Recent content by physicoo

I guess I have to relook into the question and theorems. Thanks for the advice so far :)

You're right in a way! Yes I actually did separated them into two to find both side limits. Like I said, I was referring to the thread that I had pasted earlier on. But I was unsure if it was correct since, like you said, when x approaches infinity, it is a left-sided limit, which is the reason...

May I know what's the purpose of removing the modulus? I'm not quite sure what you mean by Sorry. It's a little confusing to me. I was referring to this thread: https://www.physicsforums.com/showthread.php?t=115063 while solving my problem. But I wasn't sure if it was the same concept when...

it's through using one of the limit methods, that is to divide both the numerator and dominator by the highest power which in this case is x^2

Hi guys, I have a little problem with this question: Determine the limit (x approaching infinity) \frac{|9-3x^{2}|}{x-2x^{2}} (Sorry, I'm not sure on how to use the Latex feature) Anyways, I've combed the forum and found a couple of threads that are somewhat related to my problem...

So based on the advice, I suppose the correct answer would be D - 42000 (rounded off due to uncertainties) 'even though' the exact calculations is 41533 then?

Rounding off the values for each calculating to 2sf would result in my end result as 42000. However, I think there's a need to calculate accurately (keeping all the sfs/decimals) to determine the exact number of reflections (41533 in this case) since it determines the 'maximum' number of...

Sorry, could you explain to me what you meant by "calculate an angle to 10?" ?

The data is basically given as what the question has provided: Length of fibre - 1.0m Diameter of fibre - 20micro-m Index of refraction - 1.3 This was what I did: Sin(theta) = n(air)/n(fibre) theta = sin^-1(1/1.3) = 50.28486277 degrees From the internal reflection (and Pythogoras theorem)...

Question: An optical fibre is 1.0 metre long and has a diameter of 20micro-meter. Its ends are perpendicular to its axis. Its index of refraction is 1.30. What is the maximum number of reflections a light ray entering one end will make before it emerges from the other end? A) 18000 B)...

Think I understand now :) Thank you!

So I suppose this diffraction grating piece basically acts like a prism, 'spreading out' the light into its component colors, one of which end has the highest wavelength while the other end has the lowest wavelength?

Yup, I've read up on other sites and realized the reason why it can never exceed 90 degrees :) Thanks though! I've got a last question though it's somewhat a little more of I-just-wanna-know-why. When holding the diffraction grating to my eye and looking at light sources, I saw multiple images...

Thanks :) As for the third question, is anyone able to explain to me why the angle is unable to exceed 90 degrees? I do not understand what would happen, if it exceeds.

Yes, I've thought of that too. Which is how I derived my stand from the equation of d.sin(theta) = m(lambda). Say, in regards to the first question, if d decreases, and the wavelength and m is held constant, I find the angles increase (in fact, quite significantly). And same goes for the spread...