Final Image position for a setup with two lenses

[nish95]

Homework Statement

Focal length of convex lens is 20 cm and of concave lens is 40 cm. Find position of final image.

Relevant Equations

Lens formula (1/v - 1/u = 1/f)

First image is an object for the concave lens so won't +ve direction change from right to left?! In that case, object distance will be -ve (from concave lens towards right side). Any ideas? Solution in the book takes first image's distance to be +ve. See attached ray diagram for clarification. Thanks in advance!

 

[ehild]

Homework Statement:: Focal length of convex lens is 20 cm and of concave lens is 40 cm. Find position of final image.
Relevant Equations:: Lens formula (1/v - 1/u = 1/f)

View attachment 268761

First image is an object for the concave lens so won't +ve direction change from right to left?! In that case, object distance will be -ve (from concave lens towards right side). Any ideas? Solution in the book takes first image's distance to be +ve. See attached ray diagram for clarification. Thanks in advance!

Foe the concave lens, the object distance is negative (virtual object) : -1/20-1!v1=-1/40

 

[TSny]

Relevant Equations:: Lens formula (1/v - 1/u = 1/f)

What sign conventions are you using for v and u? Apparently, u is the object distance and v is the image distance.

 

[hutchphd]

You need to learn how to draw a proper ray trace. Hopefully your course will require that.

 

[nish95]

I am using the convention that distances measured from optical centers are +ve in the direction of propagation of the ray and -ve if they both are opposite.

 

[nish95]

You need to learn how to draw a proper ray trace. Hopefully your course will require that.

Isn't the diagram correct? I mean rays coming from the virtual object will diverge because of the concave lens right?

 

[nish95]

Foe the concave lens, the object distance is negative (virtual object) : -1/20-1!v1=-1/40

Yes, that's what I did but the book has taken it to be +ve.

 

[hutchphd]

Have you been taught how draw a ray diagram for a thin lens? What you present is not adequate. You start with a finite object (often a "fat" arrow) and then proceed...look it up.

 

[nish95]

Have you been taught how draw a ray diagram for a thin lens? What you present is not adequate. You start with a finite object (often a "fat" arrow) and then proceed...look it up.

I get what you are saying. But that doesn't change things. I mean the concave lens produces the final image nearer to itself than the first image made by the convex lens. So, magnitude of distance of final image should be less than 20 cm.

 

[TSny]

I am using the convention that distances measured from optical centers are +ve in the direction of propagation of the ray and -ve if they both are opposite.

OK, good. The image of the first lens is to the right of the second lens. So, for the second lens, should u = + 20 cm or should u = -20 cm?

 

[nish95]

OK, good. The image of the first lens is to the right of the second lens. So, for the second lens, should u = + 20 cm or should u = -20 cm?

It is to the right of the second lens. But rays have to travel backwards, that is towards the concave lens to form the final image. In that case my positive direction is right to left and distance is being measured from the optical center of the concave lens to the first image (left to right). Thus I am taking u = -20 cm, cause they are opposite.

 

[TSny]

But rays have to travel backwards, that is towards the concave lens to form the final image.

The rays travel through the concave lens from left to right to form the final image. So, for the second lens, the rays are still traveling left to right.

The image of the first lens is to the right of the second lens. This image of the first lens is the object for the second lens. So, the distance measured from the concave lens to the object of the concave lens is in the same direction as the propagation of the rays through the lens.

 

[TSny]

Look at figure 4 here . Note how the rays are always traveling left to right. There aren't any rays traveling from right to left.

 

[hutchphd]

It is to the right of the second lens. But rays have to travel backwards, that is towards the concave lens to form the final image. In that case my positive direction is right to left and distance is being measured from the optical center of the concave lens to the first image (left to right). Thus I am taking u = -20 cm, cause they are opposite.

I don't understand what you have said. Here's my method:
Light travels from left to right. Objects are positive to left of lens. Images and focus are positive to the right.

You need to do the diverging lens with the virtual object again. Light entering parallel from the left will appear to diverge from front focus. The real image will be at +40 (right of concave lens).

 

[ehild]

Look at figure 4 here . Note how the rays are always traveling left to right. There aren't any rays traveling from right to left.

Edit: removed

 

[TSny]

No, you did not do that.
View attachment 268777
what are v and v_I ? Are not v the object distance and v_I the image distance?
-20 cm is the object distance of the concave lens.

I think the OP is using a non-standard sign convention (non-standard to me, anyway).

In the equation $\large \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, $u$ is the object distance and $v$ is the image distance.

