Correct statement about a concave mirror

In summary: Sorry, the previous post was a bit of a mess. Let me try to clarify. I think the question is ambiguous because it is not clear what definition of "image" we are supposed to use. As @kuruman said, we can interpret "image" to mean both real and virtual image. In that case, options (a), (b), and (c) are all technically incorrect, as they only refer to real images. However, if we take "image" to mean only real image, then option (d) is the only incorrect option. So the correct answer depends on which interpretation we choose. This is why I think the question is not well-written.
  • #1
songoku
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Homework Statement
A person standing in front of a concave mirror. Which statement is true?
a. An image will be produced regardless where the object is
b. Image always diminished
c. Image is always inverted
d. When the person is standing between the focal point and the radius of the mirror, a sharp image of the person’s face will not be produced
Relevant Equations
1/f = 1/u + 1/v
f = focal length
u = distance object to mirror
v = distance image to mirror
Option (a) is wrong because if the person is standing at the focal point of the mirror, no image will be produced
Option (b) is wrong because concave mirror can produce magnified image
Option (c) is wrong because concave mirror can produce upright image
Option (d) is wrong because sharp image will be produced and is located beyond the radius of the mirror

So all options are wrong??

Thanks
 
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  • #2
songoku said:
Homework Statement:: A person standing in front of a concave mirror. Which statement is true?
a. An image will be produced regardless where the object is
b. Image always diminished
c. Image is always inverted
d. When the person is standing between the focal point and the radius of the mirror, a sharp image of the person’s face will not be produced
Relevant Equations:: 1/f = 1/u + 1/v
f = focal length
u = distance object to mirror
v = distance image to mirror

Option (a) is wrong because if the person is standing at the focal point of the mirror, no image will be produced
Option (b) is wrong because concave mirror can produce magnified image
Option (c) is wrong because concave mirror can produce upright image
Option (d) is wrong because sharp image will be produced and is located beyond the radius of the mirror

So all options are wrong??

Thanks
Rethink your justification for (a) being wrong in terms of the thin lens equation.
 
Last edited:
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  • #3
I agree with all your logic that leads to no correct answer. I also agree with Kuruman’s hint that places a different interpretation on a. I would add that if you take “image” to mean “real image” you get a different correct answer. I’m not saying you are supposed to take it that way, but “... image will be produced ...” isn’t the usual way people talk about virtual images.
 
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  • #4
kuruman said:
Rethink your justification for (a) being wrong in terms of the thin lens equation.
Taking u = f and putting it into the formula, I get v equals to infinity so I think there will be no image produced. Or is it still be counted as image will produce at infinity (really far) and we are still able to see it?

Cutter Ketch said:
I agree with all your logic that leads to no correct answer. I also agree with Kuruman’s hint that places a different interpretation on a. I would add that if you take “image” to mean “real image” you get a different correct answer. I’m not saying you are supposed to take it that way, but “... image will be produced ...” isn’t the usual way people talk about virtual images.
I am not sure I understand your explanation.

I take "image" to mean both real and virtual image. And what does it mean by "...image will be produced..." isn’t the usual way people talk about virtual images?

Thanks
 
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  • #5
My interpretation centers on the observation that "physics" infinity is different from "mathematical" infinity. When the position of an object u >>> f (think of a galaxy far far away), a real image forms at the focal length. By reversing the light, an object at the focal length has a real image at v >>> f.

I do not want to speak for @Cutter Ketch, but I think his interpretation is that because virtual images are not formed on a screen like real images, they don't count. When you see your enlarged face in a makeup mirror, no image is formed anywhere. The lens in your eye bends back the diverging reflected rays from the mirror and focuses them on the back of your retina. Your brain then interprets the nerve stimuli as an image.

The correct answer depends on the interpretation you choose. In my opinion this is an ambiguous question precisely for that reason.
 
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  • #6
kuruman said:
My interpretation centers on the observation that "physics" infinity is different from "mathematical" infinity. When the position of an object u >>> f (think of a galaxy far far away), a real image forms at the focal length. By reversing the light, an object at the focal length has a real image at v >>> f.

