The diagram at the right shows a lightray entering a rectangular block of unknown material and subsequently exiting the block on the opposite side. the path of the light ray through the block is shown. determine the index ofrefraction of the unknown material. perform two calculations - one for each boundary - using snell's law and the measured angles need on the two attachments ty!

Rectangular prism boundary 1 = 2.57 Rectangular prism boundary 2 = 2.57 Triangular prism boundary 2 = 1.37 Pentagram prism boundary 2 = 1.44 Note: You may get different values if you measured the angle differently. Don't be surprised if you're a few degrees different from me. The key thing in this problem is to actually physically measure the angles. If you don't measure them, you can't calculate the answer. Snell's law is sin a1/sin a2 = n2/n1 where a1 = angle of incidence of ray before boundary a2 = angle of incidence of ray after boundary n1 = index of refraction of medium before boundary n2 = index of refraction of medium after boundary The angle of incidence is the angle between the ray and the perpendicular to the boundary. For the rectangle in the first figure, I measured the angle of incidence for the first boundary to be 45 degrees and 16 degrees for the first surface. And for the 2nd surface, I got measurements of 16 degrees and 45 degrees. So let's plug those values into the formula and see what happens. First boundary: sin a1/sin a2 = n2/n1 sin 45/sin 16 = n2/1.0 0.707106781/0.275637356 = n2/1.0 2.565351779 = n2/1.0 2.565351779 = n2 So the index of refraction is 2.57 Second boundary: sin 16/sin 45 = 1.0/n1 0.275637356/0.707106781 = 1.0/n1 0.389810087 = 1.0/n1 0.389810087*n1 = 1.0 n1 = 1.0/0.389810087 n1 = 2.565351779 And it's no surprise that also shows an index of refraction of 2.57 For the right triangle on the same figure, the entry is parallel to the surface normal, so there's no data on the index of refraction for that ray. But it exits the triangle and the measurements I get is 45 degrees for the ray prior to the boundary and 75 degrees for the ray after the boundary. Assuming n = 1.0 for air outside the triangle, we get sin 45/sin 75 = 1.0/n1 0.707106781/0.965925826 = 1.0/n1 0.732050808 = 1.0/n1 0.732050808*n1 = 1.0 n1 = 1.0/0.732050808 n1 = 1.366025404 And the index of refraction of the triangle is 1.37 For the pentagon: Second boundary. Incoming angle = 36 degrees Outgoing angle = 58 degrees So: sin a1/sin a2 = n2/n1 sin 36/sin 58 = 1.0/n1 0.587785252/0.848048096 = 1.0/n1 0.693103675 = 1.0/n1 0.693103675*n1 = 1.0 n1 = 1.0/0.693103675 n1 = 1.442785597