Calculate the focal length of a thin convex lens

[Lotto]

Homework Statement

The thin, circular convex lens is placed in an opaque bracket perpendicular to the optical axis. Behind the lens is build a shade perpendicular to the optical axis at a distance $l$. By placing a spot light source in the subject focus of the lens, a circle of light with a diameter of $D$ is formed on the shade. Moving the point source within twice the distance of the lens creates a circle of a diameter $d$. Detrmine a focal length of the lens.

Relevant Equations

$\frac{1}{2f}+\frac{1}{L}=\frac 1f$

My picture is here:

The diameter $D$ is also a diameter of the lens.

If there was no shade, the image of the moved source would be by using $\frac{1}{2f}+\frac{1}{L}=\frac 1f$ at a distance $L=2f$. So by using a triangle similarity we get

$\frac{D/2}{L}=\frac{d/2}{L-l}$,

so $f=\frac{Dl}{2(D-d)}$.

Is it correct?

 

[BvU]

Homework Statement: The thin, circular convex lens is placed in an opaque bracket perpendicular to the optical axis. Behind the lens is build a shade perpendicular to the optical axis at a distance $l$. By placing a spot light source in the subject focus of the lens, a circle of light with a diameter of $D$ is formed on the shade. Moving the point source within twice the distance of the lens creates a circle of a diameter $d$. Detrmine a focal length of the lens.
Relevant Equations: $\frac{1}{2f}+\frac{1}{L}=\frac 1f$

My picture is here:

View attachment 335845
The diameter $D$ is also a diameter of the lens.

If there was no shade, the image of the moved source would be by using $\frac{1}{2f}+\frac{1}{L}=\frac 1f$ at a distance $L=2f$. So by using a triangle similarity we get

$\frac{D/2}{L}=\frac{d/2}{L-l}$,

so $f=\frac{Dl}{2(D-d)}$.

Is it correct?

Hi,

I have heard of ${1\over o}+{1\over i} = {1\over f}$ but your $\frac{1}{2f}+\frac{1}{L}=\frac 1f$ has me puzzled. Basically you write that an image of a point source is formed at a certain distance. But in your problem statement there is no image, just a bright disk.
In the first case the point source is at $f$ and the image at infinity. On the screen is a bright disk with diameter D. Independent of the screen position.

The second case has the point source at $2f$ and the image at $2f$. Since we don't know where the screen is, I don't see how the diameter $d$ of the bright disk can help determine $f$.
What is $l'$ ?
Did it occur to you that the screen can also be furtrher away than $2f$ from the lens ?

The problem statement has me puzzled: An opaque bracket ? A shade ?

$\$

 

[Lotto]

Hi,

I have heard of ${1\over o}+{1\over i} = {1\over f}$ but your $\frac{1}{2f}+\frac{1}{L}=\frac 1f$ has me puzzled. Basically you write that an image of a point source is formed at a certain distance. But in your problem statement there is no image, just a bright disk.
In the first case the point source is at $f$ and the image at infinity. On the screen is a bright disk with diameter D. Independent of the screen position.

The second case has the point source at $2f$ and the image at $2f$. Since we don't know where the screen is, I don't see how the diameter $d$ of the bright disk can help determine $f$.
What is $l'$ ?
Did it occur to you that the screen can also be furtrher away than $2f$ from the lens ?

The problem statement has me puzzled: An opaque bracket ? A shade ?

$\$

Well, we just have a point source of light shining all around and we let it shine through a thin lens. Since the lens has a circular shape, the image is also circular, I think that it has the same diameter $D$ as the image. The image occurs on a opaque shade. That $L$ is an "imagenary" distance of the image if there was no shade (when the distance from the lens is $2f$).

There is no $l'$, that is a comma at the end of the equation.

 

[berkeman]

Since the lens has a circular shape, the image is also circular

Um, the shape of the image depends on the shape of the object, not the shape of the lens, right?

https://study.com/learn/lesson/thin-lens-equation.html

 

[Lotto]

Um, the shape of the image depends on the shape of the object, not the shape of the lens, right?

View attachment 335849
https://study.com/learn/lesson/thin-lens-equation.html

Yes, but this lens is thin and its cross-section is circular. And when the point source located at the focus shines all around, the light goes through the whole lens. And since the source is at the focus, refracted beams are parallel to the optical axis, creating a circle having the same diameter as the lens. That is how I understand the problem.

I don't know if I described the assignment well.

