lenses and mirrors HELP

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Chad, you've got 666 posts, LOL.

Exam Krackers has a good way to memorize lenses. It is for single lens and mirror systems. They make you memorize 3 sentences which are:

1) Draw a diagram of the situation. Draw the lens, and then draw an observer on the side where the eye will be placed. (The eye is what is viewing the image). Memorize the following: I (eye) am positive that real is inverted. This means that any image or focal point on the same side as your eye is positive, real, and inverted. Any image on the opposite side of your eye is negative, virtual. Always.

2) The second rule is Objects are always positive when they are in front of a lens or mirror and always negative when they are behind a lens or mirror. You stand in front of a mirror to see anything, so the side your eye is on is the front side. For lenses your eye is behind the lens (think of looking through a pair of glasses or through a camera) and the back side is where your eye is located.

3) As long as the object is in front, convex mirrors and diverging lenses make negative, virtual, upright images. Concave mirrors an converging lenses make positive, inverted, real images, EXCEPT when the object is in the focal point of the concave mirror, then the image is negative, upright, and virtual. To really understand this you must get a copy of the book and look at it.
Know that f (focal length) is equal to (1/R) where R is the radius of curvature, and know the basic equations for magnification as well as 1/f = 1/di + 1/do, where di= image distance and do= object distance.

You don't need to know the lens makers equation at all...
 
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Hindiana_Jones said:
1) Draw a diagram of the situation. Draw the lens, and then draw an observer on the side where the eye will be placed.
Click to expand...

That worked really well on a paper exam. The problem with drawing on a CBT exam is that it wastes time and isn't necessary. On a paper exam, you can write right next to the question, keeping the question in view the whole time. But when using scratch paper, drawing requires going back and forth between question and the picture. Even if it's only five to ten seconds wasted looking back and forth, it adds up over the duration of the exam. For many test-takers it's not a big deal, but why not learn to do things more efficiently and in your head if you can? We (TBR) emphasize in our classes visualizing things in your head and using shortcuts. Lens and mirrors have some of the very best short cuts of any topic.

We have two ways to answer these questions in fifteen seconds or so that require no drawing at all. I'm going to walk a fine line here, because these are proprietory methods we teach in our lectures. If you do enough lens and mirror examples you'll see a pattern.

For a diverging system, think of the word "Divirtual." No matter whether it's a lens or a mirror, a diverging system generates a virtual image that is within the focal point (making it smaller).

For a converging system, lens or mirror, you need to visualize the object position in terms of distinct regions.

Region I Object beyond R leads to an image between f and R (IR/Smaller)
Region II Object at R leads to an image at R (IR/Same size)
Region III Object between f and R leads to an image beyond R (IR/Larger)
Region IV Object at f leads to an image at infinity (no image)
Region V Object between f and f/2 leads to an image beyond f (UV/Larger)
Region VI Object between f/2 and lens leads to an image between f and the lens (UV/Larger)

Draw all six of these on the same picture and then visualize this summary.
Once you do this, you'll not need to draw ray diagrams, and instead can simply visualize where the object and corresponding image correlate. It becomes easy from this point to zero in on the best answer. Consider the following questions:

1) Object at 12 cm with R =18 cm; convex len. The image is:
a) UV at 36 cm
b) IR at 12 cm
c) IR at 18 cm
d) IR at 36 cm

Sol'n: Convex lens = converging system, therefore use summary picture.
If R is 18, then f if 9, so object (at 12) is in Region III. Image forms beyond R, so it's IR and greater than 18. Only choice D fits.


2) 10 cm tall object at 20 cm with f =8 cm; concave mirror. Where is image and how tall is it?
a) 3 cm tall at 6 cm
b) 12 cm tall at 9.3 cm
c) 6.7 cm tall at 13.3 cm
d) 20 cm tall at 40 cm

Sol'n: Concave mirror = converging system, therefore use summary picture.
If f is 8, then R if 16, so object is in Region I. Image forms between f and R, so it's smaller in size (because it's closer to the mirror than the object) and found between 8 and 16. Only choice C fits both criteria.

3) Object at 50 cm with R =40 cm; convex mirror. The image is:
a) UV at -14.3 cm
b) UV at -33.3 cm
c) IR at +14.3 cm
d) IR at +33.3 cm

Sol'n: Convex mirror = diverging system, therefore think "DIVIRTUAL."
Image is UV, so C and D are out. Image must form within f, so image distance must have an absolute value less than 20. Only choice A fits.

This method works well with multiple choice questions only, but luckily the MCAT is multiple choice. It is an easy method that can be picked up with just a little practice.

Note: If you see the utility of this approach, please use this method for your personal use, but do not share it with your preparation company. We are a small company that survives because we have several of these algorithms and strategies. We spend a great deal of time developing these strategies and would prefer to not have them pilfered.

The second method involves math and is very simple and can be done in your head in five to ten seconds. If you are really troubled by this topic, I don't mind sharing it via PM. I just don't want to expose too many of our tricks to our competitors.
 
