Optics Question - How to manipulate thin lens equation?

[HinduHammer]

Can you please explain how to manipulate the thin lens equation? I am having trouble manipulating the thin lens equation and don't want to rely on conceptual understanding in case the questions asks me to find the exact value. Thanks!

A person attempts to take a picture of an object, using a lens with a focal length f. If the person then makes a second attempt, using a lens with a focal length 2f and doubles the distance between the lens and the object, the distance to the image during the second attempt will:

a. be reduced to 1/4 of the original
b. be reduced to 1/2 of the original
c. be double that of the original
d. be quadruple that of the original

---

Members don't see this ad.

 

[mysturi0us]

I think the best way to answer this question without doing anything difficult is to actually plug in numbers. and then increasing the numbers in the second attempt according to the question. Then compare the 2 answers, which for me is B. I personally thought changing it to decimal was easier, but that's just my preference. And keep your numbers simple, so you can do the math quick in your head.

 

[sps27]

v = (o* f / o - f) . This is a more convenient way to represent I/f = 1/o + 1/v. Just plug and play. I think the ans is C. Correct me if I am wrong...........

 

[BerkReviewTeach]

v = (o* f / o - f) . This is a more convenient way to represent I/f = 1/o + 1/v. Just plug and play. I think the ans is C. Correct me if I am wrong...........

You are correct, and apparently a fan of the BR method.

 

[grey_wind]

can someone explain this math of this? I'm lost...

 

[deleted388502]

Okay I don't generally answer questions on here so I hope this is the right idea lol. I think this should be but if it's wrong someone don't hesitate to correct me if I'm wrong. Let me know if I can clear anything up!


Posted using SDN Mobile

 

[sillyjoe]

View attachment 183907

Okay I don't generally answer questions on here so I hope this is the right idea lol. I think this should be but if it's wrong someone don't hesitate to correct me if I'm wrong. Let me know if I can clear anything up!


Posted using SDN Mobile

Whoopin out the notes. Impressive. Damn good handwriting.

 

[grey_wind]

Thank you so much!

 

[HinduHammer]

The correct answer was C, thanks for the responses!

v = (o* f / o - f) . This is a more convenient way to represent I/f = 1/o + 1/v. Just plug and play. I think the ans is C. Correct me if I am wrong...........

You are correct, and apparently a fan of the BR method.

What is v = (o*f/o-f) -- what does this represent, and would you mind explaining why this is easier? thanks!

 

[grey_wind]

The correct answer was C, thanks for the responses!





What is v = (o*f/o-f) -- what does this represent, and would you mind explaining why this is easier? thanks!

I just checked some of my TBR materials. Its image distance = (object distance * focal length)/(object distance - focal length)

I think its 'easier' because you aren't dealing with reciprocals, at least IMO thats why haha

 

[HinduHammer]

I just checked some of my TBR materials. Its image distance = (object distance * focal length)/(object distance - focal length)

I think its 'easier' because you aren't dealing with reciprocals, at least IMO thats why haha

Thanks alot for the quick reply! Is this valid for both mirrors and lenses?

 

[TulaneUnderdog]

Just to add more conceptual depth, the notes from that picture, although perfect for giving you the right answer, is conceptually wrong because o must be greater than f in terms of distance so that the camera lens can converge the rays onto the other side of the lens where the image should be in order for the picture to be taken. Otherwise, the image will be indiscernible (if o=f) or impossible to project (o<f) because an image that is upright and virtual (o<f) is unprojectable since the image would be located on the virtual side of the lens which is the same side as the object.

 

[deleted388502]

Just to add more conceptual depth, the notes from that picture, although perfect for giving you the right answer, is conceptually wrong because o must be greater than f in terms of distance so that the camera lens can converge the rays onto the other side of the lens where the image should be in order for the picture to be taken. Otherwise, the image will be indiscernible (if o=f) or impossible to project (o<f) because an image that is upright and virtual (o<f) is unprojectable since the image would be located on the virtual side of the lens which is the same side as the object.

Good catch. I picked random numbers to make the math easier haha, but you are indeed correct

 

[TulaneUnderdog]

Good catch. I picked random numbers to make the math easier haha, but you are indeed correct

Thank you. Sorry, that I may have sounded fastidious since I don't mean to, especially because honestly with the MCAT, plugging random numbers is the most time efficient way and will save your more points than analyzing and digesting the true meaning behind these topics. That's probably the saddest part considering this test is supposed to validate how prepared you are to be a doctor and it is endorsing less on critically thinking and more on those taking shortcuts.