Applications of Prisms and Math :: Mathematics

Missing Figures Prisms and their Applications Introduction A prism is one or several blocks of glass, through which light passes and refracts and reflects off its straight surfaces. Prisms are used in two fundamentally different ways. One is changing the orientation, location, etc. of an image or its parts, and another is dispersing light as in a refractometers and spectrographic equipment. This project will only deal with the first use. Consider an image projected onto a screen with parallel rays of light, as opposed to an image formed by the same rays that are passed through a cubic prism (assume that the amount of light that is reflected is negligible). The rays that pass through the prism will not be refracted since the angle of refraction = sin-1(sin(0)/n) = 0, or reflected, so the images will be exactly the same. More generally, if the rays enter and leave a prism at right angles (Assuming the rays only travels through one medium while passing through the prism), the only effect on the image will be the reflection of the rays off of its surfaces. Since the law of reflection I= -I’ (Angle of incidence equals the negative of the angle of reflection) is not effected by the medium, the effect of the prism will be same as that of reflective surfaces or mirrors placed in the same location as the reflective surfaces of the prism. It follows that to understand prisms it is important to understand how mirrors can be used to change the direction of rays. Mirror Location Problem 1: Consider the following example: A horizontal ray is required to undergo a 45Â º-angle change and this has to be achieved using a mirror. We need to find how the mirror should be oriented to achieve the desired change of angle. Solution: Recall the Snell’s law which deals with refraction: sinI0 /n0 = sinI1/n1 if we define the incoming and outgoing rays ray and the normal of the refractive surface as vectors and using a property of the cross-product we can say the following Q0xM1 = |Q0||M1| sinI0 = sinI0 and also Q1xM1 = |Q1||M1| sinI1 = sinI1 thus N0 (Q0xM1)= n1 (Q1xM1) If we introduce two new vectors S0 and S1 and let them equal n0 Q0 and n1Q1 respectively we will get S0x M1 = S1xM1 or (S1-S0)xM1 = 0 this implies that (S1-S0) are parallel or anti-parallel, which means that we can define a new variable Γ which is called the astigmatic constant with S1 – S0 = ΓM1 How is useful for solving our problem? Applications of Prisms and Math :: Mathematics Missing Figures Prisms and their Applications Introduction A prism is one or several blocks of glass, through which light passes and refracts and reflects off its straight surfaces. Prisms are used in two fundamentally different ways. One is changing the orientation, location, etc. of an image or its parts, and another is dispersing light as in a refractometers and spectrographic equipment. This project will only deal with the first use. Consider an image projected onto a screen with parallel rays of light, as opposed to an image formed by the same rays that are passed through a cubic prism (assume that the amount of light that is reflected is negligible). The rays that pass through the prism will not be refracted since the angle of refraction = sin-1(sin(0)/n) = 0, or reflected, so the images will be exactly the same. More generally, if the rays enter and leave a prism at right angles (Assuming the rays only travels through one medium while passing through the prism), the only effect on the image will be the reflection of the rays off of its surfaces. Since the law of reflection I= -I’ (Angle of incidence equals the negative of the angle of reflection) is not effected by the medium, the effect of the prism will be same as that of reflective surfaces or mirrors placed in the same location as the reflective surfaces of the prism. It follows that to understand prisms it is important to understand how mirrors can be used to change the direction of rays. Mirror Location Problem 1: Consider the following example: A horizontal ray is required to undergo a 45Â º-angle change and this has to be achieved using a mirror. We need to find how the mirror should be oriented to achieve the desired change of angle. Solution: Recall the Snell’s law which deals with refraction: sinI0 /n0 = sinI1/n1 if we define the incoming and outgoing rays ray and the normal of the refractive surface as vectors and using a property of the cross-product we can say the following Q0xM1 = |Q0||M1| sinI0 = sinI0 and also Q1xM1 = |Q1||M1| sinI1 = sinI1 thus N0 (Q0xM1)= n1 (Q1xM1) If we introduce two new vectors S0 and S1 and let them equal n0 Q0 and n1Q1 respectively we will get S0x M1 = S1xM1 or (S1-S0)xM1 = 0 this implies that (S1-S0) are parallel or anti-parallel, which means that we can define a new variable Γ which is called the astigmatic constant with S1 – S0 = ΓM1 How is useful for solving our problem?