# Position of the object and image in this optics problem

• PSN03
In summary, the focal length distance of A is 2F. The image formed by the lens will be reflected by the mirror and the final image will be formed at A. This is because the image formed in the end is inverted and at the same position, making the distance 2F. However, tracing a ray from pin (at 2F) through an edge of the lens, to the mirror and back, it will not hit the pin if it is un-deflected by the lens on the way back up. The image will only be formed at A if the pin is at F. Therefore, it is necessary to consider the sign conventions and carefully trace the path of the rays to accurately determine the position of the image.
PSN03
Homework Statement
A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μl is put between the lens and the mirror, the pin has to be moved to A', such that OA' = 27 cm, to get its inverted real image at A' itself. The value of μl will be ?
Relevant Equations
1/f=(μ-1)(1/R-1/R')
I know how to solve the problem but the only thing that's bothering me is what is A?
According to me A is should be 2F ie 2 times the focal length distance. I thought of it like this:
1. First there will be an image formed due to the lens.
2. The image will be formed below the lens but the mirror will reflect the image and the final image will be formed at A again.
3. Since the image formed in the end is inverted and at the same position, hence the distance is 2F.

Am I thinking in the right way? Cause I am not able to get the right answer.

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PSN03 said:
According to me A is should be 2F ie 2 times the focal length distance. I thought of it like this:
1. First there will be an image formed due to the lens.
2. The image will be formed below the lens but the mirror will reflect the image and the final image will be formed at A again.
3. Since the image formed in the end is inverted and at the same position, hence the distance is 2F.

Am I thinking in the right way? Cause I am not able to get the right answer.
If you trace a ray from pin (at 2F) through an edge of the lens, to the mirror and back, where does it go? In particular, what happens on the second pass through the lens?

jbriggs444 said:
If you trace a ray from pin (at 2F) through an edge of the lens, to the mirror and back, where does it go? In particular, what happens on the second pass through the lens?
Won't it come back to 2F?

PSN03 said:
Won't it come back to 2F?
Did you trace it? When the ray leaves the lens on the way down, does it depart at an angle? When it arrives at the lens on the way back up, does it arrive at an angle?

If un-deflected by the lens on the way back up, would the ray hit the pin?
If deflected by the lens on the way back up, would the ray hit the pin?

If ray tracing is not doing it for you, the other way to go is to re-do the algebra taking extreme pains to get the sign conventions right. A virtual image on one side of the lens is not the same as a virtual image on the other.

PSN03
jbriggs444 said:
Did you trace it? When the ray leaves the lens on the way down, does it depart at an angle? When it arrives at the lens on the way back up, does it arrive at an angle?

If un-deflected by the lens on the way back up, would the ray hit the pin?
If deflected by the lens on the way back up, would the ray hit the pin?
What I thought was first take just the lens. Let this lens form an image at say B. This B will act as object for the mirror. The reflection of B comes back to A. This was my thinking.

PSN03 said:
What I thought was first take just the lens. Let this lens form an image at say B. This B will act as object for the mirror. The reflection of B comes back to A. This was my thinking.
You have three elements in the path: The lens, the mirror and the lens again.

Where is the image after the first pass through the lens?
Where is the image after the reflection? In particular, which side of the lens is it on?
Where is the image after the second pass through the lens, paying careful attention to sign conventions.

PSN03
jbriggs444 said:
Did you trace it? When the ray leaves the lens on the way down, does it depart at an angle? When it arrives at the lens on the way back up, does it arrive at an angle?

If un-deflected by the lens on the way back up, would the ray hit the pin?
If deflected by the lens on the way back up, would the ray hit the pin?
I know the point you are telling me is take the rays after passing through lens. These ray will strike the mirror perpendicularly. These perpendicular rays will come back to the lens and hence give the object at same place. But won't this image formed by upright and not inverted? Also what is wrong in my previous way of solving the problem?

PSN03 said:
I know the point you are telling me is take the rays after passing through lens. These ray will strike the mirror perpendicularly.
Not if the pin is at 2F they won't.

jbriggs444 said:
Not if the pin is at 2F they won't.
That will happen when pin is at F

PSN03 said:
That will happen when pin is at F
Right. So work the problem that way.

I had thought that we were trying to reason past:
PSN03 said:
According to me A is should be 2F ie 2 times the focal length distance

jbriggs444 said:
Right. So work the problem that way.

I had thought that we were trying to reason past:
Thanks a lot. I got my mistake. Good day and thanks for your time.

## 1. What is the difference between the position of an object and the image in an optics problem?

In an optics problem, the position of the object refers to the location of the object in relation to the lens or mirror being used. The image, on the other hand, is the location where the light rays converge after passing through the lens or reflecting off the mirror.

## 2. How is the position of an object determined in an optics problem?

The position of an object in an optics problem is determined by measuring the distance between the object and the lens or mirror, also known as the object distance. This distance is usually represented by the letter "u" in equations.

## 3. What factors can affect the position of an image in an optics problem?

The position of an image in an optics problem can be affected by several factors, including the type of lens or mirror being used, the distance between the object and the lens or mirror, and the angle at which the light rays hit the lens or mirror.

## 4. How can I calculate the position of an image in an optics problem?

The position of an image in an optics problem can be calculated using the thin lens equation: 1/u + 1/v = 1/f, where u is the object distance, v is the image distance, and f is the focal length of the lens or mirror. By rearranging this equation, you can solve for v, which will give you the position of the image.

## 5. Is there a difference between the position of an object and the position of the image in an optics problem?

Yes, there is a difference between the position of an object and the position of the image in an optics problem. The position of the object refers to the location of the object itself, while the position of the image refers to the location where the light rays converge after passing through the lens or reflecting off the mirror. These two positions are usually not the same, unless the object is placed at the focal point of the lens or mirror.

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