Snell's Law Project (Term Paper Sample)

Name Instructor Course Date Snell's Law Project Snell's Law is also called the Law of Refraction. It explains the relationship between the angles of refraction and incidence for a wave hitting on an interface between 2 media with different refraction indices. The objective of the project was to verify Snell's Law using a mathematical model. left188788 In this case, n1 & n2 are the refractive indices of media 1& 2 in that order. While θ1 & θ2 represent incident & refracted angles respctively. (#1) Assume we have a light source at the point A (-1,3) and the x-axis is a mirrored surface. A ray of light travels from A toward the origin and ends up at B (1, √3). For a perfect mirror, i=r. In the above diagram, i=r=30⁰ Tan (30⁰) = oppadj= 1√3 (#2) Optimization Assume the light source is at A (0, 2). If the ray bounces off the x-axis and end up at B (4, 1) Let x be the coordinate on the x-axis Since I =R, we can find the total distance Travelled by the light wave. Let X be the point at which the incident Ray hits the x-axis before bouncing off. OX = √x2 + 22 BX = √(4-x)2 + 12 * The total distance is a function of x (0≤x≥4) fx= √x2 + 22 + √(4-x)2 + 12 = √x2 + 4 + √x2 - 8x + 17 f' (x)=x√(x2-4) + x-4√(x2-8x+17 = 0…………………………….. (i) Rearranging the equation (i) we get. x√(x2-4) = x-4√(x2-8x+17 Cross-multiplying the equation and squaring both sides to eliminate the root sign, we get x2(x2-8x+17)-(4-x)2(x2-4) x4- 8x3+ 17x2=(16-8x+x2) (x2-4) = 16x2−64 − 8x3 − 32x+ 4x4 17x2= 20x2 − 32x + 64 3x2 − 32x + 64 = 0…………………………… (ii) This is a quadratic equation which can be solved using the quadratic formula, x=-b±b2-4ac2a, where a=3, b=-32, c= 64 Substituting for a, b, & c in the above equation, we obtain x1,2 = 32 ± 166 = x1 =8 & x2 = 83 x=83 * 369697020447000-335009430543500 180892323116800-255460550355500 -245549048414600-3558347446681-355878846656000-279350350211900 -1938020290195-272357028114000 AM = 0− (-1) = 1, MB = 1−0 = 1 From Pythagoras Theorem, both the hypotenuses are equal (2 units each). The perpendicular height is, √22 − √12 = √3 -298184938837200 ∴tan (i) = tan ® = 1√3 = 30⁰ AB′ = AX + XB = √ (AB)2 + √(PB')2 NB: P is point at (0, -1) AP = 2 − -1 = 3 PB′ = 4 – 0 = √32 + 42 = √25 = 5 (# 4) I is the incident light ray while R is the refracted ray. Assume speeds of light in air and water are v1 & v2 respectively, Since I≠R, v1>v2. ∴, geometrical method to find x is not feasible. x is the point at the Interface hit by incident ray. * Time taken by ray to travel distance AX is DISTANCE (AX)SPEED (v1) = √(x2+1)v1 Time taken by the ray of light to travel distance XB is, DISTANCE (XB)SPEED (V2) = √(x2-4x+5)v2 Total time required to travel AX and XB is a function of f (x) = x2+1v1 + x2-4x+5v2 f ′(x) = 1v1 (x√x2+1) + 1v2 (x-2√(x2-4x+5)) In trigonometric models, we have (x√x2+1) = Sin(I) (x-2√(x2-4x+5)) = Sin(R) 1v1* Sin(I) = 1v2* Sin(R) ∴ Sin(I)V1 = Sin(R)V2 (b). by definition, v1=cn1, where v1 is the refraction index of air = 1 Also, n of water is approximately 43, so v2 = c43 = 3c4 ∴ Sin(I)cn1 = Sin(R)cv2 n1 sin (I) = n2sin (R) Letting n1= 1 & n2=43, We arrive at, v2v1 = 34 From the fig. in #4, The coordinates of the first hypotenuse AX are (0, 1) & (x, 0); x1=0, x2 = x, y1= 1, y2= 0 The coordinates of the s...