Recent content by OhBoy

Aha! Thank you! The being off by 30 degrees should of really gave it away. Thanks again for your help :)

There's a ray of light going right on the triangle's left face horizontally. It is then refracted downwards and hits the right face. It is then further refracted downwards. The 49.47 degrees should be the angle between the ray of light and the right side of the triangle.

Homework Statement A light ray traveling in the horizontal direction is incident onto a prism as shown in the figure. At what angle relative to horizontal does the light ray emerge from the second face of the shown prism if the prism has an index of refraction of 1.5 and is surrounded by air...

Gotcha. Thanks for your help

So the red light gets refracted by angle x, purple light by angle y, Angle y is less than angle x. We can get the change in vertical direction relative to when it hits left side by taking tan of angle x and multiplying it by the width. Then you can get the change vertical height of purple light...

Ouch. For some reason I was assuming part a was telling us that the front and back sides are parallel. Thought I was only working with the prism and getting the spread from the right side of the prism :S. Sorry. Light enters at 67 degrees and should leave at 67 degrees so there is no spread.

Why aren't the specifics relevant? It wants a specific angle spread. The colors are arranged in order of low frequency to high frequency. It appears to diverge to a greater degree than inside the prism.

When I just use an equilateral triangle and plug in the values to get the angle from the normal of the right side the light leaves, I got 36.8 degrees for red and 37.124 degrees for the blue end. I got the angle coming in on the right side by using the angle refracted on left side, subtracting...

Homework Statement Dispersion in a window pane. In Figure (a) below , a beam of white light is incident at angle θ1 = 67o on a common window pane (shown in cross section). For the pane's type of glass, the index of refraction for visible light ranges from 1.535 at the blue end of the spectrum...

Aha! Cheers. Thanks for your patience.

Do you think this is a good way to think about it? When in series, current is trying to increase (I and di is positive), so you want to make the term counter that so it will result in a drop in potential (making it - L di/dt). When inductor was charged with switch closed, then opened with a...

Okay I'm lost again (sorry). A negative di refers to a drop in current right? It goes in the opposite direction of the current. So the V arrow points down instead of up, meaning crossing it (along direction of current) should yield an increase in potential.But in order to get the right equation...

That's a bit unsettling... So if we have a LR in series with a battery, kirchhoffs law still predits V - IR - L di/dt? So what determines whether or not it's gaining or losing porential is in the "di" bit, so we don't even need to worry about the sign? This is a bit confusing.. So, when setting...

That's how I currently am thinking about the situation. But when I put when you just described in numbers, I get an incorrect formula. When you "walk" the circuit in the direction of the current flow thiis appears as a potential rise (+L di/dt), the resistor shows a potential drop in the...

I don't know any other way to set up the equations other than use kirchhoffs law.. If there was no inductor, the current through the resistor would pretty much instantly go to 0. But with an inductor, shouldn't it supply a potential which is why it takes a lot longer for the current to die out...