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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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