Lens & Reflector: Final Idol (Optics)

[Const@ntine]

Homework Statement

https://attachment.outlook.office.net/owa/psaraskostas@hotmail.com/service.svc/s/GetFileAttachment?id=AQMkADAwATYwMAItYWVhZi1kMjdjLTAwAi0wMAoARgAAA5PfTv0SpppHr3yc003kVrgHAAH04A%2BJ7mRGmo%2BChKo3rFcAAAIBDAAAAAH04A%2BJ7mRGmo%2BChKo3rFcAAADhQYzEAAAAARIAEABF9QJlkOpsQ7K16ghu%2FR4q&X-OWA-CANARY=bC9TQBru00KIAJfDrF7wOUA-v2MS5tQYcRcIDvCJVDVpfrxruNXN3shXZtBC7JI_1CrNLme5I60.&token=eyJ0eXAiOiJKV1QiLCJhbGciOiJSUzI1NiIsIng1dCI6ImVuaDlCSnJWUFU1aWpWMXFqWmpWLWZMMmJjbyJ9.eyJ2ZXIiOiJFeGNoYW5nZS5DYWxsYmFjay5WMSIsImFwcGN0eHNlbmRlciI6Ik93YURvd25sb2FkQDg0ZGY5ZTdmLWU5ZjYtNDBhZi1iNDM1LWFhYWFhYWFhYWFhYSIsImFwcGN0eCI6IntcIm1zZXhjaHByb3RcIjpcIm93YVwiLFwicHJpbWFyeXNpZFwiOlwiUy0xLTI4MjctMzkzMjE2LTI5MzA3NTgyNjhcIixcInB1aWRcIjpcIjE2ODg4NTI3OTEwMjIyMDRcIixcIm9pZFwiOlwiMDAwNjAwMDAtYWVhZi1kMjdjLTAwMDAtMDAwMDAwMDAwMDAwXCIsXCJzY29wZVwiOlwiT3dhRG93bmxvYWRcIn0iLCJpc3MiOiIwMDAwMDAwMi0wMDAwLTBmZjEtY2UwMC0wMDAwMDAwMDAwMDBAODRkZjllN2YtZTlmNi00MGFmLWI0MzUtYWFhYWFhYWFhYWFhIiwiYXVkIjoiMDAwMDAwMDItMDAwMC0wZmYxLWNlMDAtMDAwMDAwMDAwMDAwL2F0dGFjaG1lbnQub3V0bG9vay5vZmZpY2UubmV0QDg0ZGY5ZTdmLWU5ZjYtNDBhZi1iNDM1LWFhYWFhYWFhYWFhYSIsImV4cCI6MTUwMzA0NTE5NiwibmJmIjoxNTAzMDQ0NTk2fQ.e0JhGUznvcXpY5TcsOa_djRRybR7Uqjm66-5hbtGqnWG4V6KcwaibG_1ssNOPvI4Af73Lryh1vIFiZvwlBhhYXIK3uqpb-4dcpFcMlhjymi3OxUlHc2Do7gz-vZRqbnN-FZV51vZpqfumhhhRorE6d1ndIoEjo4jexpbG8Q29Oq-9D-pwlUZPbSTlt4_7Vflyx19FoC0MJwbVId0LValKK3Qg4fXPsqed_wAk28cMKSN3IeNctKg5NBrvFgx-9mxmLrybMAbRL-AYZSuBLmV7hGmENa3oqdOGm3Z46fZZkDBvtBLK52ATd0n5aEzLOt0GMHxxpPCQYV4xFZOnAgScw&owa=outlook.live.com&isc=1&isImagePreview=True The lens and the reflector have a distance of d = 1.00 m between them. They have a focal length (f) of +80.0 cm & -50.0 cm, respectively. An object is situated at p = 1.00 m left of the lens.

(a) What's the position of the final idol, which is created by light that has gone through the lens two times?

(b) Is the idol inverted or not?

