# Object-to-image distance

• Aelo
In summary, the lens image is 50 mm across and the overall object-to-image distance is 250 mm. The magnification is given as 10 mm/250 mm or .04.f

## Homework Statement

A 50 mm focal length thin lens in air is used to image a 250 mm x 250 mm object onto a 10 mm x 10 mm detector. What is the overall object-to-image distance (throw)? Besides the exact solution, can you think of a reasonable approximation?

## Homework Equations

Throw, T = l' - l
For a thin lens: (1/l') = (1/F) + (1/l)
For a thin lens in air: f = -F

## The Attempt at a Solution

Since f = 50 mm was given, F = -50 mm. I'm thinking that if I could find either l' or l, I could then use the second equation to find the remaining unknown, and then plug into the equation for T. I'm not sure how to use the given dimensions for the object and detector. I know of no equations that incorporate the area, and I can't determine what else they could be used to describe.

## Homework Statement

A 50 mm focal length thin lens in air is used to image a 250 mm x 250 mm object onto a 10 mm x 10 mm detector. What is the overall object-to-image distance (throw)? Besides the exact solution, can you think of a reasonable approximation?

## Homework Equations

Throw, T = l' - l
For a thin lens: (1/l') = (1/F) + (1/l)
For a thin lens in air: f = -F

## The Attempt at a Solution

Since f = 50 mm was given, F = -50 mm. I'm thinking that if I could find either l' or l, I could then use the second equation to find the remaining unknown, and then plug into the equation for T. I'm not sure how to use the given dimensions for the object and detector. I know of no equations that incorporate the area, and I can't determine what else they could be used to describe.

You get the magnification from the given sizes of object and image. The image is on the detector, so it is real. How are l and l' related ?

ehild

M = l' / l

Is it correct to think that the lateral magnification is the ratio between the horizontal length of the image and that of the object? If so, then M = 10 mm / 250 mm = .04

l' = l*M so 1/F = 1/(l*M) - 1/l = 1/(l*M) - M/(l*M) = (1 - M)/(l*M)
=> l = F*(1 - M)/M

M = l' / l

Is it correct to think that the lateral magnification is the ratio between the horizontal length of the image and that of the object? If so, then M = 10 mm / 250 mm = .04

take care of the sign of M.

ehild

Why would either 10 mm or 250 mm be negative? I understand l would be negative since it is measured from the lens to the object, so I agree that M should be negative, but I didn't think the dimensions of the object and image were given relative to anything. Is taking the ratio between horizontal lengths "one of those" formulas where I calculate the magnitude and have to come up with the sign independently?

We take the real image inverted, so its height negative.

ehild

• 1 person
Oh, it's the ratio of heights, not widths. Thank you very much for your help. :)

Oh, it's the ratio of heights, not widths. Thank you very much for your help. :)

You are welcome. ehild