Solving Thin Lens Formula Problems

[Alain12345]

I have to complete a chart using the thin lens forumla. Can someone please help me out? I know that the formula is 1/f= 1/do= 1/di, but I don't know how to use it.

Thanks.

 

[Hootenanny]

f is the focul legnth of the lense.
d_o is the distance of the object from the centre of the lense.
d_i is the distance of the image from the centre of the lense.

~H

 

[Alain12345]

yeah, I know that, but how do I find the uknown?

 

[Hootenanny]

yeah, I know that, but how do I find the uknown?

I don't know what is unknown (if that makes sense ). You haven't specified an unknown.

 

[Alain12345]

Sorry I didn't say it, but I attached a chart

 

[Hootenanny]

Sorry I didn't say it, but I attached a chart

I'm afraid I can't see your chart as it is yet to be approved by a mentor.

~H

 

[Alain12345]

oh okay then I'll just type it here

do di f
60 cm ? 20 cm
30 cm ? 15 cm
105 cm ? 70 cm
40 cm ? 20 cm
45 cm ? 15 cm
210 cm ? 70 cm
15 cm ? 20 cm​

 

[Hootenanny]

Something that has just occurred to me, you have written down the thin lense equation wrong. You written;

1/f= 1/do= 1/di

It is actually;

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}

Sorry I didnt spot that sooner.

~H

 

[Alain12345]

lol oops... I typed it out wrong, but I still don't know how to use it

 

[Hootenanny]

Well, you are given two variables (d_o and f). Just re-arrange the formula to find 1/d_i, then find the reciprical.

 

[swt211]

help

I have a test friday and I can't figure out how to get the rest of the question


I have figured out the the focal length is 4.0 , the object distance is 7.0 , the image distance is 9.3 and the magnification is -1.32X but i don't know how to get height of object and image, that's all I am given..

1/do + 1/di = 1/f , m = - di/do m=hi/ho