Distance between the object and the lens

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A diverging lens with a focal length of 6 mm produces an image with a magnification of 1/2, prompting the question of the distance between the object and the lens. Participants discuss relevant equations, including the magnification formula m = -di/do and the lens formula 1/do + 1/di = 1/f. It is noted that both equations can be used to solve for the object distance (do) and image distance (di). By rearranging the magnification formula, one can express di in terms of do and substitute it into the lens formula to find the solution. This approach effectively allows for determining the required distance between the object and the lens.
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Homework Statement



A diverging lens of 6 mm focal length produces an image of magnification m = 1/2. What
is the distance between the object and the lens?
 
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Welcome to PF!

Libra_girl said:

Homework Statement



A diverging lens of 6 mm focal length produces an image of magnification m = 1/2. What
is the distance between the object and the lens?

Hi Libra_girl ! Welcome to PF! :smile:

Hint: what equations do you know that relate the size of the object to the size of the image? :smile:
 
tiny-tim said:
Hi Libra_girl ! Welcome to PF! :smile:

Hint: what equations do you know that relate the size of the object to the size of the image? :smile:


Is it the equation: do/di = m or 1/do + 1/di = 1/f ?
 
Libra_girl said:
Is it the equation: do/di = m or 1/do + 1/di = 1/f ?

Both equations! :smile:

Except isn't it m = -di/do? :rolleyes:
 
tiny-tim said:
Both equations! :smile:

Except isn't it m = -di/do? :rolleyes:


yeah it is sorry :smile:
 
Last edited:
Libra_girl said:
yeah it is sorry :smile:

tiny-tim said:
Both equations! :smile:

Except isn't it m = -di/do? :rolleyes:

Ok so I know f and I know m, now I am solving for do, but I don't know the di?
 
Hi Libra girl,

You now have two equations in two unknowns. You can solve the magnification formula for one of the variables and plug it into the other formula. What does that give?
 

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