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Saturday, December 26, 2009

Edition 6: IIT entrance guidance


In this article we will discuss some fundamental limits imposed by nature. You definitely have studied the related subjects for IIT-JEE but may not have analyzed the subjects from the point of view that we will discuss.
Problem 1 : Mathematical upper limit on growth
The other day I went to a bank to make some fixed deposit for my father. The bank offered 8% flat interest rate for a period of one year. Upon consulting with my father, I decided to check out the rate at another bank. They too offered 8% annual interest rate however, they offered to deposit & compound the interest every six months thus affectively raising the annual yield beyond 8%. While a third bank offered the same annual rate, they offered to deposit and compound the interest every 4 months and a fourth bank did the same thing every 3 months, thus pushing up the annual yield higher and higher. Perplexed, I returned home and wanted to find out what if there exists a bank which deposits and compounds the interest monthly -> weekly -> daily -> hourly -> minutely -> secondly -> micro-secondly - > …… -> instantaneously ! Would I get infinite annual yield this way?
To solve the problem, let's express the yield in terms of mathematics for the first bank. Suppose the interest rate is r % annually. So if you deposit Rs 1.00, after a year you will get
Net Amount = 1.00 + 1.00 * ( r/100) ; we will denote the quantity (r/100) with x. So we can say
Net Amount = 1 + x .
For the second bank, after six months the Net Amount = 1 + x/2 ; and after another six months the
Net Amount = ( 1 + x/2 ) ( 1+ x/2) = (1 + x/2) ^ 2 ( this is > 1 + x, prove yourself)
For a bank which deposits every 4 month, the
Net Amount = (1+x/3) ^ 3. (this is greater than (1+x/2) ^ 2, prove yourself)
So in general the Net Amount = (1 + x/n) ^ n, where the bank deposits & compounds the money in 'n' equal terms within a year.
So the bank which deposits instantaneously, the
Net Amount = (1 + x/n) ^ n , where n -> infinite.
= exp ^ x ; where exp is the exponential .
This is a finite quantity and therefore it is a limit imposed by nature. This is applicable to quantities which grow/decay instantaneously like growth a bacteria culture, decay of radioactive mass. You can prove the above results with calculus as well. Basically the growth of a quantity Y (net Amount), with respect t X (time), is proportional to the present value of Y (because of compounding)
dY/dX = Y => Y = exp ^X (ignoring the constants)
Problem 2 : fundamental law of propagation of light
Now let's discuss another limit imposed by nature on the propagation of light. This is time it is a minimum limit (physical) and not a maximum.
We have studied refraction in Ray Optics. The law of refraction (Snell's law) says
Refractive index = sine ( i ) / sine ( r ) , which is a constant , given the two media. My question to you is why is this so? What is achieved by nature by following this trigonometric calculation? Typically Sine and Cosine functions occur only in laws involving vector quantities. How does the poor light ray know about trigonometry? Huh !
Let's look at the attached diagram to understand the law of refraction. Light ray starts from point A (0,h) and reaches point B(d, h) after refracting at point C (x, 0).
We have chosen the co-ordinates of these points very carefully to simplify the mathematics. Velocity of light in the media above X axis is V1 while it's velocity below X axis is V2.
We have drawn three lines from A -> B
1. The left most one passes through point 'P' : such a path will be taken if V1 < v1 =" V2"> V2
To generalize the situation we will use the path through C, whose co-ordinate is (x,0). The incident angle and refracted angles are 'I' and 'r' respectively at point C.
Our next objective is to measure the time taken by light to travel between A and B.
The distance between A and C will be travelled in time = [sqrt (h^2 + x^2)] / V1
The distance between C and B will be travelled in time = [ sqrt {h^2 + (d - x)^2}] / V2
This follows from measuring the distances using Pythagoras' rule and dividing the same by velocity in that medium.
So the total time , which is a function of x, is
T(x) = [sqrt (h^2 + x^2)] / V1 + [ sqrt {h^2 + (d - x)^2}] / V2
Now we are interested to find out the maxima and minima of this function T, with respect to x.
To do so, we take first order derivative, T'(x) , and equate to zero. We get
x / [ V1 * sqrt (h^2 + x^2 ) ] = (d x) /
[ V2 * sqrt {h^2 + (d-x)^2 }]
l V1/V2 = [ x / sqrt (h^2 + x^2 ) ] / [(d-x) / sqrt {h^2 + (d-x)^2 }]
Notice that Sine(i) = x/ sqrt (h^2 + x^2 ) and Sine(r) = (d-x)/ sqrt {h^2 + (d - x)^2} Therefore we can write that the local minima or maxima occurs when
Sine(i) / Sine(r) = V1/V2.
I want you to carry on the second order derivative of T(x) and prove that it is indeed a minimum.
So , we have concluded that, light ray will take a path in which the time taken to reach its destination is minimum. And that's the secret behind Snell's law.
Why is this so ? the answer to this question requires reference to very advanced Quantum Theory put forward by Richard Feynman. In short it is “Quantum particles (like photons) can sniff around the vicinity of its path and find out the one which takes minimum time. A macro object like a bullet can't sniff around and hence it would have taken a straight path from A to B. The capability to 'sniff round' empowers light ray at the surface of separation (C) with the knowledge that its velocity would change in the neighborhood of point C and hence it alters it's path. When there is no change in velocity it keeps travelling in the same straight path just like any other macro object will do!”.

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