I do not want to speak for @Cutter Ketch, but I think his interpretation is that because virtual images are not formed on a screen like real images, they don't count. When you see your enlarged face in a makeup mirror, no image is formed anywhere. The lens in your eye bends back the diverging reflected rays from the mirror and focuses them on the back of your retina. Your brain then interprets the nerve stimuli as an image.

The correct answer depends on the interpretation you choose. In my opinion this is an ambiguous question precisely for that reason.
I see. So option (a) is, at least, the best option out of those four options. The reflected ray from concave mirror will be diverging and parallel but our eyes will still see image of the object.

Thanks
 
  • #7
songoku said:
Option (d) is wrong because sharp image will be produced and is located beyond the radius of the mirror
Are we to take the mirror as spherical or parabolic?
 
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  • #8
haruspex said:
Are we to take the mirror as spherical or parabolic?
Spherical
 
  • #10
haruspex said:
Are we to take the mirror as spherical or parabolic?

EEK! More possible interpretations! How sharp is “sharp”? I think I hate this question!
 
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  • #11
songoku said:
I take "image" to mean both real and virtual image. And what does it mean by "...image will be produced..." isn’t the usual way people talk about virtual images?

@kuruman explained my meaning perfectly. “formed” doesn’t guarantee anything, but it kind of hints at one way the questioner might have been thinking. Just another possible interpretation to highlight that this question is a little ambiguous.
 
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  • #12
Uhm ... I somehow misused multi-quote (had a remnant from earlier) and made it look like someone else (Eitan Levy) said something. That was me. Sorry about that. Trying to edit the post now.
 
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  • #13
haruspex said:
So how are you so sure the image would be sharp?
https://en.m.wikipedia.org/wiki/Spherical_aberration
Wow I never considered about spherical aberration. So yeah there is possibility the image formed is not "that sharp" (whatever that means 😅)

This question is so confusing, I can not know what to interpret from the options (a) and (d)
 
  • #14
songoku said:
Wow I never considered about spherical aberration. So yeah there is possibility the image formed is not "that sharp" (whatever that means 😅)

This question is so confusing, I can not know what to interpret from the options (a) and (d)
Option d seems so carefully constructed that I suspect it is the intended answer. Not that I know why a is wrong.
 
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  • #15
haruspex said:
Option d seems so carefully constructed that I suspect it is the intended answer. Not that I know why a is wrong.
Despite being intended, d is wrong too. One radius away from the mirror is the position of maximum freedom from spherical aberration.
 
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  • #16
If this is for a typical intro physics course, I seriously doubt that spherical aberration is supposed to come into play here. Has songoku’s course even discussed spherical aberration?
 
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  • #17
jtbell said:
If this is for a typical intro physics course, I seriously doubt that spherical aberration is supposed to come into play here. Has songoku’s course even discussed spherical aberration?
Now that I think about it after all this discussion I would agree with @haruspex that the likely candidate for correct answer is (d) but not because of spherical aberration. Note that (a), (b) and (c) all use the general word "image" while (d) specifically mentions "image of a person's face". A person's face is not flat; the tip of the nose and the eyes are at different points on the optical axis. So ...
 
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  • #18
kuruman said:
Now that I think about it after all this discussion I would agree with @haruspex that the likely candidate for correct answer is (d) but not because of spherical aberration. Note that (a), (b) and (c) all use the general word "image" while (d) specifically mentions "image of a person's face". A person's face is not flat; the tip of the nose and the eyes are at different points on the optical axis. So ...
Soryy, not following. Why would that matter?
 
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  • #19
jtbell said:
If this is for a typical intro physics course, I seriously doubt that spherical aberration is supposed to come into play here. Has songoku’s course even discussed spherical aberration?
If the student is not expected to know anything about spherical aberration then it is strange that option d is offered, whether it is correct or not.
 
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  • #20
haruspex said:
Soryy, not following. Why would that matter?
Where would one have to place a flat screen to produce "a sharp image of the person's face"? At the point where a sharp image of the tip of the nose is produced or at the point where a sharp image of the eyes is produced? A person's face is not flat.
 
