# Thin Lens Formula Problems (1 Viewer)

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#### 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.

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#### Hootenanny

Staff Emeritus
Gold Member
f is the focul legnth of the lense.
do is the distance of the object from the centre of the lense.
di 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

Staff Emeritus
Gold Member
Alain12345 said:
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 :tongue:

#### Hootenanny

Staff Emeritus
Gold Member
Alain12345 said:
Sorry I didn't say it, but I attached a chart :tongue:
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

Staff Emeritus
Gold Member
Something that has just occured 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

Staff Emeritus
Gold Member
Well, you are given two variables (do and f). Just re-arrange the formula to find 1/di, then find the reciprical.

#### swt211

help

I have a test friday and I cant 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 dont know how to get height of object and image, thats all I am given..

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

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