Will real images always point downwards?

[0192837465]

Will real images ALWAYS point downward?

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[RockyMD]

no, real images only point down (and is real) when the 'do' is greater than the focal length. if the 'do' is less than the focal length, the image is virtual and upright.

 

[mdhopefully23]

Yes Real images are ALWAYS inverted and Virtual images are ALWAYS upright.

IR UV--that's how you remember it

hope this helps.

 

[bluemonkey]

Yes Real images are ALWAYS inverted and Virtual images are ALWAYS upright.

IR UV--that's how you remember it

hope this helps.

That's not actually the case. Things are more complicated when you have multiple lens/mirror systems where the image of one serves as the object of another...in this case you essentially have a "virtual object."

 

[mmh309]

Yes Real images are ALWAYS inverted and Virtual images are ALWAYS upright.

IR UV--that's how you remember it

hope this helps.

this is not always the case. If object distance is less than the focal length, then image distance becomes negative. A negative image distance for a converging lens (which usually produces a real image) means that the image is on the same side as the object. Hence the image is virtual. So real images are not always inverted, like a previous poster was also saying.

 

[con1988]

REAL image is ALWAYS inverted. VIRTUAL image is ALWAYS upright.

Convex mirror and diversing lens ALWAYS make negative, virtual, upright image.

Concave mirrow and convering lense ALWAYS make postive, real, inverted image, EXCEPT when the object is within the focal point, then the image is VIRTUAL, INVERTED, and negative.

hope that helps.

 

[BloodySurgeon]

A negative image distance for a converging lens (which usually produces a real image) means that the image is on the same side as the object. Hence the image is virtual.

Exactly. It is virtual, not real.

Will real images ALWAYS point downward?

If pointed downwards you mean inverted then yes. I use inverted because a downward arrow can be inverted to an upward arrow through a real image.

conversing lens/mirror: o > f i = real and inverted
...............................o < f i = virtual and upright.

diverging lens/mirrors: all images are virtual and upright

 

[bozz]

this is not always the case. If object distance is less than the focal length, then image distance becomes negative. A negative image distance for a converging lens (which usually produces a real image) means that the image is on the same side as the object. Hence the image is virtual. So real images are not always inverted, like a previous poster was also saying.

real images are always inverted and virtual images are always upright (single lens/mirror systems)

your explanation doesn't really contradict anything the other poster said

the main difference is (and this is where most people make mistakes):

1)Lenses .... real images are on the other side of the object

2) Mirrors ... real images are on the same side as the object

Just to clarify, anyone on this thread, ask yourself this question:

A flat mirror ... is the image virtual or real? You'd be surprised how many people get this question wrong (I did too... 10 days before the MCAT).

 

[BerkReviewTeach]

That's not actually the case. Things are more complicated when you have multiple lens/mirror systems where the image of one serves as the object of another...in this case you essentially have a "virtual object."

I sincerely hope everyone reads your post, so I am quoting it and thus bumping it.

SUMMARY For a single lens or single mirror system, it's UV (upright/virtual) and IR (inverted/real). That is the always everyone is memorizing. BUT striaght memorization is dangerous on this exam, because they like to perturb the standard system.

As Bluemonkey has pointed out, you have systems with multiple lenses and/or mirrors, in which case the aforementioned memorized rule can break down. For instance, with a dual converging lens system where the object is outside the first focal length, the first image is IR. If that IR image is outside of the focal length of the second converging lens, then the second (and final) image will be inverted again, resulting in an upright/real image. This is how some projectors work.

The moral to the story is be very careful with what you memorize!!! It's a thinking test. You can bet that there are a few test writers out there who read the popular review books and create systems that violate the always true statements those review materials make.

 

[HippocratesX]

can somebody show us a picture of what this multiple lens/mirror system looks like when this is the case?

 

[BerkReviewTeach]

can somebody show us a picture of what this multiple lens/mirror system looks like when this is the case?

Hope this helps. I modified an example from a BR optics handout and colorized it.

 

[DrMattOglesby]

so we have CONCAVE and CONVEX mirrors...

and we have CONVERGING and DIVERGING lenses?

i just want to make sure im using the right vocabulary.

which apparently i wasnt!
we have concave and convex in reference to the lens or the mirror...
and converging and diverging in reference to what a lens might do to the light!
im slowly losing my mind guys...

 

[da8s0859q]

I sincerely hope everyone reads your post, so I am quoting it and thus bumping it.

