Recent content by cgaleb

I believe its just a single answer problem. The wording "find the value of the initial angle of incidence for which total internal reflection can take place" is a bit tricky. But I feel great having finally sovled the intial angle of incidence, regardless. THANK YOU so much for all of your help!

Alright, I plugged in the work just like you said, and came up with \Theta i =51.1, \Theta 2 =37.8, and \Theta 3 =52.2. Hoping that this is right... If so, what do I do next, I'm still confused about my professor's tan\Theta = sin \Theta / cos \Theta Which # would I use for \Theta in...

Oh, guess I'll add this as it may help N(glass)sin\Theta3=n(liquid)sin90o So, sin \Theta3=sin(90o-\Theta2) Therefore 1.65 (1-sin\Theta2)=1.3 (1) and sin\Theta2=.21 So, \Theta2=12.12 Is my logic sound?

Yeah-you are right about the typo. The professor said that \Theta3=90-\Theta2, so I took 90o-12.12 (my \Theta2=77.88. All of my angles add up to 180o. (o=degrees) 90o+12.12o+77.88o=180. We covered all of this stuff 3 chapters today, and he really only touched on the topics, giving us...

Homework Statement In the figure shown below, a light ray enters a glass slab (with index of refraction nglass = 1.65 ) at point A and then undergoes total internal reflection at point B. The medium that surrounds the glass slab is a liquid whose index of refraction, nliquid =...

Yeah, I actually came across that a few minutes ago and tried to plug in what I knew. which would give me UE=qV, or (4x10^-6)(4.8x10^5)=1.92 Joules. I plugged in the answer for V that I got before, and used the given charge. Does this sound right?

Homework Statement Four point charges are individually brought from infinity and placed at the corners of a square whose sides are 0.30 m each. Each charge has the identical value + 4.0 mC. What is the electric potential energy of these four charges? Homework Equations V=kq/r (or...