Hi/ho problem -- Lens magnification

[premed_love]

Homework Statement

An arrow h0 = 2.20 cm long is located 74.0 cm from a lens, which has a focal length

= 30.6 cm. If the arrow is perpendicular to the principal axis of the lens, as shown in the figure below, what is its lateral magnification, defined as hi/h0?

I am given: ho=2.20 cm. do 74.0 cm. focal length is 30.6 cm

Homework Equations

1/f=1/di=1/do

The Attempt at a Solution

I used 1/f=1/di=1/do= 1/30.6=1/di+1/74.0 I get 52.175 with this equation which is my Di.
Then, I hi/2.20 cm=74.0/52.175 =3.12 from there I have to do hi/ho 3.12/2.20=1.418

I am completely wrong! Help![/B]

 

[TSny]

Not completely wrong. At least you got a number. (If you had ended up with an elephant, then yes, completely wrong.)

Your calculation for d_i looks good to me, although when you wrote your formula for the lens equation you have too many equal signs.

The calculation of the magnification is incorrect. Double-check your formula that relates h_i and h_o to d_i and d_o. You don't actually need to find the image height. You should be able to relate the magnification directly to the object and image distances.

 

[premed_love]

I am still confused! Because if I relate it directly, then magnification is di/do therefore I get 1.418 which is still wrong

 

[TSny]

I am still confused! Because if I relate it directly, then magnification is di/do therefore I get 1.418 which is still wrong

What value did you use for d_i? For d_o?

 

[premed_love]

74.0/52.175

 

[TSny]

d_i goes on top. What was your value for d_i?

 

[premed_love]

I was given do in the problem which was 74.0 cm, you may reread the problem to see if I am interpreting that correctly. I found di to be 52.175 which you said my math was right.

 

[TSny]

Yes, I believe your value for d_i (52.175 cm) is correct. The magnification is $\frac{d_i}{d_o}$. So, what number goes on top?

 

[premed_love]

Yes, I believe your value for d_i (52.175 cm) is correct. The magnification is $\frac{d_i}{d_o}$. So, what number goes on top?

IN that case: it would be 52.175/74.0 = .70506 and that is wrong. Thank you for your help.

 

[TSny]

Are you supposed to include a - sign if the image is inverted?

 

[premed_love]

I was struggling with the problem. I showed you my work. However, if you are also confused then I can not do anything about that. You tell me how to do di/do which I have done. I am not sure whether your supposed to show a sign.

 

[TSny]

The usual convention is to define the height of the image as negative if the image is inverted. Then the magnification formula is m = -d_i/ d_o. However, I did not know what convention you use in your class. In your first post you calculated an image height as positive and in post #3 you did not include a - sign. So, I wasn't sure of the sign convention you are using . Did you check your notes from class or your textbook? Anyway, hopefully the discussion has been beneficial. If not, please accept my apology.

 

[premed_love]

The usual convention is to define the height of the image as negative if the image is inverted. Then the magnification formula is m = -d_i/ d_o. However, I did not know what convention you use in your class. In your first post you calculated an image height as positive and in post #3 you did not include a - sign. So, I wasn't sure of the sign convention you are using . Did you check your notes from class or your textbook? Anyway, hopefully the discussion has been beneficial. If not, please accept my apology.

Yes, I checked. I did my work correctly and used the negative sign three times! However, I was getting a wrong answer due to sig figs. That was frustrating because I never thought I had to be careful of sig figs!

 

[TSny]

Yes, that's frustrating. At least now you know to watch the sig figs.