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3. The formula is given by n = 1 sin(C) n = 1 sin (C), n n is the refractive index of the denser medium, C is the critical angle. From this formula, it seems to be that we are substituting the angle of refraction as the angle of incidence, therefore sin(90) sin(C) sin (90) sin (C), but why can we do this? Why not sin(C) sin(90) sin (C) sin (90 ...
When light moves from water to air, it is refracted away from the vertical. For some angle (the critical angle) the angle of refraction is 90 degrees. This is the situation where n1> n2 n 1> n 2. For angles larger than the critical angle, light is reflected instead. The opposite situation is when light moves from air to water.
The more conventional formula for calculating the refractive index is: n21 = sinx1 sinx2 n 21 = sin x 1 sin x 2 so basically the refractive index of n2 n 2 with respect to n1 n 1 equals sinx1 sinx2 sin x 1 sin x 2. It works for rays originating from both rarer and denser mediums. Share. Improve this answer. Mar 14, 2018 at 21:10.
0. When light moves from a medium with a high refraction index to a medium with a low refractory index (water to vacuum), that there exists a "critical angle" at which no more refraction occurs. What is happening conceptually (chemically, physically, thermodynamically, electromagnetically, etc) at the critical angle that is stopping the light ...
The angle of tipping must be one that cancels out the torques of the weight and the applied force. With some geometry, we have the following results. ∠EQR = θ. Hence EQ = h sinθ, and RQ = hcosθ sinθ. QB = RB − RQ = r − hcosθ sinθ. Also, ∠PQB = θ. Therefore, PB = QBsinθ, and QP = QBcosθ. PB = (r − hcosθ sinθ)sinθ = rsinθ ...
The critical angle is given by: c = arcsin (n2 - n1) For a typical optical fibre, it says on the web that refractive index (n2) for cladding is higher than that of the glass core (n1) but it's only a few percent higher. For instance, let's calculate c (critical angle) for this example. c = arcsin (1.45/1.55) = 1.21 rad = 69°.
2. As far as I know, the critical angle exists only if light passes from a medium with a greater index of refraction to one with a smaller index of recreation. However, when it comes to Snell's window, a critical angle (48.6°) exists, even though light is passing from a medium with an index of refraction of 1 (air) to a medium with an index of ...
I have been studying the relation between Brewster's angle and the critical angle, and I am left with the ...
1 Answer. You have found the critical angle θc θ c at which the block begins to slide. That gives you the coefficient of static friction μs = tanθc μ s = tan θ c. Kinetic friction comes into consideration when the block is actually moving. Before the block can move, the force mg sin θ m g sin θ acting down the incline must be at least ...
0. When the first medium is air, then sin(c) = 1/n sin (c) = 1 / n where c c is the critical angle and n n is the refractive index of n2/n1 n 2 / n 1 (where n2 n 2 is second medium and n1 n 1 is in air). So, 1/(n2/n1) 1 / (n 2 / n 1) would be n1/n2 n 1 / n 2. However, when calculating with the first medium being not air, I find that you ...