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Chapter 21: Problem 16

The fraction of nonreflected radiation that is transmitted through a \(5-\mathrm{mm}\) thickness of a transparent material is \(0.95\). If the thickness is increased to \(12 \mathrm{~mm}\), what fraction of light is transmitted?

Short Answer

Expert verified
Answer: The absorption coefficient (µ) can be found using the formula µ = -ln(F1) / t1, where F1 is the initial transmitted fraction (0.95) and t1 is the initial thickness (5 mm). By plugging in the values, we get µ = -ln(0.95) / 5.

Step by step solution

01

Find the absorption coefficient

First, we will use the Beer-Lambert Law to find the absorption coefficient (µ) of the transparent material. The formula for Beer-Lambert Law is: Transmitted Fraction = e^(-µ * Thickness) We know the initial thickness (t1 = 5 mm) and Transmitted Fraction (F1 = 0.95), so we can solve for the absorption coefficient using: F1 = e^(-µ * t1) Rearrange the formula to isolate µ: µ = -ln(F1) / t1 Now, substitute the known values and solve for µ: µ = -ln(0.95) / 5

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Most popular questions from this chapter

At the end of Section \(21.14\) it was noted that the intensity of light absorbed while passing through a \(16-\mathrm{km}\) length of optical fiber glass is equivalent to the light intensity absorbed through a \(25-\mathrm{mm}\) thickness of ordinary window glass. Calculate the absorption coefficient \(\beta\) of the optical fiber glass if the value of \(\beta\) for the window glass is \(10^{-4} \mathrm{~mm}^{-1}\).

In your own words, briefly describe the phenomenon of photoconductivity.

Zinc selenide has a band gap of \(2.58 \mathrm{eV}\). Over what range of wavelengths of visible light is it transparent?

(a) In your own words, briefly describe the phenomenon of luminescence. (b) What is the distinction between fluorescence and phosphorescence?

Briefly explain why the magnitude of the absorption coefficient ( \(\beta\) in Equation 21.18) depends on the radiation wavelength.
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