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Fermat's Principle a. Two poles of he ights and n are
Chapter 7, Problem 59E(choose chapter or problem)
a. Two poles of heights m and n are separated by a horizontal distance d. A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when \(\theta_{1}=\theta_{2}\) (see figure).
b. Fermat's Principle states that when light travels between two points in the same medium (at a constant speed), it travels on the path that minimizes the travel time. Show that when light from a source A reflects off of a surface and is received at point B, the angle of incidence equals the angle of reflection, or \(\theta_{1}=\theta_{2}\) (see figure).
Questions & Answers
QUESTION:
a. Two poles of heights m and n are separated by a horizontal distance d. A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when \(\theta_{1}=\theta_{2}\) (see figure).
b. Fermat's Principle states that when light travels between two points in the same medium (at a constant speed), it travels on the path that minimizes the travel time. Show that when light from a source A reflects off of a surface and is received at point B, the angle of incidence equals the angle of reflection, or \(\theta_{1}=\theta_{2}\) (see figure).
ANSWER:
Solution Step 1 (a)Consider that two poles of height and are separated by a horizontal distance apart. A rope is stretched from the top of one pole to the ground and then the top of the other pole. Consider the following figure In the above figure, let,AB = m ,CD = n ,BD = d angle APB = a1d angle CPD = 2 Now the length of the rope is given by AP +CP Let BP = x then DP = dx Then, in right triangleABP By Pythagoras Theorem, 2 2 2 AP = AB +BP = m +x 2 AP = m +x2 2 Similarly, in right triangle CDP By Pythagoras Theorem, CP = CD +DP 2 2 = n +(dx) CP = n +(dx) 2 Therefore, the length of the rope is given by, say L L = AP +CP = m +x + 2 n +(dx) 2