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Snell's Law Project (Term Paper Sample)
The task was to investigate the Snell\'s Law and verify it mathematically. This was a physics term paper. source..
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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