverse equation.

Transverse means perpendicular to the axis, so transverse magnification affects the height of the image. Longitudinal magnification affects the length.

 

[questions_uk]

Ok thanks your for your dude.

 

[Redbelly98]

Thanks for the reply.

Just a question about the diagram, does this seem more correct?

http://www.freewebs.com/questionsuk/diagram.JPG (you might need to refresh)

Test

There are some problems with that diagram:

• Rays that are parallel to the central optical axis on one side of the lens, should pass through the central axis at 1 focal length away on the other side of the lens.
  
• A ray is undeflected only if it passes through the exact center of the lens (i.e., it crosses the central axis in the middle of the lens).

 

[questions_uk]

There are some problems with that diagram:

• Rays that are parallel to the central optical axis on one side of the lens, should pass through the central axis at 1 focal length away on the other side of the lens.
  
• A ray is undeflected only if it passes through the exact center of the lens (i.e., it crosses the central axis in the middle of the lens).

Thanks for the reply although am not sure what you mean about the first point. "central optical axis"? and what do you mean by the 1 focal length away bit?

 

[Doc Al]

Thanks for the reply although am not sure what you mean about the first point. "central optical axis"? and what do you mean by the 1 focal length away bit?

The optic axis is just a line drawn perpendicular to the lens through its center. One focal length away means one focal length distance from the lens.

Please review the link I gave you in post #7 about how to draw ray diagrams. (The instructions on that link are similar to what Redbelly98 points out.)

I should have looked at your diagram more closely--it's quite a ways off. Thanks to Redbelly98 for catching that.

If you draw the diagram accurately, you should be able to determine the location of the images and compare it to the locations you calculated with the thin lens formula.

 

[questions_uk]

Hi there. The diagram has been edited: http://www.freewebs.com/questionsuk/diagram2.JPG

It is just supposed to be a "sketch" by the way.

 

[Redbelly98]

Thanks for the reply although am not sure what you mean about the first point. "central optical axis"? and what do you mean by the 1 focal length away bit?

Just to add to Doc Al's explanation. The "central optic axis" is the long horizontal line in your sketch. "1 focal length away" are the f0 and f1 points in your sketch, each 100 mm (the focal length) from the lens.

Sketch a ray diagram to find the position of thе respective images.

Not to sound too obvious or snarky, but to answer the question correctly would require drawing the diagram and rays properly. One suggestion: try using some graph paper so that you can maintain a consistent scale. I noticed in your figure that "300 mm" to the right of the lens is only about 1.5 times the "100 mm" distance indicated there. The idea is to see where rays converge to form the image, which can then be reasonably estimated only if it is a scale drawing.

Doc Al's link gave a good example of a ray sketch; more specifically:
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/raydiag.html#c2

Regards, and good luck

Mark

 

[questions_uk]

Thanks.

 

[questions_uk]

Hi there. Just a quick question regarding an equation. In order to find the transverse magnification of an object 10 mm high and 5 mm long, it gives the equation:

fo / xo = xi / fi = - yi / yo giving xo x xi / fo x fi

Now which part of that is supposed to be used to calculate the transverse magnification?

Am not sure why some variables are equated to other variables e.g. xo x xi = fo x fi

Thanks.

 

[Doc Al]

Hi there. Just a quick question regarding an equation. In order to find the transverse magnification of an object 10 mm high and 5 mm long, it gives the equation:

fo / xo = xi / fi = - yi / yo giving xo x xi / fo x fi

Now which part of that is supposed to be used to calculate the transverse magnification?

Am not sure why some variables are equated to other variables e.g. xo x xi = fo x fi

The definition of transverse magnification is:
m ≡ yi/yo

It can be shown (using similar triangles) that:
m ≡ yi/yo = -si/so (Where si is the image distance; so is the object distance. Both measured from the lens.)

If you measure distances from the focal points (xi & xo are the image and object distances measured from the focal points), then you can show (using the thin lens equation) that:
xi * xo = f*f

m ≡ yi/yo = -xi/f = -f/xo

Using that last term is the easy way to calculate the transverse magnification.

 

[questions_uk]

Thanks.