Challenge Questions

Thanks for beautiful picture. You and jim are correct. Here is full explanation (with not so beautiful picture)

http://www.glenbrook.k12.il.us/gbssci/phys/Class/refln/u13l2d.html

Ok, thats the equation to find the answer, but for me, the answer would be 3'2", but as every one is different, the ultimate answer would be different for every individual...

But the question (by the wording) wasn't "find the equation to calculate the length of the mirror" but " What is the minimum length of the mirror?" So the formula above is really only 1/2 the answer

So, the Min length for me is 3'2" (using the formula above)

Half the height of a person is correct. But there is one more condition: The top edge of the mirror should be at the level of the eye of the observer. Otherwise one will not see his / her full figure.

No gsuhas, the top edge of the mirror should be at a level half-way between your eyes and the top of your head.

Cheers.......Codey

Yes, you are right master. Thanks for correction.

Looks like just a bit under half =0.49998

If she stood straight up instead of the flirtatious stand I could get better measurements.

You and Jim give good answers, but as I don't look like that I'm going to make a different assumption - that I position myself to minimise the size of the mirror that is needed. If I lie on the floor and bend my head back, the longest length of the mirror will be about
16/17 x (width_of_my_shoulders/2 - the_separation_of_my_eyes).

The 16/17 is because the rays continue to diverge in the distance that my shoulders are behind my eyes.

OK, so I've assumed that my eyes are 4' away from the mirror, and my body is at greater distances, but neither can your exemplar place all of herself 4' away from the mirror. If I interpret the challenge as placing my centre of gravity 4' from the mirror, the length of the mirror could be even smaller - possibly as little as
5/6 x (width_of_my_shoulders/2 - the_separation_of_my_eyes), or about 6".
If you are somewhat taller you could calculate an even smaller fraction - but you would probably have broader shoulders.

P.S. I don' think the following is in the spirit of the challenge, but I didn't find anything to say that you can't interpose a strong lens between yourself and the mirror. However, that reduces the answer to the minimum size that allows you to resolve whether all of you is visible - say 0.1"? (a concave lens would give an erect image, a convex lens would also work, but your image would be upside-down)

Wanting to do this experimentally, I went to the garage (still not finished cleaning it), found a mirror that is 34 inches tall, propped it on the garage door, and took a picture of myself. Last time I measured my height (10 years ago) I was 6' 1". I had to stand more than four feet away from the mirror to get my reflection in the picture without figuring out how to hang the mirror at the right height. Looks like half my height if about right (must have shrunk a little).

Now I'm going to have to figure out how to take a picture while lying on my back with my head down.

Sorry. Make that a 37 inch tall mirror. typo

Nice one Fyz.

I'ld like to see you try shaving like that

Apologies - my P.S. doesn't answer to the wording of the challenge [I misinterpreted the "full-size" phrase as meaning entire (blame that speed-reading course?)]

depends on the curvature of the mirror.

Flat mirror, I think your height + 4'.

Curved, very small, 2 / 3".

Dear Mirror Mirror: CR4 Challenge (09/30/08) Answer: My Height (or the Individuals Height that is making the Check). On Plane Mirror in our Hall, Standing 4 ft. back, My Height- 5' 10" shows just below a mark 5' 10" up from the floor, on the wall beside the mirror.. Guest, Carl.

The minimum length of the mirror is my half height

One-half of your height!

The guest in post #3 has the correct answer. You would need the extra 4 inches as that is the approx height from eyes to top of head. Secondly there was a person who asked how high the mirror was off the ground, it isn't specified in the question, but the question asks for the length of the mirror, not its height and as such that I believe is immaterial.

Good question though with the usual plethora of good answers

Half your height.

Codey

You are 4 ft. in front of a plane mirror, and you notice that you can see a full-size image of yourself in the mirror. What is the minimum length of the mirror.

Since incidence = reflectance, the bottom of the mirror would have to be one half the distance between your eyes and your feet from the floor.

minimum length of the mirror is 4ft

Holy Crow, Guest, that puts you at eight feet tall! I hope you never leave the house without a basketball in your hand...

In keeping with the correct answer of half your hieght, and taking into account the proctor has already told us how tall we are, I surmise the total length of the mirror shall be 2 ft.

I like it. Much better than the obvious interpretation. But why assume that the 4' is parallel with the surface of the mirror?

This might be a trick question, because it states you could see a full-size image, not a full-length image of your self. So if it's a plane mirror (no warpage), regardless of the dimensions, any image being reflected would be indeed FULL-SIZE.

Very clever indeed. I like the way you play with the wording to substantiate your answer.

I am sitted. My sitted height is 3 ft only. The mirror required is 1 1/2 ft. I can see my full figure (sitted)

I think the word 'challenge' is going a bit far this time.

We did this at school at about age 13!

Now, let us have challange in challange.

You are standing straight in front of a vertical mirror, in which you see another vrtical mirror and your full image in the mirror in mirror.

What are the minimum heights needed for each of the mirror?

Not enough information. For example, if you are right up against one of the second mirror it would have to be as tall as you are.

There are two mirrors, Mirror 1 & Mirror2, kept vertical facing each other.

We see reflection of mirror 1 along with the full figure reflection of ourself in mirror 2.

What should be the minimum sizes of two mirrors?

As described, you cannot see your reflection in the second mirror at all, as your body is in the way.

If you are sufficiently transparent* that you can see the second reflection, the smallest the mirrors could be is if the mirror you are looking directly into is close to your eye (compared with the distance to the other mirror). The size of the closer mirror would limited by the size of and any necessary movement of your eye - otherwise it be as small as you like. The other mirror would be half your height.

Reading your text to mean that you are looking directly into mirror2 at its reflection of the reflection of yourself in mirror 1, the formula (ignoring changes due to possible angling of the mirrors and the effect of your eye movement) is:

V1 = H.(L2+2.L1)/(L1+L2)/2
V2 = H.L1/(L1+L2)/2
where H is your height, V1 and V2 are the vertical dimensions of the mirrors, and L1 and L2 are the distances from you to each of the mirrors.

You can see that the difference between the dimensions of the mirrors is half your height.

*or you angle the mirrors very slightly away from "facing each-other" so that you can see past yourself

P.S Strictly speaking, I suppose this should be marked off topic?

It's not that far off-topic, but it sure makes me think of Versailles...

Very strictly speaking it is off topic. But certainly it is on topic as compared to many posts, we see where it is totally out from original thread. But is it not challanging question?

You are right, we can not see our own reflection in second mirror, as our body will obstruct it. But if I tilt one of the mirrors, I can see the image, just like how we see our many more reflections in hair cutting saloon.

I could not solve my own problem. Otherwise I would have sent it as next CR4 challange, where we are supposed to put the challange and answer together.

I am trying to understand the solution you have give.

Thanks

All I did was measure the entire (folded) paths from eye to each mirror and thence to yourself standing. But I carelessly swapped L1 and L2, so the algebraic answer as presented was incorrect

Light path distance to mirror you are looking directly at = L2
Light path distance to other mirror = 2*L2+L1
Light path distance back to yourself (seen via both mirrors): 2*(L1+L2)

Then use equal ratios in similar triangles, so
H/(2*(L1+L2)) = V2/L2 = V1/(2*(L2+L1))
. where H is your height, V1 and V2 are the vertical dimensions of the mirrors

Thus the answer should have read:
V2 = H.L2/(L1+L2)/2
V1 = H.(L1+2.L2)/(L1+L2)/2