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# Snell's Law Project (Term Paper Sample)

Instructions:
The task was to investigate the Snell\'s Law and verify it mathematically. This was a physics term paper. source..
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 Î¸1 & Î¸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, âˆš3). For a perfect mirror, i=r. In the above diagram, i=r=30â° Tan (30â°) = oppadj= 1âˆš3 (#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 = âˆšx2 + 22 BX = âˆš(4-x)2 + 12 * The total distance is a function of x (0â‰¤xâ‰¥4) fx= âˆšx2 + 22 + âˆš(4-x)2 + 12 = âˆšx2 + 4 + âˆšx2 - 8x + 17 f' (x)=xâˆš(x2-4) + x-4âˆš(x2-8x+17 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (i) Rearranging the equation (i) we get. xâˆš(x2-4) = x-4âˆš(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) = 16x2âˆ’64 âˆ’ 8x3 âˆ’ 32x+ 4x4 17x2= 20x2 âˆ’ 32x + 64 3x2 âˆ’ 32x + 64 = 0â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (ii) This is a quadratic equation which can be solved using the quadratic formula, x=-bÂ±b2-4ac2a, where a=3, b=-32, c= 64 Substituting for a, b, & c in the above equation, we obtain x1,2 = 32 Â± 166 = x1 =8 & x2 = 83 x=83 * 369697020447000-335009430543500 180892323116800-255460550355500 -245549048414600-3558347446681-355878846656000-279350350211900 -1938020290195-272357028114000 AM = 0âˆ’ (-1) = 1, MB = 1âˆ’0 = 1 From Pythagoras Theorem, both the hypotenuses are equal (2 units each). The perpendicular height is, âˆš22 âˆ’ âˆš12 = âˆš3 -298184938837200 âˆ´tan (i) = tan Â® = 1âˆš3 = 30â° ABâ€² = AX + XB = âˆš (AB)2 + âˆš(PB')2 NB: P is point at (0, -1) AP = 2 âˆ’ -1 = 3 PBâ€² = 4 â€“ 0 = âˆš32 + 42 = âˆš25 = 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 Iâ‰ R, v1>v2. âˆ´, 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) = âˆš(x2+1)v1 Time taken by the ray of light to travel distance XB is, DISTANCE (XB)SPEED (V2) = âˆš(x2-4x+5)v2 Total time required to travel AX and XB is a function of f (x) = x2+1v1 + x2-4x+5v2 f â€²(x) = 1v1 (xâˆšx2+1) + 1v2 (x-2âˆš(x2-4x+5)) In trigonometric models, we have (xâˆšx2+1) = Sin(I) (x-2âˆš(x2-4x+5)) = Sin(R) 1v1* Sin(I) = 1v2* Sin(R) âˆ´ 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 âˆ´ 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...
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