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Wow BerkReviewTeach, very impressive. I'll be buying your books in the coming weeks. The EK system is about memorization as well, you don't really need to draw anything out, but it helps.
 
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Hindiana_Jones said:
Wow BerkReviewTeach, very impressive. I'll be buying your books in the coming weeks. The EK system is about memorization as well, you don't really need to draw anything out, but it helps.
Click to expand...

I in no way meant that as a knock on anyone's method. It's a basic difference in approach to a CBT versus paper exam. If you draw enough ray diagrams during your practice, they start to look the same. Once this happens, the summary picture method becomes very easy. So you are right that it's a memorization, but not of facts... of pathways to an answer.
 
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you guys should be pretty darn fast with fractions ( for the other sections) so writing it out (1/o +1/i = 1/f) should not take much longer
 
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BerkReviewTeach said:
That worked really well on a paper exam. The problem with drawing on a CBT exam is that it wastes time and isn't necessary. On a paper exam, you can write right next to the question, keeping the question in view the whole time. But when using scratch paper, drawing requires going back and forth between question and the picture. Even if it's only five to ten seconds wasted looking back and forth, it adds up over the duration of the exam. For many test-takers it's not a big deal, but why not learn to do things more efficiently and in your head if you can? We (TBR) emphasize in our classes visualizing things in your head and using shortcuts. Lens and mirrors have some of the very best short cuts of any topic.

We have two ways to answer these questions in fifteen seconds or so that require no drawing at all. I'm going to walk a fine line here, because these are proprietory methods we teach in our lectures. If you do enough lens and mirror examples you'll see a pattern.

For a diverging system, think of the word "Divirtual." No matter whether it's a lens or a mirror, a diverging system generates a virtual image that is within the focal point (making it smaller).

For a converging system, lens or mirror, you need to visualize the object position in terms of distinct regions.

Region I Object beyond R leads to an image between f and R (IR/Smaller)
Region II Object at R leads to an image at R (IR/Same size)
Region III Object between f and R leads to an image beyond R (IR/Larger)
Region IV Object at f leads to an image at infinity (no image)
Region V Object between f and f/2 leads to an image beyond f (UV/Larger)
Region VI Object between f/2 and lens leads to an image between f and the lens (UV/Larger)

Draw all six of these on the same picture and then visualize this summary.
Once you do this, you'll not need to draw ray diagrams, and instead can simply visualize where the object and corresponding image correlate. It becomes easy from this point to zero in on the best answer. Consider the following questions:

1) Object at 12 cm with R =18 cm; convex len. The image is:
a) UV at 36 cm
b) IR at 12 cm
c) IR at 18 cm
d) IR at 36 cm

Sol'n: Convex lens = converging system, therefore use summary picture.
If R is 18, then f if 9, so object (at 12) is in Region III. Image forms beyond R, so it's IR and greater than 18. Only choice D fits.


2) 10 cm tall object at 20 cm with f =8 cm; concave mirror. Where is image and how tall is it?
a) 3 cm tall at 6 cm
b) 12 cm tall at 9.3 cm
c) 6.7 cm tall at 13.3 cm
d) 20 cm tall at 40 cm

Sol'n: Concave mirror = converging system, therefore use summary picture.
If f is 8, then R if 16, so object is in Region I. Image forms between f and R, so it's smaller in size (because it's closer to the mirror than the object) and found between 8 and 16. Only choice C fits both criteria.

3) Object at 50 cm with R =40 cm; convex mirror. The image is:
a) UV at -14.3 cm
b) UV at -33.3 cm
c) IR at +14.3 cm
d) IR at +33.3 cm

Sol'n: Convex mirror = diverging system, therefore think "DIVIRTUAL."
Image is UV, so C and D are out. Image must form within f, so image distance must have an absolute value less than 20. Only choice A fits.

This method works well with multiple choice questions only, but luckily the MCAT is multiple choice. It is an easy method that can be picked up with just a little practice.

Note: If you see the utility of this approach, please use this method for your personal use, but do not share it with your preparation company. We are a small company that survives because we have several of these algorithms and strategies. We spend a great deal of time developing these strategies and would prefer to not have them pilfered.

The second method involves math and is very simple and can be done in your head in five to ten seconds. If you are really troubled by this topic, I don't mind sharing it via PM. I just don't want to expose too many of our tricks to our competitors.
Click to expand...

Developing tricks is good. But why don't you guys employ more efficient way to sell materials? You lost me as a customer just because of that? Thus mentioning "survival" and spending a great deal of time into developing these strategies sound quite contradictory.
 
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I loved this website for optics, here are some links for both types of mirrors and lenses:

http://www.physicsclassroom.com/Class/refln/u13l3e.cfm
http://www.physicsclassroom.com/Class/refrn/u14l5db.cfm
http://www.physicsclassroom.com/Class/refln/U13L4c.cfm
http://www.physicsclassroom.com/Class/refrn/u14l5eb.cfm

I pretty much memorized the diagrams at the bottom of each website cause it shows what happens to the image (both height and distance from lens) as the object approaches the lens from far away.

