Snell’s law or the Law of Refraction is .Consider a light ray which travels from fixed points P to Q. The ray will choose the quickest route (through the point X, which is not fixed) and so minimize the time of travel. The light ray has velocity CA in medium A, and encounters an interface with a medium B along the x axis. The velocity of the light in medium B is CB.
a) Find an expression in terms of the variable x, for the distance travelled from P to Q, via X.
Using Pythagorean Theorem;
Total distance travelled through Medium A and Medium B;
b) Hence, find an expression for the time taken to travel from P to Q.
Time taken (T);
Light travels at a constant velocity, v, in each medium.
Medium A;
Substituting CA;
Time through Medium A;
Medium B
Substituting CB;
Time through Medium A;
Total time (T) taken for the array to travel from P to Q;
c) By minimising this time, prove Snell’s Law.
Minimizing T;
Rewriting T;
Minimizing the Equation for T;
Simplifying the derivative;
From the diagram;
Substituting and ;
Question 2
A cable has the following constants: R = 10 Ω, L = 0.1´10-3 H, G = 1´10-6 siemens and C = 0.1´10-9 F . If w = 10000 rad.s-1 find:
a) The characteristic impedance, Z0 , where Z0 =
Substituting the values of R, L, G, and C in Z0;
Converting into polar form;
The Characteristic impedance Z0 will be the square root of the magnitude and the division of the phase angle by 2.
b) The propagation coefficient, γ, given by g = (R + jwL)(G + jwC) where Re(g)³ 0. The units of γ are m-1.
Substituting the values of R, L, G, and in γ gives;
Converting into polar form;
c) The wavelength, λ, where l = , where b = Im(g).
The propagation constant b is the value of the imaginary part of g.
d) The velocity of propagation, v =
Question 3
A common design for cooling towers is modelled by a branch of a hyperbola with equation x2 = a2 + by2, where a and b are constants, rotated about an axis. The design is used because
of its strength and efficiency. Not only is the shape stronger than either a cylinder or a cone, it takes less material to build. This shape also maximises the natural upward draft of hot air, without the need for fans.
How much concrete is required to build a cooling tower that has a base 140 m wide that is 135 m below the vertex if the top of the tower is 95 m wide and 40 m above the vertex? Assume the walls are a constant thickness of 15 cm.
Solution
Volume of the outer hyperbola is given by;
Substituting x2 = a2 + by2
Where y1 is the distance of the top from the vertex and y2 is the distance of the base base from the vertex.
The coordinates for the base are (-70, -135) and (70, -135). The coordinates for the top are (-47.5, 40) and (47.5, 40).
Substituting the values for a and b;
Volume of the inner hyperbola is given by;
Substituting x2 = (a-15)2 + by2
Required volume of concrete;
