Thread: Magnification question - beginner student

1. Hello,

I'd be very grateful if anyone could help me.

I'm a student, studying foundation Biology alone by distance learning. I've just started learning the very basics of microscopy. I've so far learned the equation required to calculate the "magnification" of an object: ("magnification = image size/object size"). I've also learned the equations for calculating the "size of an object" when given the "size of an image" and the "magnification of an object" ("size of object = size of image/magnification") and the "size of an image" when given the "size of an object" and "magnification" ("size of image = size of object x magnification").
The book I'm studying from has an exercise on magnification, where I must answer a number of questions. Each question must initially be interpreted, and then the answer calculated using one of the above equations (whichever one I think is appropriate).

I've been able to answer several of them without any problems, however one of them has confused me.

This is:

"A publisher wants to put a photograph of a weightlifter in her book. The athlete is 1.85 m tall. The photograph needs to be shrunk 50 times. What is the height of the weightlifter in the photograph?"

How would I interpret this question and what steps would I take to calculate the answer? Which of the 3 equations would I use ("magnification = image size/object size", "object size = image size/magnification" or "image size = object size x magnification")? My initial thought was that I should be calculating "image size" and using that equation, as my interpretation of the question was that the "1.85 m" value was the object size and the "50" value was something to do with the magnification. I, hence, thought that a value for image size needed to be found.
However, doing the equation "image size = object size x magnification" ("image size = 1.85 m x 50") didn't get me anything like the correct answer (I have a copy of all the answers to the questions in the back of my exercise book, albeit without any helpful explanations on the correct way to calculate them). I have obviously made a big mistake, which I'm sure has a lot to do with how I interpreted the question in the first place.
I wonder whether I am perhaps correct to use the equation "image size = object size x magnification" but MUST FIRST do something else with the "1.85" and "50" values (work out something else/do another calculation first). The 3 equations are the only material I've been given to help me answer this question. I'm unsure what to do with them.

Could anybody please help me out by talking me though how I should have initially interpreted the question and then give me a step-by-step in working it out? What calculations must be made?

The answer to the question at the back of my exercise book is: 37 mm. How do I calculate and get to this?

I'm also aware that these equations require you to convert all the values into the same units using standard form. Must I do anything like that here?

I would be incredibly appreciative if anyone can help. I know this is very basic stuff for the forum, which I know is generally focused on much more in depth Biology. It's just really worrying me that I can't work out this question. I very much want to grasp this so that I can answer other similar questions with ease in the future (especially in exams). I have dyscalculia (a sort of math dyslexia) which seriously impedes me from grasping simple mathematical concepts and answering questions. This is the reason, I believe, I am struggling so much with this question! I'm a self-study student, so I don't have anyone to turn to for help with this microscope topic!

Thank you very much for your time and help.  2.

3. If you magnify something 50 times then the magnification is 50, but if you shrink it 50 times then the magnification is 1/50.  4. Hello,

Your initial thought was absolutely right, you should be calculating the image size and using that equation. As you alluded to, it's easy to tell that the '50' part relates to magnification, and that the athlete is the object, in this case. By process of elimination, you can interpret the question as asking you to find out what the image size is. The only mistake was that the object size was multiplied by the wrong number. The object size was multiplied by +50, which would be correct if the image was being magnified by 50 times. However, in this case the image is being shrunk by 50 times. Therefore, the object size needs to be multiplied by -50. There are different ways of working this out; all methods give the same answer:

Using the Equation

Image size = Object size x Magnification

Image size = 1.85m x -50

Image size = 0.037m

Whilst this is correct, the units are still in metres, rather than millimetres. In order to convert into millimetres, remember that there are 100cm in 1m. Remember also that there are 10mm in 1cm. So if you wanted to find the answer in cm:

Image size = 0.037m x 100 = 3.7cm

And then to convert into mm:

Image size = 3.7cm x 10 = 37mm.

You can choose to convert the units at the beginning if you prefer, sometimes that is easier.

Intuitively, you may see that the Image size in this case is going to be 50 times smaller than the Object size. So, you can divide the Object size by 50, to give the Image size (if you don't want to use the equation):

Object size = 1.85m = 185cm = 1850mm

Image size = 1850mm/50 = 37mm.

I hope that this helps, let us know how you get on.

Tri.   5. Therefore, the object size needs to be multiplied by -50. There are different ways of working this out; all methods give the same answer:

Using the Equation

Image size = Object size x Magnification

Image size = 1.85m x -50

Image size = 0.037m

Please ignore this part, it's completely wrong! Sorry!  6. The only proper way to do this is would to use a reference.

The only reliable reference is the person's length.

I would make a scale bar of that.

Then you can magnify whatever you want. That's what you do in science.

And the answer to the question is of course.

"A publisher wants to put a photograph of a weightlifter in her book. The athlete is 1.85 m tall. The photograph needs to be shrunk 50 times. What is the height of the weightlifter in the photograph?"

1.85m. The athlete's height doesn't change because his picture was taken. :P  7. I cant thank you all enough for your responses. Your help has made such a great deal of difference to me, and is truly valuable and appreciated.
Tridimity, thank you so much for taking your time to write such an excellently informative and articulate reply. Youve explained everything so clearly and considerately, and have really put me into perspective with this! I get so muddled with math of all sorts, which can be very worrying and cause a lot of difficulty progressing with many topics in Biology.

