A strange optics (or math) problem

[mAtUxAz]

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...
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!

 

[Integral]

Think a bit about what the sign means.

Then try multiplying through by -x² rather then just x².

 

[DrewD]

Are you sure that your book is correct?

 

[mAtUxAz]

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!

 

[nasu]

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?

 

[Integral]

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.

 

[rudolfstr]

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.

 

[Rap]

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.