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 to change the orientation, location, etc. of an image or its parts, and another to disperse light as in a refractometer 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 passing through a cubic prism (assume that the amount of reflected light is negligible). Rays passing 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 rays enter and exit a prism at right angles (assuming that the rays pass through only 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 the reflecting surfaces or mirrors placed in the same place. like 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 must undergo an angle change of 45º and this must be achieved. using a mirror. We need to find how the mirror should be oriented to achieve the desired change in angle.Solution:Remember Snell's law which deals with refraction: sinI0 /n0 = sinI1/n1if we define the radius of the incoming and outgoing rays and the normal of the refractive ray. surface as vectors and using a property of the cross product, we can say the following: Q0xM1 = |Q0||M1| sinI0 = sinI0and alsoQ1xM1 = |Q1||M1| sinI1 = sinI1so 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 = S1xM1 or (S1-S0)xM1 = 0 this implies that (S1-S0) are parallel or anti-parallel which means we can define a new variable Γ which is called the astigmatic constant with S1 – S0 = ΓM1How is this useful to solve our problem?