Missing figuresPrisms and their applicationsIntroductionA prism is one or more blocks of glass, through which light passes, refracts and reflects on its straight surfaces. Prisms are used in two fundamentally different ways. One is changing the orientation, position, etc. of an image or its parts, and another is dispersing the light as in refractometers and spectrographic equipment. This project will only cover first use. Consider an image projected onto a screen with parallel light rays, as opposed to an image formed by the same rays passing through a cubic prism (we assume that the amount of reflected light is negligible). The rays passing through the prism will not be refracted since the angle of refraction = sin-1(sin(0)/n) = 0, nor reflected, so the images will be exactly the same. More generally, if rays enter and exit a prism at right angles (assuming that the rays travel only through one medium as they pass through the prism), the only effect on the image will be the reflection of the rays off its surfaces. Since the law of reflection I= -I' (the angle of incidence is equal to the negative of the angle of reflection) is not affected by the medium, the effect of the prism will be the same as that of reflecting surfaces or mirrors positioned in the same position as the reflecting 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 the rays. Position of mirrors Problem 1: Consider the following example: A horizontal ray must undergo an angle change of 45º and this must be achieved using a mirror. We need to find how the mirror needs to be oriented to achieve the desired change in angle. Solution: Let's remember Snell's law which deals with refraction: sinI0 /n0 = sinI1/n1if we define the incoming and outgoing ray and the surface normal of the refractive ray as vectors and using a cross product property we can say the followingQ0xM1 = |Q0|| M1| sinI0 = sinI0and alsoQ1xM1 = |Q1||M1| sinI1 = sinI1therefore N0 (Q0xM1)= n1 (Q1xM1)If we introduce two new vectors S0 and S1 and leave them equal to n0 Q0 and n1Q1 respectively we will obtain S0x M1 = S1xM1o(S1-S0)xM1 = 0this implies that (S1-S0) they are parallel or antiparallel which means we can define a new variable Γ which is called astigmatic constant with S1 – S0 = ΓM1How is this useful to solve our problem?