Am I right in understanding that, in addition to the 3 equations (calculating magnification, object size, and image size) I mentioned in my post, there is actually another equation which one would use when working out image size in cases where the object is shrunk, rather than magnified:
image size = object size/the number the image is to be shrunk by (in this case 50)?'
Would the number 50 still be referred to as magnification?
If this is the case, would the equation for working out image size when the object is shrunk be image size = object size/magnification?

Would it be correct to jump straight into this equation for a question like this (where the object is shrunk), and then convert the units at the end?
Harold14370 also mentioned that magnification is 1/50, when the object is shrunk. With this in mind, would it also be ok to initially divide 1 by 50 (= 0.02) and then go ahead and do the normal image size equation image size = object size x magnification (1.85m x 0.02) to get 0.037m?
Is this appropriate? Is one way preferable to use over the other (e.g. the instant equation: image size = object size/the number the image is to be shrunk by)?

Also, how would I know which unit is best to answer in? (Sorry if this seems a silly question!) The answer in my book to the question "A publisher wants to put a photograph of a weightlifter in her book. The athlete is 1.85 m tall. The photograph needs to be shrunk 50 times. What is the height of the weightlifter in the photograph? is in millimeters, but Im not sure I would have known the right unit to convert to and answer in had I not seen the book answer first! Is there a way of knowing?

Thank you very much again for your support. It is very kind of you to take your time to help me!

Spuriousmonkey, you really made me laugh! I get so confused with math, that, before all your help, that is honestly exactly the kind of answer I would have given! :P This has just reminded me of when a friend of mine was once asked to do a stem and leaf diagram in a test and drew a picture of a plant as a joke because they didnt know they were doing!  8. Hullo again,

Yes, there are other ways to work through this problem, besides your 3 equations. As mentioned, you can just divide the object size by 50, in this case - I don't know how to explain this any better, it is just the intuitive thing to do - because the image size is going to be 50 times smaller than the object size. So, technically, you could use the equation:

image size = object size/the number the image is to be shrunk by (in this case 50)?'

But the number 50 would not be referred to as the magnification, in this case. If 50 were the magnification, the image size would be 50 times larger than the object size. So, the following equation would be incorrect:

image size = object size/magnification?

Would it be correct to jump straight into this equation for a question like this (where the object is shrunk), and then convert the units at the end?
Harold14370 also mentioned that magnification is 1/50, when the object is shrunk. With this in mind, would it also be ok to initially divide 1 by 50 (= 0.02) and then go ahead and do the normal image size equation image size = object size x magnification (1.85m x 0.02) to get 0.037m?
(*)

Yes, this is an excellent way of doing it. The magnification in this case, is (1/50).

In every day life, I guess it wouldn't really matter which method you used, so long as you got the right answer. However, depending on which level you are at in your education, examiners tend to prefer that you show all of your workings (often 'method marks' are granted for using the correct equation or generally for approaching the question in a logical way). In the interests of your passing exams then, it would probably be best to stick to the original equation with which you were provided(*).

In terms of knowing which units to answer in, ordinarily they will tell you which units they would like you to answer in, especially if the final answer is worth marks in an exam. If they don't explicitly tell you which units to answer in, the best I can recommend is to use common sense - if they hypothetically ask you to calculate a fictional person's commuting time, do not answer in milliseconds!

Best wishes,
T.
:wink:  9. That is an inexcusably misleading and confusing statement of the problem.

It's one step away from saying the picture needs to be 50 times as less, at which point you would be better off teaching math in a foreign language.

Meanwhile, it engenders stuff like this:
it is just the intuitive thing to do - because the image size is going to be 50 times smaller than the object size.
.  10. I think your confusion comes from the phrase "shrunk 50 times" which is awkward and a mathematically crude way of stating the problem. When we multiply something by 50 it is like adding the number 50 times. When we divide by 50, we are not doing anything "50 times," we are simply dividing by 50.

As far as the units are concerned, it would have been just as correct to state the answer as 0.037 meters. However, if you are measuring the size of a photo in a book, you would more likely be using a ruler marked in millimeters, so that would be a more convenient unit of measure for the purpose.  11. Dear iceaura,

How best could the problem be stated?

it is just the intuitive thing to do - because the image size is going to be 50 times smaller than the object size.

What was so wrong about saying this?

Thanks  12. If you have a solution at 50M concentration and you need it to be at 1M, then the original solution is 50 times more concentrated than you desire, so you do a 1 in 50 dilution. Isn't this size issue analogous?

And isn't the object size 50 times larger than the image size? That's just the inverse way of saying what I previously said.  13. I couldn't log on for a few days, but I'd just like to thank you very much for all your help. It is greatly appreciated and has made such a lot of difference! I've carefully taken everything into account and tested my understanding by having a go at some similar problems. I'm very pleased and relieved to find that, because of your help, I can now work through them with understanding and ease - a great difference from before! :-D  14. It's quite a big book though if they person in the book is 1.85m tall.  Posting Permissions
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