Recent content by kuruman

Good one!

I am not sure that AI is capable of generating questions that have never been asked before. Asking such questions and then answering them is commonly known as "research." I believe that AI is good at the answering part but not at the asking part. AI is also good at stealing ideas without...

I forgot to add that the statement above is nonsensical and contradicts the familiar observation that when you look at your face in a mirror, your virtual face looks back at you behind the mirror. A virtual image appears where light has no business being and that is why it is called "virtual"...

Yes it does. Do you see the similarity between the answer to my question and the correct answer to your problem?

Figure (a) on the right shows a glass of water with a source of light at the bottom. The distance from the surface of the water ito the bottom of the glass is ##d##. The eye sees the source at an apparent distance below the surface ##d'=d/n## where ##n## is the index of refraction of water...

Ok, I will do that. Post #16 concludes with the statement I agree that you have shown that PE = (1/3) AE and that a similar argument yields QF = (1/3) AF. You have not provided the reasoning that relates these equations to what the problem is asking you to show, namely that $$~BP = PQ =...

I agree with the first half of the sentence, but not the second. OP has shown that ##\mu=\frac{2}{3}AE## and that ##\lambda=\frac{1}{3}BD.## OP needs to show that this is a necessary and sufficient condition for the trisection of BD. An easier approach would be to draw diagonal AC and note that...

Why is that? The problem is asking you to "show that two lines from one vertex of the parallelogram to the midpoints of the two opposite sides trisect the diagonal." According to your drawing, The chosen vertex is A and the two lines from it are AE and AF. The diagonal is BD. Trisect means...

The same as what? For the third Newton ring, you are given that D = 0.730 mm when there is air in the gap and you are asked to find what D becomes when there is water in the gap. So D changes depending on what is placed in the gap. That's because the wavelength changes from...

I don't think you should worry about ##\lambda_{lens}##. The number of wavelengths you can fit in the lens part of the path is the same regardless of whether there is air or water in the gap. What does this ##D## value in the formula represent? I think you need to understand that first.

Assume thin film interference at near normal incidence where Snell's law effects are negligible. What happens to the wavelength of the light when it travels through water instead of air? This Wikipedia article has all you need.

Y'all missed the point of "Hitchhiker's Guide." Clearly, $$\text{lim}_{x\rightarrow\infty} \frac{1}{x}=42.$$

Sure. Shown below is a quadrilateral in which one side (green) is equal to the sum of the other three. There is no circle that passes through all 4 points where the color changes. If you shorten the green segment by ##\epsilon## keeping the other three at the same length, there will be a...

OK, but your equation $$\sum_{i=1}^m \sin^{-1} \frac{L_i}{d}=\pi$$ is derived by assuming that the polygon is already inscribed in a circle. Therefore the statement cannot be a conclusion because it is an assumption in your argument. The equation is a necessary and sufficient condition for...

But there is no circumcircle in the case that you yourself proposed in post #42: If you draw the 4 perpendicular bisectors they will not all intersect at the same point. Also, what is this parameter ##d## and how is it related to the number of sides ##m##?