Hope it helps 👍 !!!
 
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BerkReviewTeach said:
That worked really well on a paper exam. The problem with drawing on a CBT exam is that it wastes time and isn't necessary. On a paper exam, you can write right next to the question, keeping the question in view the whole time. But when using scratch paper, drawing requires going back and forth between question and the picture. Even if it's only five to ten seconds wasted looking back and forth, it adds up over the duration of the exam. For many test-takers it's not a big deal, but why not learn to do things more efficiently and in your head if you can? We (TBR) emphasize in our classes visualizing things in your head and using shortcuts. Lens and mirrors have some of the very best short cuts of any topic.

We have two ways to answer these questions in fifteen seconds or so that require no drawing at all. I'm going to walk a fine line here, because these are proprietory methods we teach in our lectures. If you do enough lens and mirror examples you'll see a pattern.

For a diverging system, think of the word "Divirtual." No matter whether it's a lens or a mirror, a diverging system generates a virtual image that is within the focal point (making it smaller).

For a converging system, lens or mirror, you need to visualize the object position in terms of distinct regions.

Region I Object beyond R leads to an image between f and R (IR/Smaller)
Region II Object at R leads to an image at R (IR/Same size)
Region III Object between f and R leads to an image beyond R (IR/Larger)
Region IV Object at f leads to an image at infinity (no image)
Region V Object between f and f/2 leads to an image beyond f (UV/Larger)
Region VI Object between f/2 and lens leads to an image between f and the lens (UV/Larger)

Draw all six of these on the same picture and then visualize this summary.
Once you do this, you'll not need to draw ray diagrams, and instead can simply visualize where the object and corresponding image correlate. It becomes easy from this point to zero in on the best answer. Consider the following questions:

1) Object at 12 cm with R =18 cm; convex len. The image is:
a) UV at 36 cm
b) IR at 12 cm
c) IR at 18 cm
d) IR at 36 cm

Sol'n: Convex lens = converging system, therefore use summary picture.
If R is 18, then f if 9, so object (at 12) is in Region III. Image forms beyond R, so it's IR and greater than 18. Only choice D fits.


2) 10 cm tall object at 20 cm with f =8 cm; concave mirror. Where is image and how tall is it?
a) 3 cm tall at 6 cm
b) 12 cm tall at 9.3 cm
c) 6.7 cm tall at 13.3 cm
d) 20 cm tall at 40 cm

Sol'n: Concave mirror = converging system, therefore use summary picture.
If f is 8, then R if 16, so object is in Region I. Image forms between f and R, so it's smaller in size (because it's closer to the mirror than the object) and found between 8 and 16. Only choice C fits both criteria.

3) Object at 50 cm with R =40 cm; convex mirror. The image is:
a) UV at -14.3 cm
b) UV at -33.3 cm
c) IR at +14.3 cm
d) IR at +33.3 cm

Sol'n: Convex mirror = diverging system, therefore think "DIVIRTUAL."
Image is UV, so C and D are out. Image must form within f, so image distance must have an absolute value less than 20. Only choice A fits.

This method works well with multiple choice questions only, but luckily the MCAT is multiple choice. It is an easy method that can be picked up with just a little practice.

Note: If you see the utility of this approach, please use this method for your personal use, but do not share it with your preparation company. We are a small company that survives because we have several of these algorithms and strategies. We spend a great deal of time developing these strategies and would prefer to not have them pilfered.

The second method involves math and is very simple and can be done in your head in five to ten seconds. If you are really troubled by this topic, I don't mind sharing it via PM. I just don't want to expose too many of our tricks to our competitors.
Click to expand...

@BERKREVIEW: is this in your books? i have no interest in selling your secrets to the competition, but if i were a higher-up at any of the companies, i would hire a student to take your classes and report back to me or buy your books (if they're in the books). ofc, the secrets would have to be modified to avoid copyright infringement suits, but i'm sure companies do it (EK has a variation of this, idk which of you came up with it first).

nonetheless, thanks 🙂
 
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PreMedDude said:
I loved this website for optics, here are some links for both types of mirrors and lenses:

http://www.physicsclassroom.com/Class/refln/u13l3e.cfm
http://www.physicsclassroom.com/Class/refrn/u14l5db.cfm
http://www.physicsclassroom.com/Class/refln/U13L4c.cfm
http://www.physicsclassroom.com/Class/refrn/u14l5eb.cfm

I pretty much memorized the diagrams at the bottom of each website cause it shows what happens to the image (both height and distance from lens) as the object approaches the lens from far away.

Hope it helps 👍 !!!
Click to expand...
Great links, the diagram on the first link is a slightly more thorough drawing of what BerkReviewTeach was describing.
 
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I just draw out the EK method and formulas on scratch paper before the test. I can draw a functional version of the diagram in their physics book in about 30 seconds. Of course, I'm drawing it from memory, so I don't really have to draw it. It's already in my head, but I think the drawing speeds up the process and reduces the chance of a mistake. If a question comes up that pertains to it, I just find the lens and object location that fits the question.
 
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