 

[berkeman]

Yes, but this lens is thin and its cross-section is circular. And when the point source located at the focus shines all around, the light goes through the whole lens. And since the source is at the focus, refracted beams are parallel to the optical axis, creating a circle having the same diameter as the lens. That is how I understand the problem.

Ah, that helps. So there is no image, since the object is placed one focal length away from the lens.

 

[berkeman]

Except the problem statement says to put the point source at the focal length of the lens, and then to determine the focal length of the lens? Sorry, I'm still confused...

 

[Lotto]

Except the problem statement says to put the point source at the focal length of the lens, and then to determine the focal length of the lens? Sorry, I'm still confused...

We don't know that $f$, that is our aim to determine. It is crucial that when we place the object at $2f$ from the lens, the diameter decreases to $d$. So then I used a triangle siimilarity to determine the focal lenght.

 

[BvU]

Well, we just have a point source of light shining all around and we let it shine through a thin lens. Since the lens has a circular shape, the image is also circular, I think that it has the same diameter $D$ as the image.

In case 1 we have the point source at the focal point of the lens. The exit beam is parallel and forms a bright disk on the screen, independent of the distance between screen and lens.

The image occurs on a opaque shade. That $L$ is an "imaginary" distance of the image if there was no shade (when the distance from the lens is $2f$).

I translate 'opaque shade'= screen. I agree $L= 2f$

There is no $l'$, that is a comma at the end of the equation.

So what is $l$ ?

$\$

 

[Lotto]

Aaah, I am so sorry. I have forgotten to say that the distance of the lens from the shade is $l$.

 

[Lotto]

No, wait, I have said that. Sorry again.

 

[BvU]

Is the value of $l$ known ?

 

[Lotto]

Is the value of $l$ known ?

Yes, we know it.

 

[BvU]

Ok, that means there is enough information to solve, but do you agree there are two possibilities ?

$\$

 

[Lotto]

Ok, that means there is enough information to solve, but do you agree there are two possibilities ?

$\$

Not really. What two possibilities do you mean?

 

[BvU]

the diameter of the bright disk is zero at $2f$; diameter $d>0$ can be either side of that:

 

[Lotto]

Ah, right. So the second solution is $f=\frac{Dl}{2(D+d)}$?

 

[BvU]

That's what I got, too. Well done !

$\$

 

[kuruman]

Ah, right. So the second solution is $f=\frac{Dl}{2(D+d)}$?

I am not so sure about this second case calculation. I do agree, however, with the first case calculation when the screen is between the lens and the image $$f=\frac{DL}{2(D-d)}.\tag{1}$$ (I use $L$ instead of $l$ for the lens-screen distance to distinguish it from $1$ in the figure below.)

Here is my reasoning.
First, the focal length of the lens cannot depend on where one places the screen. It is fixed.

Second, when the screen is placed at the position of the lens, we have $L=0$ and $d=D$. Equation (1) for the focal length becomes $f=\frac{0}{0}$, undetermined as one would expect. This is not the case with the second calculation which predicts that $f=0.$

Third, the figure below, drawn to scale, shows the geometry. The source is at S, the image at P and the screen between the lens and the image. If the screen is placed to the right of P, the red triangle should be flipped twice, once about the dashed vertical line going through P and then about the optical axis to keep line AP straight after it crosses the optical axis. This double reflection is an inversion about P that does not affect the proportionalities of the similar triangles that went into the first calculation. Therefore, the expression for $f$ should be independent of screen placement.

It is much easier to use analytic geometry to do this as I indicated in an earlier similar post. There is only one line that matters and that is AP. Taking the origin of coordinates at P, its equation is $$y=-\frac{\frac{1}{2}D}{2f}x=-\frac{D}{4f}x.\tag{2}$$ Case I: Screen between lens and image.
The coordinates of the intersection point of ray AP and the screen are $\{-(2f-L),\frac{1}{2}d\}.$ Substitute in equation (2) to get $\dfrac{1}{2}d=+\dfrac{D}{4f}(2f-L).$ Solve for $f$.

Case II: Image between lens and screen.
The coordinates of the intersection point of ray AP and the screen are $\{(2f-L),-\frac{1}{2}d\}.$ Substitute in equation (2) to get $-\dfrac{1}{2}d=-\dfrac{D}{4f}(2f-L).$ Solve for $f$.

 

[BvU]

Case $l_1<2f \qquad \frac d D = \frac{2f-l_1}{2f}$ done.

Case $l_2>2f\qquad \frac d D = \frac{l_2-2f}{2f}$

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