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AlexandervonHumboldt2
 one year ago
Prism Tutorial
re posted from previous time because it was accidently deleted
thx to ashley for retriving my tutorial
AlexandervonHumboldt2
 one year ago
Prism Tutorial re posted from previous time because it was accidently deleted thx to ashley for retriving my tutorial

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AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3Prism  an optical element made of transparent material (for example, optical glass) in the form of a geometric body  a prism having a flat polished face through which light enters and exits. The light is refracted in a prism. The most important characteristic of the prism is the refractive index of the material from which it is made The simplest type of prism is a triangular prism, it is the body that represents the geometric shape prism with two triangular bases and three side faces in the form of rectangles. The figure shows a section of a triangular prism plane parallel to its base. Legend: \[\Delta\] ,  the angle of deflection, \[\omega \] ,  refractive angle of the prism, \[a _{1}, b_{2}\]  the angle of refraction of the two beams, respectively. In this figure, the material of the prism  an optically denser medium than its surroundings, because the angle of incidence of the incoming beam refraction angle is greater than he. That is, the relative refractive index of the material  is greater than one, which we denote by \[n\] ,. The simplest formula for the deflection angle is obtained, assuming that the refractive angle of the prism and the angle of incidence of the incoming beam is small . Then it will be small and the angle \[a_{2}\] thus there will be small angles \[b_{1}, b_{2}\]. According to the law of refraction: \[a_{1}\approx \sin a_{1}=n*\sin b_{1}\approx n*b_{1}\] \[a_{2}\approx \sin a_{2}=n* \sin b_{2}\approx n*b_{2}\]\ Given that the sum of the angles of a quadrilateral is \[2\Pi \] and taking into mind \[\omega=b_{1}+b_{2}\] \[\pi\delta+\pi\omega+a_{1}+a_{2}=2\pi\] \[\delta=a_{1}+a_{2}\omega \approx n*(b_{1}+b_{2})\omega=n*\omega\omega=(n1)*\omega \] Thus, at a low angle of incidence of the incoming beam we have an approximate formula for the angle of deflection: \[\delta \approx (n1)*\omega\] This formula is more important because it can help to bring the dependence of the focal length of a thin lens of the radii of its surfaces, with a thin lens is replaced by a triangular prism and a formula is applied to the deflection angle. In the case of arbitrary angle of the refracting prism and the angle of incidence of the incoming beam, and if the absolute refractive index of the prism material is equal to \[n_{2}\] and its surroundings \[n_{1}\] . Similar considerations may obtain a formula \[\delta=a_{1}\omega+\arcsin(\frac{ n_{2} }{ n_{1} }*\sin \omega*\sqrt{(1(\frac{ n_{1} }{ n_{2} })^2*\sin^2a_{1}}\cos \omega *\sin a_{1})\] diagram https://upload.wikimedia.org/wikipedia/commons/thumb/6/6f/Lom_hranol.svg/587pxLom_hranol.svg.png

Zale101
 one year ago
Best ResponseYou've already chosen the best response.1Awesome tutorial. Keep up the good work :) @AlexandervonHumboldt2

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0yes i agree with astro

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3thx zale there is also a sphere tutorial opened in calculus1

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1my question is that the post refers to figures... where are they..? and is the source wikipedia..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1oh, and why is a Physics tutorial in Maths..? just a thought..

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3i thought it is a math tutorial

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well it talks about refected light... and snell's law... I think... that's physics to me... but then all I know is gravity sucks...

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3it is a 1/2*(math+physics) tutorial

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so what is the source of the information..?

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3what do you mean?

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3just wikipedia had same illustration i needed and i decided to use it,, but i dont use it as my own...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well were did you get the information from, and isn't the normal thing to cite references...

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3any idiot knows by the URL that the pic of from wiki

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so did you take the wikipedia and rewrite it..?

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3i learned info from wiki and my own knowledge but this is not a plaguate!

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3i didn;t copy and pasted!

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well if you are going to publish, such as on Open Study, the laws of liable, and plagarism still apply.. so people need to be aware of cite references... and they would in any academic paper...

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.3maybe you are right.
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