A strange optics (or math) problem

  • Context: Undergrad 
  • Thread starter Thread starter mAtUxAz
  • Start date Start date
  • Tags Tags
    Optics Strange
Click For Summary

Discussion Overview

The discussion revolves around a physics problem related to optics and the calculation of illumination on two screens from a point source. Participants are attempting to reconcile their solutions with the answer provided in a textbook, which is approximately 0.4m. The conversation includes various interpretations of the problem and the mathematical relationships involved.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion over their solution yielding approximately -0.4m instead of the textbook's ~0.4m.
  • Another participant suggests considering the implications of the sign in their calculations.
  • Some participants question the correctness of the textbook answer, with one asserting that their calculations support the answer being correct.
  • There is a discussion about the distances to the screens and how they relate to the illumination ratios, with one participant providing specific distances and ratios.
  • Concerns are raised about the applicability of the equations used, particularly regarding the geometry of the screens and the assumptions made about the source.
  • One participant proposes that the problem may have been misunderstood regarding the notations and assumptions about the screens being small compared to the distances involved.
  • A later reply presents alternative solutions based on different interpretations of the problem, suggesting that the answer of 0.4m is incorrect.

Areas of Agreement / Disagreement

Participants do not reach a consensus; multiple competing views remain regarding the correctness of the textbook answer and the validity of the calculations presented.

Contextual Notes

Participants express uncertainty about the assumptions made in the problem, the definitions of terms used, and the geometric considerations of the setup. There are unresolved mathematical steps and differing interpretations of the problem's conditions.

mAtUxAz
Messages
2
Reaction score
0
Hello everyone, I was reading a physics book and found an exercise. This time I faced with a really strange one! I can't find out what I am doing wrong! I am really good at physics and math, but sometimes I fail as many people do... :smile:
The answer in the book says ~0.4m, but I am unable to get it (instead of it, I get ~ -0.4m)
I am attaching a Word document with my solution.

I will be looking forward to getting the right solution!
 

Attachments

Last edited:
Science news on Phys.org
Think a bit about what the sign means.

Then try multiplying through by -x2 rather then just x2.
 
Are you sure that your book is correct?
 
DrewD said:
Are you sure that your book is correct?

No, the answer is right. When I put in the 0.4m as R (or 0.16 as R^2) I get correctly - the E1 screen illumination is two times better than E2. Looks like there is a mistake in my own equations, but I can't find where, as it looks like that it must be like that!
 
If the distance to the first screen is 0.41m, then the distance to the second one is 1.41 m.
The ratio of their squares will be about 2/0.16.
How is this equal to 2?

Maybe you misunderstood something about the notations in the original problem?
 
mAtUxAz said:
No, the answer is right. When I put in the 0.4m as R (or 0.16 as R^2) I get correctly - the E1 screen illumination is two times better than E2. Looks like there is a mistake in my own equations, but I can't find where, as it looks like that it must be like that!


The equation you used does not apply to a flat screen, the distance from the top edge of the flat screen to the source is different from the center to see this apply Pythagoras. They assume spheres, so -.41m refers to the same sphere as .41m. x is the radius of a sphere with center at the source. You never said what your second distance is. I find the relationship holds for spheres of .41 and .58. That is -.41 and 1+(-.41) as well as 2.14 and 3.14

Again x=-.41 indicates the same sphere as x= .41 so those are the same solution.
 
Integral said:
The equation you used does not apply to a flat screen, the distance from the top edge of the flat screen to the source is different from the center to see this apply Pythagoras. They assume spheres, so -.41m refers to the same sphere as .41m. x is the radius of a sphere with center at the source. You never said what your second distance is. I find the relationship holds for spheres of .41 and .58. That is -.41 and 1+(-.41) as well as 2.14 and 3.14

Again x=-.41 indicates the same sphere as x= .41 so those are the same solution.
I doubt it would go to so difficult maths. It's more likely that they want to just approximate equal distance.
 
If the problem has been properly presented, and E1 and E2 represent the energy per unit area on the screens, and the screens are very small compared to the distances from the source (or spherical), and the source is a point source, the two possible answers are x=1-\sqrt{2} and 1+\sqrt{2} or about -0.414 and 2.414 for the position of the first screen and x+1=0.586 and 3.414 for the second screen. From the diagram, x is positive, so x=2.414 and x+1=3.414. The answer x=0.4 is wrong.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 37 ·
2
Replies
37
Views
5K
  • · Replies 171 ·
6
Replies
171
Views
11K
  • · Replies 17 ·
Replies
17
Views
9K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K