Homework Equations

1/p + 1/q = 1/f

Signs:

>For Lenses:

p: + when the object is in front of the lens, - if it's behind
q: + when the idol is behind the lens, - if it's in the front
h': + when the idol is standing/not inverted, - if it's inverted
R1 & R2: + when the centre of curvature is behind the lens, - when it's behind the lens
f: + when the lens is convergent, - when the lens is divergent

>For Reflectors:

p: + when the object is in front of the reflector, - if it's behind
q: + when the idol is in front of the reflector, - if it's behind
h': + when the idol is standing/not inverted, - if it's inverted
f & R: + when the reflector is hollow, - when the reflector is curved
M: + when the idol is standing/not inverted, - when the idol is inverted

The Attempt at a Solution

Gotta be honest here, I've tried this about 3 times but I always screw up at the end. Anyway, let's take it from the beginning:

Lens:

1/p₁ + 1/q₁ = 1/f₁ <=> 1/1m + 1/q₁ = 1/0.8m <=> q₁ = + 4.00 m [Behind the lens]

M₁ = -q₁/p₁ = - 4 [Bigger than the object, inverted]

So now, the idol is 4 meters behind the lens. Therefore,now that the idol is acting like the object for the reflector, it's p₂ should be 3.00 m (d + p₂ = q1 <=> 1m + p₂ = 4m <=> p₂ = 3.00 m). It is, however, behind the reflector, so it should be p₂ = - 3.00 m.

And that's where I get stuck. In my book, I have only two examples, one for curved, and one for hollow reflectors, and they're just the basic stuff. Now here, the reflector is curved, but since the idol/"new object" is facing the hollow side, shouldn't it work as a hollow reflector? I mean, I can just take the 1/p + 1/q = 1/f formula again, this time for the reflector, but I don't get the proper result.

1/p₂ + 1/q₂ = 1/f₂ <=> 1/-3m + 1/q₂ = 1/-0.5m <=> q₁ = - 0.6 m

Now,according to the book and all, the final idol should be 0.6 meters "behind the reflector". Problem is, what's the "behind" here? For a curved reflector, it's behind the hollow side, but for a hollow reflector it's the curved side.

Plus, there's this whole "light that has gone through the lens two times" thing. What does it mean? The book's answer is "160 cm to the left of the lens".

Any help is appreciated!

 

[mjc123]

I can't see the picture you posted, so I can't answer the question, but on a point of terminology, we talk about the "image", not the "idol" (which suggests a statue of a pagan god). Otherwise your English is very good.

 

[Const@ntine]

I can't see the picture you posted, so I can't answer the question, but on a point of terminology, we talk about the "image", not the "idol" (which suggests a statue of a pagan god). Otherwise your English is very good.

Hmm, it shows up for me, but here it is again (this time hosted ovet at imgur):

Yeah, I messed up on the idol/image thing. I was going to go with reflection, but I was reading the exercise while posting it, so I got a bit carried away with the "literal translation".

 

[alejandromeira]

According with my sign criterium there are a problem with the data. focal distande for the mirror is f=f'=2/r. Then in a convex mirror the center or mirror is +r and also f=+50cm. (because after the center of figure distances are + in my sign criterium)
I have made the problem with this data (+50cm), an then the final position of the image is s'=+60cm, the sing + indicate that is behind the mirror. Is normal, because an convex mirror make only virtual images behind the mirror.
I think can be a problem with the criterium of signs. Notes that if the final idol has a positive sing because is behind the mirror, then the radius of mirror necessarily has to be positive because is also behind the mirror. Ok, I'm not sure if I have help you.

Another thing, evidently 160cm after the first lens, is 60cm after the mirror.

Review your criteria for signs!
Ok sorry for my little Englis, I'm not native.

 

[ehild]

The Attempt at a Solution

Lens:

1/p₁ + 1/q₁ = 1/f₁ <=> 1/1m + 1/q₁ = 1/0.8m <=> q₁ = + 4.00 m [Behind the lens]

M₁ = -q₁/p₁ = - 4 [Bigger than the object, inverted]

So now, the idol is 4 meters behind the lens. Therefore,now that the idol is acting like the object for the reflector, it's p₂ should be 3.00 m (d + p₂ = q1 <=> 1m + p₂ = 4m <=> p₂ = 3.00 m). It is, however, behind the reflector, so it should be p₂ = - 3.00 m.