SUMMARY For a single lens or single mirror system, it's UV (upright/virtual) and IR (inverted/real). That is the always everyone is memorizing. BUT striaght memorization is dangerous on this exam, because they like to perturb the standard system.

As Bluemonkey has pointed out, you have systems with multiple lenses and/or mirrors, in which case the aforementioned memorized rule can break down. For instance, with a dual converging lens system where the object is outside the first focal length, the first image is IR. If that IR image is outside of the focal length of the second converging lens, then the second (and final) image will be inverted again, resulting in an upright/real image. This is how some projectors work.

The moral to the story is be very careful with what you memorize!!! It's a thinking test. You can bet that there are a few test writers out there who read the popular review books and create systems that violate the always true statements those review materials make.

Period, end of story, close this bad boy.

OT, but I'm pissed that there weren't any optics questions on my 05/23 administration. I know that stuff fairly well. That's probably why the AAMC left them out.

 

[bozz]

REAL image is ALWAYS inverted. VIRTUAL image is ALWAYS upright.

Convex mirror and diversing lens ALWAYS make negative, virtual, upright image.

Concave mirrow and convering lense ALWAYS make postive, real, inverted image, EXCEPT when the object is within the focal point, then the image is VIRTUAL, INVERTED, and negative.

hope that helps.

#3) Virtual Upright and Negative

 

[Vihsadas]

I sincerely hope everyone reads your post, so I am quoting it and thus bumping it.

SUMMARY For a single lens or single mirror system, it's UV (upright/virtual) and IR (inverted/real). That is the always everyone is memorizing. BUT striaght memorization is dangerous on this exam, because they like to perturb the standard system.

As Bluemonkey has pointed out, you have systems with multiple lenses and/or mirrors, in which case the aforementioned memorized rule can break down. For instance, with a dual converging lens system where the object is outside the first focal length, the first image is IR. If that IR image is outside of the focal length of the second converging lens, then the second (and final) image will be inverted again, resulting in an upright/real image. This is how some projectors work.

The moral to the story is be very careful with what you memorize!!! It's a thinking test. You can bet that there are a few test writers out there who read the popular review books and create systems that violate the always true statements those review materials make.

Yes, yes and yes!

Remember, you have been taught in the context of the MCAT that real images are always inverted and virtual images are always upright. In terms of what you have learned, for single lens problems, this is true...but remember that the MCAT may well throw you a passage with a short explanation of a system where this rule breaks down and then ask you to apply what you just learned.. It is a thinking test!!

 

[MedGrl@2022]

Hope this helps. I modified an example from a BR optics handout and colorized it.

Can you repost this?

 

[sdm33]

D.U.V ( Diverging Upright and Virtual) Eg: Convex Mirror or Diverging Lenses, it does not matter if the object inside or outside the focal point.

Now for Converging , it can be a little tricky ,
C . I . R ( Converging Inverted Real) Except if object inside the focal point (C . U . V) image distance (-)

Also remember that in Diverging, Distance of image and focal length are negative ( most common mistake that students make)
distance of object is always positive

 

[MedGrl@2022]

I NEED A VOTE... Do people use ray diagrams when solving MCAT optic problems? EK tells me not too but I can't seem to memorize the stuff (at least yet). Plus, MCAT is a thinking exam and memorization can be dangerous.

Tell me what works best. I am going to try to see what Berkley Review and other test prep companies recommend too. 🙂

Thanks future Doctors. 🙂

 

[sdm33]

You don't have too, if you see them in a passage you should at least have the basic feel, here is a list that might help
Step-by-Step Method for Drawing Ray Diagrams

The method for drawing ray diagrams for concave mirror is described below. The method is applied to the task of drawing a ray diagram for an object located beyond the center of curvature (C) of a concave mirror. Yet the same method works for drawing a ray diagram for any object location.

1. Pick a point on the top of the object and draw two incident rays traveling towards the mirror.

Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the principal axis. Place arrowheads upon the rays to indicate their direction of travel.




2. Once these incident rays strike the mirror, reflect them according to the two rules of reflection for concave mirrors.

The ray that passes through the focal point on the way to the mirror will reflect and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray that traveled parallel to the principal axis on the way to the mirror will reflect and travel through the focal point. Place arrowheads upon the rays to indicate their direction of travel. Extend the rays past their point of intersection.


3. Mark the image of the top of the object.

The image point of the top of the object is the point where the two reflected rays intersect. If your were to draw a third pair of incident and reflected rays, then the third reflected ray would also pass through this point. This is merely the point where all light from the top of the object would intersect upon reflecting off the mirror. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point. (See note below.)




