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Chapter 9: Problem 84

An insulated window consists of two parallel panes of glass with a small spacing between them. Suppose that each pane reflects a fraction \(p\) of the incoming light and transmits the remaining light. Considering all reflections of light between the panes, what fraction of the incoming light is ultimately transmitted by the window? Assume the amount of incoming light is 1.

Short Answer

Expert verified
Answer: The fraction of the incoming light transmitted through the window can be found using the formula \(T = \frac{(1-p)^2p}{1-p^2}\), where p is the fraction of light reflected by each pane of glass.

Step by step solution

01

Single Transmission and Reflection

First we will calculate the fraction of the incoming light that is transmitted through the first pane and then reflected back by the other pane. The incoming light is transmitted by the first pane at a fraction \(1-p\), and reflected back by the second pane at a fraction \(p\), so the total amount of incoming light fraction that is reflected back by the first interval is \((1-p)p\).
02

Accounting for Multiple Reflections

The light will continue to bounce between the two panes and will transmit the infinite series of reflections. The reflected light after each transmission and reflected follows a geometric sequence \((1-p)p, (1-p)p^3, (1-p)p^5, ...\).
03

Determine the Sum of the Infinite Geometric Sequence

The sum of an infinite geometric sequence can be represented by the formula: \(S = \frac{a}{1 - r}\) Where \(S\) is the sum of the sequence, \(a\) is the first term of the series, and \(r\) is the common ratio between the terms. In our case, \(a=(1-p)p\) and \(r=p^2\). So the sum of the sequence is: \(S = \frac{(1-p)p}{1 - p^2}\)
04

Calculate the Fraction of the Incoming Light Transmitted

Now we can calculate the fraction of the incoming light that is actually transmitted through the window by accounting for the sum of reflections. The fraction of the incoming light transmitted after the multiple reflections is the product \(1-p\) and \(S\) (since each transmission also reduces the incoming light by a fraction \(1-p\)). Fe: \(T = (1-p)S\) Replacing S in the above equation, the transmitted light fraction will be: \(T = (1-p)\times\frac{(1-p)p}{1 - p^2} = \frac{(1-p)^2p}{1-p^2}\) So the fraction of the incoming light that is ultimately transmitted by the window is: \(T = \frac{(1-p)^2p}{1-p^2}\)

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Most popular questions from this chapter

Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using. the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}},\) for \(n=0,1,2,3, \ldots .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}}\) where \(p > 0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty}(-1)^{k} k$$

Suppose a ball is thrown upward to a height of \(h_{0}\) meters. Each time the ball bounces, it rebounds to a fraction r of its previous height. Let \(h_{n}\) be the height after the nth bounce and let \(S_{n}\) be the total distance the ball has traveled at the moment of the nth bounce. a. Find the first four terms of the sequence \(\left\\{S_{n}\right\\}\) b. Make a table of 20 terms of the sequence \(\left\\{S_{n}\right\\}\) and determine a plausible value for the limit of \(\left\\{S_{n}\right\\}.\) $$h_{0}=20, r=0.5$$

It can be proved that if a series converges absolutely, then its terms may be summed in any order without changing the value of the series. However, if a series converges conditionally, then the value of the series depends on the order of summation. For example, the (conditionally convergent) alternating harmonic series has the value $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2.$$ Show that by rearranging the terms (so the sign pattern is \(++-\) ), $$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2} \ln 2.$$

Explain the fallacy in the following argument. Let \(x=\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots .\) It follows that \(2 y=x+y\), which implies that \(x=y .\) On the other hand, $$x-y=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots>0$$ is a sum of positive terms, so \(x>y .\) Thus, we have shown that \(x=y\) and \(x>y\).
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