# Ray OPtics Problem 2

1. Feb 20, 2013

Hi friends,

The problem is as follows,

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-prn1/s480x480/11130_2937865262994_1085043361_n.jpg

Well,
Friends when the lower mirror would be at position 'A', which is moving in the upward direction, the img of the source 'S' will be at position (I') as shown in the figure below. And and after some time when the mirror will be at position 'B' the same ray will make the img of the source and the new position (I').
https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash3/60754_2937792661179_1384286388_n.jpg

Hence the answer of the question should be option (A). But in the book the answer of the problem is given options (B) & (D)

Please friends help me in this problem.
Thank you very much in advance.

2. Feb 20, 2013

### Simon Bridge

You have only drawn what happens to one ray in the point source - draw in the others.
OR you can try it with an LED.

3. Feb 21, 2013

### Simon Bridge

... that was a little terse.
Imagine the mirror in position A in your diagram, put a detector at the place the ray ends up. The detector goes off - fine.

Move the mirror to position B, will the detector go off?
The same angle ray won't set it off, but that is not the only ray coming from the source.
Is there another ray that will set the detector off?

Think about the spot of light - where is it centered? How does the brightness fall off towards the edge? Is there anywhere you could put the detector and not detect any light?

4. Feb 21, 2013

### PeterO

The two rays you have drawn, while correct, are not helping with the solution.

You need 4 other rays.

A ray from S to the left end of mirror in position A - then trace on to the wall
A ray from S to the left end of mirror in position B - then trace it to the wall.

Then do two rays - one each to the right hand ends of the mirror in position A and B.

5. Feb 21, 2013

### Basic_Physics

Consider what happens to the incident angle when the normal is kept at the same position on the mirror.