Mirror optics

[ashtonjam]

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33. For a convergent mirror with f = 0.30 cm, at what
position should an object be placed to generate an
image that is three times the object in magnitude of
size?
A. At the 0.20 cm mark
B. At the 0.40 cm mark
C. At the 0.50 cm mark
D. At the 0.60 cm mark

Answer is B

For some reason, both A and B seems like possible answers and the only difference is that one image is virtual while the other is real.

e.g. For choice A, if object distance is 0.2 cm:
Image distance = of/(o-f) = (0.2 * 0.3)/(0.2 - 0.3) = 0.06/-0.10 = -0.6 cm.
Magnification = -i/o = -(-0.6 cm)/(0.2 cm) = 3. That would fit the criteria for the magnification being equal to 3 but it's a virtual image.


e.g. For choice B, if object distance is 0.4 cm:
Image distance = of/(o-f) = (0.4 * 0.3)/(0.4 - 0.3) = 0.12/0.1 = 1.2 cm.
Magnification = -i/o = -(1.2 cm)/(0.4 cm) = -3.

So it seems like there are two cases in which magnification magnitude is 3, but only one is the answer? The answer key seems to eliminate choice A right away just for being a virtual image, which confuses me.

33. Choice B is the best answer. For a convergent mirror to generate a real image, the object must be beyond the focal point, so
choice A can be eliminated immediately. With f = 0.3 cm, R must be 0.6 cm. If the object is placed at the radius of curvature,
then the image will also form at R, resulting in an image with the same magnitude of size as the object. Choice D is
eliminated. Only math can make the final distinction between choices B and C. If the object is at 0.4 cm, then the image will
form at (0.4 x 0.3)/(0.4 - 0.3) = 0.12/0.10 = 1.2 cm. Magnification is found using m = -(dj/d0) = -(1.2/0.4) = -3. Because m is
-3, the image is three times as large as the object and it is inverted. The best answer is choice B.

 

[IlyaR]

.

 

[sps27]

33. For a convergent mirror with f = 0.30 cm, at what
position should an object be placed to generate an
image that is three times the object in magnitude of
size?
A. At the 0.20 cm mark
B. At the 0.40 cm mark
C. At the 0.50 cm mark
D. At the 0.60 cm mark

Answer is B

For some reason, both A and B seems like possible answers and the only difference is that one image is virtual while the other is real.

e.g. For choice A, if object distance is 0.2 cm:
Image distance = of/(o-f) = (0.2 * 0.3)/(0.2 - 0.3) = 0.06/-0.10 = -0.6 cm.
Magnification = -i/o = -(-0.6 cm)/(0.2 cm) = 3. That would fit the criteria for the magnification being equal to 3 but it's a virtual image.


e.g. For choice B, if object distance is 0.4 cm:
Image distance = of/(o-f) = (0.4 * 0.3)/(0.4 - 0.3) = 0.12/0.1 = 1.2 cm.
Magnification = -i/o = -(1.2 cm)/(0.4 cm) = -3.

So it seems like there are two cases in which magnification magnitude is 3, but only one is the answer? The answer key seems to eliminate choice A right away just for being a virtual image, which confuses me.

I think the keyword 'real' is missing from the question stem. Is this from TBR, cause I remember having run into this before and choosing A for ans because there is only 1 instance for the converging mirror when the image is erect and for some reason I assumed erect after reading the word magnitude. Anyways, I just let this slide by....

 

[ashtonjam]

Yes it's from TBR Optics chapter 10. Well if they left out the word 'real' then that makes me feel better 🙂

 

[sps27]

Yes it's from TBR Optics chapter 10. Well if they left out the word 'real' then that makes me feel better 🙂

You may already know this but I did not find this covered either in Kaplan or TBR (don't know about TPRH or EK) so this is just an fyi and for some reason I remember this particular discreet - How many images will be formed if two plain mirrors are placed at right angles to each other? It is quite simple and I did not know it at that time but, # images = (360/angle) -1.

 

[ashtonjam]

I didn't know that actually, cool. It makes sense since the plane mirrors would be forming a corner and you would see 3 reflections of yourself in addition to your real self.