Snell’s Law Project - 550 Words

Snell's Law Project (Term Paper Sample) Content: 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 ÃŽÂ ¸1 ÃŽÂ ¸2 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, à ¢Ã‹â€ Ã… ¡3). For a perfect mirror, i=r. In the above diagram, i=r=30à ¢Ã‚ Ã‚ ° Tan (30à ¢Ã‚ Ã‚ °) = oppadj= 1à ¢Ã‹â€ Ã… ¡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 = à ¢Ã‹â€ Ã… ¡x2 + 22 BX = à ¢Ã‹â€ Ã… ¡(4-x)2 + 12 * The total distance is a function of x (0à ¢Ã¢â‚¬ °Ã‚ ¤xà ¢Ã¢â‚¬ °Ã‚ ¥4) fx= à ¢Ã‹â€ Ã… ¡x2 + 22 + à ¢Ã‹â€ Ã… ¡(4-x)2 + 12 = à ¢Ã‹â€ Ã… ¡x2 + 4 + à ¢Ã‹â€ Ã… ¡x2 - 8x + 17 f' (x)=xà ¢Ã‹â€ Ã… ¡(x2-4) + x-4à ¢Ã‹â€ Ã… ¡(x2-8x+17 = 0à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦.. (i) Rearranging the equation (i) we get. xà ¢Ã‹â€ Ã… ¡(x2-4) = x-4à ¢Ã‹â€ Ã… ¡(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) = 16x2à ¢Ã‹â€ Ã¢â‚¬â„¢64 à ¢Ã‹â€ Ã¢â‚¬â„¢ 8x3 à ¢Ã‹â€ Ã¢â‚¬â„¢ 32x+ 4x4 17x2= 20x2 à ¢Ã‹â€ Ã¢â‚¬â„¢ 32x + 64 3x2 à ¢Ã‹â€ Ã¢â‚¬â„¢ 32x + 64 = 0à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦ (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 = 0à ¢Ã‹â€ Ã¢â‚¬â„¢ (-1) = 1, MB = 1à ¢Ã‹â€ Ã¢â‚¬â„¢0 = 1 From Pythagoras Theorem, both the hypotenuses are equal (2 units each). The perpendicular height is, à ¢Ã‹â€ Ã… ¡22 à ¢Ã‹â€ Ã¢â‚¬â„¢ à ¢Ã‹â€ Ã… ¡12 = à ¢Ã‹â€ Ã… ¡3 -298184938837200 à ¢Ã‹â€ Ã‚ ´tan (i) = tan  ® = 1à ¢Ã‹â€ Ã… ¡3 = 30à ¢Ã‚ Ã‚ ° ABà ¢Ã¢â€š ¬Ã‚ ² = AX + XB = à ¢Ã‹â€ Ã… ¡ (AB)2 + à ¢Ã‹â€ Ã… ¡(PB')2 NB: P is point at (0, -1) AP = 2 à ¢Ã‹â€ Ã¢â‚¬â„¢ -1 = 3 PBà ¢Ã¢â€š ¬Ã‚ ² = 4 à ¢Ã¢â€š ¬Ã¢â‚¬Å" 0 = à ¢Ã‹â€ Ã… ¡32 + 42 = à ¢Ã‹â€ Ã… ¡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 Ià ¢Ã¢â‚¬ °Ã‚  R, v1v2. à ¢Ã‹â€ Ã‚ ´, 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) = à ¢Ã‹â€ Ã… ¡(x2+1)v1 Time taken by the ray of light to travel distance XB is, DISTANCE (XB)SPEED (V2) = à ¢Ã‹â€ Ã… ¡(x2-4x+5)v2 Total time required to travel AX and XB is a function of f (x) = x2+1v1 + x2-4x+5v2 f à ¢Ã¢â€š ¬Ã‚ ²(x) = 1v1 (xà ¢Ã‹â€ Ã… ¡x2+1) + 1v2 (x-2à ¢Ã‹â€ Ã… ¡(x2-4x+5)) In trigonometric models, we have (xà ¢Ã‹â€ Ã… ¡x2+1) = Sin(I) (x-2à ¢Ã‹â€ Ã… ¡(x2-4x+5)) = Sin(R) 1v1* Sin(I) = 1v2* Sin(R) à ¢Ã‹â€ Ã‚ ´ 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 à ¢Ã‹â€ Ã‚ ´ 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...