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  • #21
kuruman said:
Where would one have to place a flat screen to produce "a sharp image of the person's face"? At the point where a sharp image of the tip of the nose is produced or at the point where a sharp image of the eyes is produced? A person's face is not flat.
What flat screen? I read it as producing a sharp 3D image viewable by the person.
 
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  • #22
haruspex said:
What flat screen? I read it as producing a sharp 3D image viewable by the person.
That's not how I read it. Part (d) talks about a sharp image being produced not viewed by the person standing in front of the mirror. I used the flat screen as an indicator of the point where the image that is "produced" appears sharp.

Edit: Your interpretation assumes that the image is produced by a two-component optical system, the mirror and the lens of the observer's eye.
 
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  • #23
kuruman said:
Edit: Your interpretation assumes that the image is produced by a two-component optical system, the mirror and the lens of the observer's eye.
No, I mean a 3D virtual image in space. It is sharp in the same sense that one person’s view of another's face is sharp, i.e. as you turn your attention to different points each appears sharp in turn.
This is not true of an image subject to significant spherical aberration.

Anyway, as @jbriggs444 points out, a key question for d is whether spherical aberration is particularly bad within the given range. JB says not; I have not checked but I am inclined to accept that.
 
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  • #24
I will not belabor the point past this post which will be my last one on this thread. A 3D image as you describe appears sharp because the lens of the observer's eye normally adjusts its focal length for that to happen. In my case, the tired old lenses that I was born with can no longer accomplish this feat. Depending on how far objects are from my eyes, I have to angle my head appropriately in order to achieve the correct angle of incidence on the bifocal lenses that I wear. :oldsmile:
 
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  • #25
kuruman said:
I will not belabor the point past this post which will be my last one on this thread. A 3D image as you describe appears sharp because the lens of the observer's eye normally adjusts its focal length for that to happen. In my case, the tired old lenses that I was born with can no longer accomplish this feat. Depending on how far objects are from my eyes, I have to angle my head appropriately in order to achieve the correct angle of incidence on the bifocal lenses that I wear. :oldsmile:
Sure, but I would quantify the sharpness of an image formed by any lens or mirror to be in relation to the 3D image it forms, independently of any subsequent processing. Spherical aberration blurs that.
 
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  • #26
It's a badly written question as evidenced by the discussion here, but as suggested by jtbell its a question you might come across in an introductory physics course. As such, if it were an exam question the candidate would be expected to reach an answer within less than one minute. The nature of the image depends, amongst other things, on where the object is, but you always get an image, whether it's of poor quality or not. Therefore I think the expected answer is a. Also, I suspect that the person who wrote the question would think that most candidates could reach that answer using general knowledge only. It's a pity that more care wasn't taken when writing the question.
 
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  • #27
jtbell said:
If this is for a typical intro physics course, I seriously doubt that spherical aberration is supposed to come into play here. Has songoku’s course even discussed spherical aberration?
No, haven't covered about spherical aberration.

I think I understand the explanation about this question. Thank you very much for all the help
 

Related to Correct statement about a concave mirror

1. What is a concave mirror?

A concave mirror is a curved mirror with a surface that curves inward. This type of mirror is also known as a converging mirror because it causes parallel light rays to converge at a point.

2. How does a concave mirror form an image?

A concave mirror forms an image by reflecting light rays that strike its surface. The reflection follows the law of reflection, which states that the angle of incidence (incoming light ray) is equal to the angle of reflection (reflected light ray).

3. What is the difference between a real image and a virtual image?

A real image is an image that can be projected onto a screen, whereas a virtual image cannot be projected. In a concave mirror, a real image is formed when the object is placed beyond the focal point, while a virtual image is formed when the object is placed between the focal point and the mirror.

4. How does the size of the image compare to the size of the object in a concave mirror?

The size of the image depends on the distance of the object from the mirror. When the object is placed beyond the focal point, the image is smaller than the object. When the object is placed between the focal point and the mirror, the image is larger than the object.

5. Can a concave mirror produce an upright image?

Yes, a concave mirror can produce an upright image. This occurs when the object is placed between the focal point and the mirror. However, the image will be smaller than the object and will be virtual, meaning it cannot be projected onto a screen.

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