4. Repeat the process for the bottom of the object.

The goal of a ray diagram is to determine the location, size, orientation, and type of image that is formed by the concave mirror. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the mirror as the image of the top of the object. At this point the entire image can be filled in.

 

[MedGrl@2022]

D.U.V ( Diverging Upright and Virtual) Eg: Convex Mirror or Diverging Lenses, it does not matter if the object inside or outside the focal point.

Now for Converging , it can be a little tricky ,
C . I . R ( Converging Inverted Real) Except if object inside the focal point (C . U . V) image distance (-)

Also remember that in Diverging, Distance of image and focal length are negative ( most common mistake that students make)
distance of object is always positive

Does this hold true for double lens systems? Are virtual images always negative? And real images always positive for both single and double lens problems?

Is there a way to tell JUST from the thin lens equation whether an image is virtual and/or real?

 

[MedGrl@2022]

Does this hold true for double lens systems? Are virtual images always negative? And real images always positive for both single and double lens problems?

Is there a way to tell JUST from the thin lens equation whether an image is virtual and/or real?

Also, D.U.V ad CIR/CUV is for objects that are in front and positive what if the object is behind and negative, like in a double lens system?

 

[sdm33]

Does this hold true for double lens systems? What you mean?

 

[BerkReviewTeach]

Can you repost this?

I'm out of storage room for images. In adding newer ones, I was forced to remove older ones. If I can find that image I'll try to make it take less storage room, but finding might be challenging.

If you do examples 10.8a through 10.10b, you'll see the BR way for getting these questions right quickly and without having to draw a ray diagram.

 

[brainyhack]

I'm out of storage room for images. In adding newer ones, I was forced to remove older ones. If I can find that image I'll try to make it take less storage room, but finding might be challenging.

If you do examples 10.8a through 10.10b, you'll see the BR way for getting these questions right quickly and without having to draw a ray diagram.

I went through my old physics midterms and found a problem which perfectly demonstrates the case where a virtual and inverted image is formed in a two lens system.
Attached below is the image.
P.S. Sorry if it is messy -___-

 

[AFLATPEG]

Also, D.U.V ad CIR/CUV is for objects that are in front and positive what if the object is behind and negative, like in a double lens system?

if you can get ahold of the TBR physics book, it helps. i learned the "short-cut" way they do it for a SINGLE lens/mirror system. drawing the rays doesn't work for me and it takes too long.

Diverging lens/mirror = "SUV" (smaller, upright, virtual)

Converging lens/mirror = "UV" (upright, virtual) or "IR" (inverted, real) for when object is placed greater than the focal point. if object is placed inside the focal point, then you have "LUV" (larger, upright, virtual).

use the lens equation to figure out more specifically if you need to. i try to avoid math as much as possible.


for the multiple lens system, i just reason it out to see where the first image lands in respect to the 2nd lens. (because your first image is the object for your 2nd lens)

 

[sdm33]

Keep it Real lol

 

[dudewheresmymd]

if you can get ahold of the TBR physics book, it helps. i learned the "short-cut" way they do it for a SINGLE lens/mirror system. drawing the rays doesn't work for me and it takes too long.

Diverging lens/mirror = "SUV" (smaller, upright, virtual)

Converging lens/mirror = "UV" (upright, virtual) or "IR" (inverted, real) for when object is placed greater than the focal point. if object is placed inside the focal point, then you have "LUV" (larger, upright, virtual).

use the lens equation to figure out more specifically if you need to. i try to avoid math as much as possible.


for the multiple lens system, i just reason it out to see where the first image lands in respect to the 2nd lens. (because your first image is the object for your 2nd lens)

Could there be a question in which we have to pick a correct ray diagram for a particular lens? I know how to do it the TBR way without diagrams, but I didn't review how to draw the diagrams themselves lol.

 

[AFLATPEG]

Could there be a question in which we have to pick a correct ray diagram for a particular lens? I know how to do it the TBR way without diagrams, but I didn't review how to draw the diagrams themselves lol.

well i guess we're both screwed on that, aren't we? LOL
no, it's easier to pick out the correct rays diagram than drawing them yourself. if you know where your image lands, you can see where the rays intersect, etc.

 

[dudewheresmymd]

well i guess we're both screwed on that, aren't we? LOL
no, it's easier to pick out the correct rays diagram than drawing them yourself. if you know where your image lands, you can see where the rays intersect, etc.

OK sounds good. Keep me posted if you see any on real AAMCS/self-assessment. I saw like 8-9 of them in EK 1001...but that's EK 1001....