# Finding the distance of reflection in a mirror

1. Aug 18, 2015

### Barclay

1. The problem statement, all variables and given/known data
A student wants to find the position of an image in a plane mirror. She sets up the plane mirror with a pin placed vertically a few centimetres in front of it. To find the image the student looks into the mirror at an angle a little further along the mirror. She can see the image of the object pin in the mirror and places two "sighting pins" in line with the image in the mirror.
The student repeats the process from a doughty slightly different angle, again putting in two sighting pins.

Q1. How can the student use the position of the sighting pins to find the position of the image?
Q2. How can the position of the sighting pins be used to show that the angle of incidence equals the angle of reflection?
Q3. How can the student checked that she has marked the position of the image correctly?
Q4. Would this method for finding the image work with a curved mirror?

2. Relevant equations

3. The attempt at a solution
I haven't got a clue to the answers. I know no one will just give me the answers and I'm happy to deduce the answers myself if someone can guide me. Thank you.

Pythagoras comes to mind but did not help.

Last edited: Aug 18, 2015
2. Aug 18, 2015

### BvU

Hello again,
Do you have the means to make a drawing of the situation and post it ?
My perception so far is that the pin is vertical and parallel to the mirror, so a top view would be a good start.

3. Aug 18, 2015

### Barclay

Hello BvU. I'm unable to get a picture posted but I can describe it ..... just a mirror with a bit of vertically orientated string sitting a few centimetres in front of it

4. Aug 18, 2015

### BvU

Well, it's not exactly Disney quality, but this is what I imagine:

Observer sees pin in mirror and places two pins,
then moves a bit and repeats. Right ? Try to guess four reasonable pin positions and we'll pick up from there.
(It all being geometry then)

Q1 should be easy to answer !

5. Aug 18, 2015

### Barclay

Thanks BvU. that's a great picture. Just as in the book.

I can't see the geometry. There are no numbers given so I can only use x, y , z etc but nothing is making sense. I don't think I even understand the question properly. What is a sighting pin?

"She can see the image of the object pin in the mirror and places two "sighting pins" in line with the image in the mirror". I have no idea what this means. Perhaps it means that the sighting image blocks the image of the original pin. There are 5 pins altogether. Just a mess of lines in my attempt at a geometrical diagram.

6. Aug 18, 2015

### BvU

And here's the next step, the first two pins. I also drew the sight line. Your turn to draw another sight line

Start to see some geometry ?

7. Aug 19, 2015

### Barclay

Hello BuV,
All I can think of is that the student should get a ruler and measure the distance from the new pin from the mirror then double it. Also with ruler measure the distance from the new pin to the string and double it. This gives us two lengths of a triangle so with Pythagoras work out the distance of the image.

But the question isn't asking for distances. It just wants to know how to use the pins to find the image of the string. Stuck

8. Aug 19, 2015

### BvU

When I look at the drawing, it doesn't work for me. Don't see a string anywhere either.

In the drawing there is now one original pin and two sighting pins. Your exercise wants two more sighting pins. Surely you understand the principle ? If you want to measure a distance to something you can't reach (because it is on the other side of a canal or a mirror), then you draw two straight lines in the direction of the something from two different starting points and where the two lines cross is where the something is located. And two points are enough to uniquely define a straight line ! You don't have to jump into the canal or the mirror and still can say exactly what the position of the something is !

Your turn to place two more pins (didn't I ask that before ... ? )

9. Aug 19, 2015

### Barclay

BuV ... that string that you can't see is the red pin that you have drawn. The book shows the pin as a string (like a candle wick).
BuV I don't think I can go any further with the question because I don't understand the question.

The question is "How can the student use the position of the sighting pins to find the position of the image?"

I don't know what they mean.

Last edited: Aug 19, 2015
10. Aug 19, 2015

### BvU

I draw the top view, so you see the head of the pin. The word string isn't meaningful to me, unless it's the shaft of the pin, sorry.

They want you to move the observer eye a bit and place two more pins on the new line to the image.
If you then remove the mirror (leaving a line where it was on the table for Q2) you can place a pencil on the crossing point: that's where the image was.

Take a mirror and five pins and see it work !

Last edited: Aug 20, 2015
11. Aug 20, 2015

### BvU

To get you going again (how come you were so tenacious with the resistance graph and gave up on this one ?) :

Is what I meant with moving the observer eye a bit and placing two more pins. Then you take away the mirror (leaving a pencil line where it was located) and you have the five pins lefte over on the table (or whatever):

Now what can the student (why is it always a female in your exercises ?) do to reconstruct the position of the image ?

Please tell me it's easy now ...

--

12. Aug 20, 2015

### Barclay

Th student draws lines up to the mirror the takes the mirror away. Then continues to draw the lines further along and where they cross is the point of the IMAGE..

Voila?

13. Aug 26, 2015

### BvU

Sorry, I missed this alert. Yes. Simple, isn't it ?
So how are (were) you doing with Q2 and onwards ?

14. Aug 26, 2015

### Barclay

Oh dear. I thought I had finished. Took me so long to get question one correct that completely forgot there were more questions.

Just had a quick look at the questions and will ponder over them and definitely post tomorrow. Thanks for your continued interest

15. Aug 27, 2015

### Barclay

Q2. How can the position of the sighting pins be used to show that the angle of incidence equals the angle of reflection?

Hello BuV, Draw a line from the point where the sighting line meets the mirror to the red pin. With a protractor measure the angles from the normal (the perpendicular line to the mirror) and it van be seen that the angle of incidence = angle of reflection

Q3. How can the student checked that she has marked the position of the image correctly?

Place a pin a the pont of the image and then look at the image from te position of the original red in. The new pin should be directly in front (perpendicular to the mirror)

Q4. Would this method for finding the image work with a curved mirror?

Depends how far the red pin is from the mirror because if too far the image become distorted by the curvature of the mirror

16. Aug 27, 2015

### BvU

Q2: measuring is one thing. Doing some geometry to actually show it probably gets you more brownie points...
Q3: through two red pinheads there is always a straight line. Doesn't feel like a thorough check to me...
Q4: Draw a diagram and conclude that it never works, no matter how far...

We're not through with this one yet ...

17. Aug 28, 2015

### Barclay

Its getting too tough now. The answers I gave were wishy washy because I've reached the end of my knowledge and was just clutch at straws

18. Aug 29, 2015

### BvU

I'm sorry to see you struggling with this. Is there a context of lectures, notes or textbook or are you exploring this all on your own ?

What doesn't make this exercise easy is that we don't really have listed what is known to us and what the exercise asks us to show/prove. So I'm just guessing in that area, just like you seem to be.
For Q3 I was thinking of the perpendicular bisector of the line red pin - reconstructed image. It has to coincide with the line marking location of the removed mirror. But I think we also need that bisector for Q2 (you know, similar triangles, congruent triangles).
For Q4 the result of Q2 is important: if those angles are equal, then it's clear that if the angle of the reflecting surface changes, then the sight lines won't cross the optical axes in one single point.

19. Aug 30, 2015

### Barclay

I must apologise to you BvU. I have not read this chapter thoroughly. Just skimmed it. I had some free time and thought I would be able to answer the questions with a little help from the forum. These questions are in a text book with no good answers and I thought I would be able to develop the answers with the forums help. Certainly the first answer you gave me was much more helpful than the book. But its too intense after that. I need to read the chapter more thoroughly. So sorry BuV

20. Aug 31, 2015

### BvU

No need to apologize. You learn from this (I hope) and in a way I also learn. So we're even and we're good ! Good luck and I look forward to next thread.