Mirror question

[SephirothXR]

Focal length = 5m
Object is placed 4m from CONVERGING mirror

So using 1/f = 1/o + 1/i,
(1/5) = (1/4) + (1/i) or .2 = .25+ (1/i), and I got i to be -20. So if I were to describe this image, it would be 20m to the right of the mirror, upright, and virtual, correct?

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[Labminion]

That's what I got.

 

[SephirothXR]

Then I'm leaning towards the fact that the answer explanation in my book is wrong. Hmm....

 

[Labminion]

What book are you using? What did they say was the correct answer?

 

[Ferocity]

Are you using TBR to study? I memorized all the scenarios of what happens for diverging and converging mirrors.

 

[SephirothXR]

What book are you using? What did they say was the correct answer?

Just some random practice test in the bookstore, they said it was in front of the mirror, but I know for fact that's wrong..

 

[whiteshadodw]

Are you using TBR to study? I memorized all the scenarios of what happens for diverging and converging mirrors.

Really? if you just memorize the rules, you don't have to memorize situations. and its far more convenient.

Just some random practice test in the bookstore, they said it was in front of the mirror, but I know for fact that's wrong..

i got image distance of -20 m as well. virtual and upright. UV-IR, doesn't make sense to me, but i guess it makes sense.

 

[MShopes]

Just some random practice test in the bookstore, they said it was in front of the mirror, but I know for fact that's wrong..

I agree. For converging mirror, the image distance is positive, real, inverted when it is in front of the mirror! Since the image distance is negative, then it would be negative, virtual, upright (That is true because the object moved within the focal length so the image must be negative, virtual, and upright). A negative image therefore would be behind the mirror since the positive image is in front of a mirror. The opposite holds true for converging lens. The image is positive for converging lens when it is behind the lens (or on the side of where our eyes are looking) while it is negative when it is in front of the lens (or on the opposite side of where our eyes are looking at the lens)

Examkrackers makes a beautiful drawing of the four cases of two mirrors and two lenses describing all possible scenarios with positive and negative values. Also, their rules at the last page are golden.

I think also that memorizing the rules without being able to visualize everything is not a good way to solve problems. Remember that the MCAT might not just give you numbers like this and you simply plug and chug but instead ask you a concept problem that involves qualitative answer rather than just certain numbers. I found the optics topic to be the easiest topic ever in physics. It simply gets in my head very fast!

 

[Danlee07]

you're right... I always tell people i teach mirrors.. just do the ray diagrams for in front of, at, and behind the focal point for converging/diverging mirror and lens. And one past the center, and in front of the mirror.