And that's where I get stuck. In my book, I have only two examples, one for curved, and one for hollow reflectors, and they're just the basic stuff. Now here, the reflector is curved, but since the idol/"new object" is facing the hollow side, shouldn't it work as a hollow reflector? I mean, I can just take the 1/p + 1/q = 1/f formula again, this time for the reflector, but I don't get the proper result.

1/p₂ + 1/q₂ = 1/f₂ <=> 1/-3m + 1/q₂ = 1/-0.5m <=> q₁ = - 0.6 m

Now,according to the book and all, the final idol should be 0.6 meters "behind the reflector". Problem is, what's the "behind" here? For a curved reflector, it's behind the hollow side, but for a hollow reflector it's the curved side.

The mirror reflects on the curved (convex) side. The back is not reflecting. So consider the first image as virtual object for the mirror, with negative object distance.

Plus, there's this whole "light that has gone through the lens two times" thing. What does it mean? The book's answer is "160 cm to the left of the lens".

Any help is appreciated!

The light reflected from the mirror hits the lens and go through it second time. The mage by the mirror will be new object for the lens. Find its distance from the lens and calculate the distance of the new image from the lens.

 

[Const@ntine]

According with my sign criterium there are a problem with the data. focal distande for the mirror is f=f'=2/r. Then in a convex mirror the center or mirror is +r and also f=+50cm. (because after the center of figure distances are + in my sign criterium)
I have made the problem with this data (+50cm), an then the final position of the image is s'=+60cm, the sing + indicate that is behind the mirror. Is normal, because an convex mirror make only virtual images behind the mirror.
I think can be a problem with the criterium of signs. Notes that if the final idol has a positive sing because is behind the mirror, then the radius of mirror necessarily has to be positive because is also behind the mirror. Ok, I'm not sure if I have help you.

Another thing, evidently 160cm after the first lens, is 60cm after the mirror.

Review your criteria for signs!
Ok sorry for my little Englis, I'm not native.

Yeah, different books from different countries tend to have different +/- classifications, which make such exercises a hurdle.

The mirror reflects on the curved (convex) side. The back is not reflecting. So consider the first image as virtual object for the mirror, with negative object distance.

The light reflected from the mirror hits the lens and go through it second time. The mage by the mirror will be new object for the lens. Find its distance from the lens and calculate the distance of the new image from the lens.

So, taking into account what I've written,I continue with:

Lens': |p₃| = d + |q₂| = 1.6 m (image is on the hollow side of the reflector)

Now, if I understood correctly what you said (when you say mirror you mean what I called reflector, right?), it's as if the image from the mirror/reflector acts as the new object for the lens and is now in front of it, so I should take p₃ as +1.6 m? That said:

1/p₃ + 1/q₃ = 1/f_l <+> q₃ = 1/(1/+0.8m - 1/1.6m) = +1.6 m

I'm a bit lost though. I can't figure out what's the front and what's the back of the reflector or the lens. My book only has exercises with one lens or one reflector, and just differentiates between left and right. Is the way the light moves the positive/front side, and the other the negative/back side? In my case doesn't the light just go from left to right? Or does the direction of the light "change" because each time the image becomes the new object, and thus its side becomes the positive one?

 

[ehild]

Yeah, different books from different countries tend to have different +/- classifications, which make such exercises a hurdle.
So, taking into account what I've written,I continue with:

Lens': |p₃| = d + |q₂| = 1.6 m (image is on the hollow side of the reflector)

Now, if I understood correctly what you said (when you say mirror you mean what I called reflector, right?), it's as if the image from the mirror/reflector acts as the new object for the lens and is now in front of it, so I should take p₃ as +1.6 m? That said:

1/p₃ + 1/q₃ = 1/f_l <+> q₃ = 1/(1/+0.8m - 1/1.6m) = +1.6 m

I'm a bit lost though. I can't figure out what's the front and what's the back of the reflector or the lens. My book only has exercises with one lens or one reflector, and just differentiates between left and right. Is the way the light moves the positive/front side, and the other the negative/back side? In my case doesn't the light just go from left to right? Or does the direction of the light "change" because each time the image becomes the new object, and thus its side becomes the positive one?