Also, for this equation, real objects have negative object distance while virtual objects have positive object distance! Note, for example, that the OP correctly used an object distance $u = - 30$ cm for the real object of the convex lens.

 

[TSny]

Yes, but the paths of the light rays are reversible. In reality, the rays travel from left to right, but you get the final image of the concave lens at the same position as if a real object was placed on the right of the lens, from where the light rays traveled from right to left.

If a real object is placed 20 cm to the right of the concave lens, then you get a virtual image 13 `1/3 cm to the right of the concave lens. But, the actual final image of this system is a real image located more than 20 cm to the right of the concave lens.

 

[ehild]

If a real object is placed 20 cm to the right of the concave lens, then you get a virtual image 13 `1/3 cm to the right of the concave lens. But, the actual final image of this system is a real image located more than 20 cm to the right of the concave lens.

You are right, I was confused by the signs.

 

[TSny]

I was confused by the signs.

Me too.

 

[hutchphd]

That's why I always do the blasted ray trace. I can't keep the signs straight in Newton's Laws!

 

[TSny]

@nish95

If you use the equation $\large \frac 1 v - \frac 1 u = \frac 1 f$, then you can get the correct sign for the object distance $u$ as follows:

If the object is on the same side of the lens as the side where the light exits the lens, then $u$ is positive.
If the object is on the same side of the lens as the side where the light enters the lens, then $u$ is negative.

Consider the convex lens in your system. The light travels from left to right through the lens. So, the light enters the lens on the left side and exits on the right side. The object is on the left side. So, the object is on the same side of the lens as the side where the light enters the lens. So, $u$ is negative. ($u = -30$ cm for your problem.)

Consider the concave lens. The light still travels from left to right through this lens. However, the object of this lens is now on the right side of the lens. That is, the object is on the side of the lens where the light exits the lens. So, $u$ is positive for this case.

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Almost all textbooks that I've seen write the equation as $\large \frac 1 v + \frac 1 u = \frac 1 f$ with both terms on the left written with a positive sign. Then the sign convention for $u$ is opposite to the sign convention given above.

 

[hutchphd]

Almost all textbooks that I've seen write the equation as $\large \frac 1 v + \frac 1 u = \frac 1 f$ with both terms on the left written with a positive sign. Then the sign convention for $u$ is opposite to the sign convention given above.

Me, too.
And just to reiterate positive values for object on source side and focus and image on "not source" side for a lens.

 

[TSny]

And just to reiterate positive values for object on source side

Yes, if the equation is written as $\frac 1 v + \frac 1 u = \frac 1 f$.

 

[TSny]

Doing some web searching, I found that there exists "Cartesian sign conventions" for lenses. This is what the OP is using. For example, see this video.

 

[hutchphd]

Makes my head hurt. I'll just draw a little ray trace. I have no spare memory neurons at this point ...

 

[nish95]

@nish95

If you use the equation $\large \frac 1 v - \frac 1 u = \frac 1 f$, then you can get the correct sign for the object distance $u$ as follows:

If the object is on the same side of the lens as the side where the light exits the lens, then $u$ is positive.
If the object is on the same side of the lens as the side where the light enters the lens, then $u$ is negative.

Consider the convex lens in your system. The light travels from left to right through the lens. So, the light enters the lens on the left side and exits on the right side. The object is on the left side. So, the object is on the same side of the lens as the side where the light enters the lens. So, $u$ is negative. ($u = -30$ cm for your problem.)

Consider the concave lens. The light still travels from left to right through this lens. However, the object of this lens is now on the right side of the lens. That is, the object is on the side of the lens where the light exits the lens. So, $u$ is positive for this case.

--------

Almost all textbooks that I've seen write the equation as $\large \frac 1 v + \frac 1 u = \frac 1 f$ with both terms on the left written with a positive sign. Then the sign convention for $u$ is opposite to the sign convention given above.

I find the convention that you are using really simple. Don't know why my book has to complicate things!

Anyways, thanks for all the help. @TSny @hutchphd @ehild

Btw, what does OP stand for ? Also, @TSny which textbooks do you guys use for high school physics?

 

[TSny]

I find the convention that you are using really simple. Don't know why my book has to complicate things!

Anyways, thanks for all the help. @TSny @hutchphd @ehild

Btw, what does OP stand for ? Also, @TSny which textbooks do you guys use for high school physics?

You are welcome. I sugggest that you stick with the sign conventions that you are using in your course for now. I don't have any recommendations for textbooks at the high school level.

Enjoy your study of physics.