please include a diagram and the equation with variables before you plug in your numbers so that I can follow along.
i see other posts of similar problems saying you add the angle of incidence with the 37.5° but i dont understand WHY people are doing that... i dont see how theyre similar triangles and i dont understand what angle the 37.5° is speaking to. is it the top left corner or? thank you(:
thank you(:
Please Include A Diagram And The Equation With Variables Before You Plug In Your Numbers So That I Can Follow Along I S 1 (26.47 KiB) Viewed 46 times
As shown below, light from a vacuum is incident on a shard of Shawtonium (a newly discovered compound). The backside of the shard is up against an unknown material. When the light strikes the backside of the shard, total internal reflection occurs. The light then emerges from the side of the shard and resumes traveling through a vacuum. The index of retraction of Shawtonium is 2.5. Determine the speed of light in Shawtonium, 6, & the upper bound of unknown 66° 37.5° unknown vacuum. shard 0 Vchan 120000000 m ✔ upper bound of nie 0-
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