Recent content by sally143

Here the angle of incidence, i = 53 degree.whenever reflection takes place the angle of reflection will be equal to the angle of incidence.then the angle of reflection = 53 degree. The angle between the reflected and the refracted ray = 90 degree Therefore the angle between the refracted ray...

I guess i assumed, because i was up at 2:30am, and was confused :), sorry

N2 and THeta 2, s (1.003)water sin 90? ANd N.1 sing theta1, = N1 sin ().1 = (1)sin THeta2, solve for n1?? i need the steps I am confused, trying to draw it out, but i get angle of incident comng in, perpendicular . and angle of refraction on the right side. Both having angles of 90...

Im confused, and not sure. But Incident angle is =53 degrees, refracted angle i think is =90degrees. i know that, , sin (theta).critical = n2/n1. Water is 1.33 air is 1.0003 Not sure on where to start.

Homework Statement A beam of light in air makes an incident angle with the normal at 53 with an unknown substance. Part of the light is relected and part is refracted into the substance. The reflected ray and the refracted ray make an angle of 90 degrees. a) What is the refracted angle...