BR physics question.

[stevvo111]

Question is on passage VII in section 10 of TBR. I'm confused by this darn chart they give us?!

For trials D and E, what does the (-) signify? Does it mean the original object was to the right of the diverging lens? If it was then, the image formed would be (+) and to the right of the lens, and this image would serve as the object for the second lens which would place the image on lens 1.

I'm wondering because I always though with diverging lenses the focal length was negative, and the object was placed in a positive location, thus the resulting image would be "small upright and virtual" with i= (-)...


Can someone help me make sense of this?

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[drspacheman]

I don't really understand this either, here's the diagram and table if anyone is interested.

 

[akim6890]

Question is on passage VII in section 10 of TBR. I'm confused by this darn chart they give us?!

For trials D and E, what does the (-) signify? Does it mean the original object was to the right of the diverging lens? If it was then, the image formed would be (+) and to the right of the lens, and this image would serve as the object for the second lens which would place the image on lens 1.

I'm wondering because I always though with diverging lenses the focal length was negative, and the object was placed in a positive location, thus the resulting image would be "small upright and virtual" with i= (-)...


Can someone help me make sense of this?

Ok, it's really easy to get all mixed up with signs especially when they give you both types of lenses in one system. I think the confusion comes from the focal length sign. That has nothing to do with either side. The focal length is always negative for a diverging lens.

Remember for any lens, for the object, positive is on the side the light is coming from and negative is the side of the lens the light is going to. In this case for the object, positive is on the left and negative is on the right.
For the image, it's the opposite. Positive is on the side of lens light is going to (right) and negative where it's coming from (left). Always consider what's real and what's virtual for +/-.

So for D, the object distance with the negative sign is on the right side and since diverging lenses only produce virtual images, it will be on the same side giving the positive image distance. From here like you said, the +5 image distance is less than the focal length of the concave lens and also gives a virtual image back where lens 1 is giving a negative image distance. My kaplan book mentions: "Although the object of a single lens is on the V side, this does not make the object virtual. Objects are real, with a positive object distance, unless they are placed in certain multiple lens systems (a scenario which, by the way, is rarely encountered on the MCAT.)"

I think if you think about it more intuitively, it's easier. The only way for that combination of lenses to have a negative 1st object and positive 1st image as well as a positive 2nd object is for the 1st object to be in between the two lenses. I know...so confusing.. hope that helped.

 

[stevvo111]

Ok, it's really easy to get all mixed up with signs especially when they give you both types of lenses in one system. I think the confusion comes from the focal length sign. That has nothing to do with either side. The focal length is always negative for a diverging lens.

Remember for any lens, for the object, positive is on the side the light is coming from and negative is the side of the lens the light is going to. In this case for the object, positive is on the left and negative is on the right.
For the image, it's the opposite. Positive is on the side of lens light is going to (right) and negative where it's coming from (left). Always consider what's real and what's virtual for +/-.

So for D, the object distance with the negative sign is on the right side and since diverging lenses only produce virtual images, it will be on the same side giving the positive image distance. From here like you said, the +5 image distance is less than the focal length of the concave lens and also gives a virtual image back where lens 1 is giving a negative image distance. My kaplan book mentions: "Although the object of a single lens is on the V side, this does not make the object virtual. Objects are real, with a positive object distance, unless they are placed in certain multiple lens systems (a scenario which, by the way, is rarely encountered on the MCAT.)"

I think if you think about it more intuitively, it's easier. The only way for that combination of lenses to have a negative 1st object and positive 1st image as well as a positive 2nd object is for the 1st object to be in between the two lenses. I know...so confusing.. hope that helped.

Makes a lot of sense, I ended up figuring it out through logic a while ago, but definitely helpful to review this.

My only question is, who the heck ever puts an object between two lenses? This never happens, ever. (at least with most lens systems I've learned about/encountered)

 

[drspacheman]

For Trial E, why is the image formed at the same place the object is? I thought the image was always formed between the lens and the focal length?