# PhysicsOptics: Lenses

#### MermaidWonders

##### Active member
An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance and magnification?

Can someone please explain this with a diagram of the different possibilities (or, if not, just give a detailed explanation on how one should approach this)?

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance and magnification?

Can someone please explain this with a diagram of the different possibilities (or, if not, just give a detailed explanation on how one should approach this)?
Attempt?
Which formula applies?

#### MermaidWonders

##### Active member
Attempt?
Which formula applies?
Well, formulas $\frac{1}{{d}_{o}} + \frac{1}{{d}_{i}} = \frac{1}{f}$ and $m = -\frac{{d}_{i}}{{d}_{o}}$ probably apply.... I just have trouble finding ${d}_{o}$ and ${d}_{i}$ given that the object and its image are 2.4 m apart....

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Well, formulas $\frac{1}{{d}_{o}} + \frac{1}{{d}_{i}} = \frac{1}{f}$ and $m = -\frac{{d}_{i}}{{d}_{o}}$ probably apply.... I just have trouble finding ${d}_{o}$ and ${d}_{i}$ given that the object and its image are 2.4 m apart....
Good! It means that we have the set of equations:
\begin{cases} d_o+d_i=2.4 \text{ m} \\
f = 0.55 \text{ m}\\
\frac 1{d_o} + \frac 1 {d_i} = \frac 1f \\
m=\left| \frac{d_i}{d_o}\right|
\end{cases}
Can you solve it? #### MermaidWonders

##### Active member
Good! It means that we have the set of equations:
\begin{cases} d_o+d_i=2.4 \text{ m} \\
f = 0.55 \text{ m}\\
\frac 1{d_o} + \frac 1 {d_i} = \frac 1f \\
m=\left| \frac{d_i}{d_o}\right|
\end{cases}
Can you solve it? I keep thinking that ${d}_{o} - {d}_{i}$ is 2.4 m.... I didn't know they add up to be 2.4 m.... Why?

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
I keep thinking that ${d}_{o} - {d}_{i}$ is 2.4 m.... I didn't know they add up to be 2.4 m.... Why?
Normally the object and image are at different sides of the lense, meaning their distances add up.
However, it is possible that we have a virtual image that is on the same side of the lense as the object.
In that case we treat the image distance as a negative distance.
Either way, to do the math, we treat them as adding up to a total distance between object and image.

#### MermaidWonders

##### Active member
Normally the object and image are at different sides of the lense, meaning their distances add up.
However, it is possible that we have a virtual image that is on the same side of the lense as the object.
In that case we treat the image distance as a negative distance.
Either way, to do the math, we treat them as adding up to a total distance between object and image.
Oh, I see. I'll try that. 