First the light goes from left to right through the lens, but after it reflects from the mirror (or reflector, as you referred it), it goes from right to left.

In this case you can consider the right side as the front of the lens, and a real image will appear on the other side, to the left from the lens. How far?

 

[Const@ntine]

I'm a bit lost though. I can't figure out what's the front and what's the back of the reflector or the lens. My book only has exercises with one lens or one reflector, and just differentiates between left and right. Is the way the light moves the positive/front side, and the other the negative/back side? In my case doesn't the light just go from left to right? Or does the direction of the light "change" because each time the image becomes the new object, and thus its side becomes the positive one?

First the light goes from left to right through the lens, but after it reflects from the mirror (or reflector, as you referred it), it goes from right to left.
View attachment 209351
In this case you can consider the right side as the front of the lens, and a real image will appear on the other side, to the left from the lens. How far?[/QUOTE]

Okay, I think I got it. So:

Lens: >The light goes from left to right. Left is the front, right is the back.
>Thus, p₁ = + 1.00 m

1/p₁ + 1/q₁ = 1/f_l <=> 1/1m + 1/q₁ = 1/0.8m <=> q₁ = +4.00 m

>Positive q, so the image is on the back of the lens (the right), 4 meters away from it.

Mirror:

>Light now goes from right to left, so right is the front, left is the back.
>|p₂| = |q₁| - |d| = 3.00 m
>I'll treat the mirror as hollow, so I'll turn f_m from "-50.0 cm" to "+50.0 cm".
>Now the "object" is in front of the hollow mirror, so p₂ = + 3.00 m

1/p₂ + 1/q₂ = 1/f_m <=> 1/3m + 1/q₂ = 1/0.5m <=> q₂ = + 0.6 m

>q₂ > 0 so the image is in the front of the hollow mirror (to the right).

Lens2:

>Light goes from right to left, so right is the front, left is the back.
>|p₃| = |d| + |q₂| = 1.6 m
>The new "object" is in the front of the lens, so p3 = +1.6 m

1/p₃ + 1/q₃ = 1/f_l <=> 1/1.6m + 1/q₃ = 1/0.8 m <=> q₃ = + 1.6 m

>q₃ > 0 so it's on the back of the lens.

Final Answer: The final image is 1.6 m on the left of the lens (same as the book's).

(b) + (c) Now I have to answer whether the finalimage is inverted or not, so:

Lens:

M₁ = -q₁/p₁ = - (4 m)/1m = - 4.00 (Inverted)

Mirror:

M₂ = -q₂/p₂ = - (+0.6 m)/(+3 m) = - 0.2 (Inverted as per the image from the lens, so not inverted as per the original object)

Lens2:

M₃ = -q₃/p₃ = - (1.6 m)/(1.6 m) = - 1 (Inverted as per the image from the mirror, and also inverted as per the original object)

So, all together: M =M₁M₂M₃ = - 0.8 (Inverted)

Therefore, the final image is inverted as per the original object (same as the book's answer).

 

[alejandromeira]

Ok. I have not fully understood the problem, and then I only calculated the virtual mirror idol. This virtual mirror is in +160cm = 2F at the "rigth" of a convergent lens. As now the rays of light go to the left, the idol must be in -160cm=-2F, this is 160cm on the lef of the lens; (objet in 2F make the image in -2F)
I have made the calculations and give me this result.
Also I have made the magnification beta=beta1·beta2·beta3=-4/5=-0.8 then is inverted.

Ok I try help you best I can do. I'm not native, sorry for my English.

 

[Const@ntine]

Ok. I have not fully understood the problem, and then I only calculated the virtual mirror idol. This virtual mirror is in +160cm = 2F at the "rigth" of a convergent lens. As now the rays of light go to the left, the idol must be in -160cm=-2F, this is 160cm on the lef of the lens; (objet in 2F make the image in -2F)
I have made the calculations and give me this result.
Also I have made the magnification beta=beta1·beta2·beta3=-4/5=-0.8 then is inverted.

Ok I try help you best I can do. I'm not native, sorry for my English.

These are the book's answer as well so you've done it right as well (I mean I hope, that's what I got)!

Thanks for the help everyone!