/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 A thin film of thickness \(t\) a... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 41

A thin film of thickness \(t\) and index of refraction \(1.33\) coats a glass with index of refraction \(1.50 .\) Which of the following thickness \(t\) will not reflect normally incident light with wavelength \(640 \mathrm{~nm}\) in air? (A) \(120 \mathrm{~nm}\) (B) \(240 \mathrm{~nm}\) (C) \(300 \mathrm{~nm}\) (D) \(480 \mathrm{~nm}\)

Short Answer

Expert verified
None of the given thicknesses (A) $120 \mathrm{~nm}$, (B) $240 \mathrm{~nm}$, (C) $300 \mathrm{~nm}$, and (D) $480 \mathrm{~nm}$ will result in destructive interference and hence will not prevent the reflection of normally incident light with a wavelength of $640 \mathrm{~nm}$ in air.

Step by step solution

01

Calculate the effective wavelength in the thin film

Since the wavelength of light changes when it passes through a medium with different indices of refraction, we first need to calculate the effective wavelength in the thin film. To do this, we will use the formula: effective wavelength = wavelength in air / index of refraction of thin film
02

Analyze Option (A)

For option (A), t = 120 nm. Using the formula mentioned above, let's check if this thickness will result in a path difference equivalent to an odd multiple of half the wavelength: (2 * 120 * 1.33) = (2n + 1) * (640 / (2 * 1.33)) We can simplify this equation to: 160 = 2n + 1 Solving for n, we find that n = 79.5. Since n is not an integer, the path difference will not result in destructive interference, and this thickness will not fulfill the condition of no reflection.
03

Analyze Option (B)

For option (B), t = 240 nm. Using the formula mentioned above, let's check if this thickness will result in a path difference equivalent to an odd multiple of half the wavelength: (2 * 240 * 1.33) = (2n + 1) * (640 / (2 * 1.33)) We can simplify this equation to: 320 = 2n + 1 Solving for n, we find that n = 159.5. Since n is not an integer, the path difference will not result in destructive interference, and this thickness will not fulfill the condition of no reflection.
04

Analyze Option (C)

For option (C), t = 300 nm. Using the formula mentioned above, let's check if this thickness will result in a path difference equivalent to an odd multiple of half the wavelength: (2 * 300 * 1.33) = (2n + 1) * (640 / (2 * 1.33)) We can simplify this equation to: 400 = 2n + 1 Solving for n, we find that n = 199.5. Since n is not an integer, the path difference will not result in destructive interference, and this thickness will not fulfill the condition of no reflection.
05

Analyze Option (D)

For option (D), t = 480 nm. Using the formula mentioned above, let's check if this thickness will result in a path difference equivalent to an odd multiple of half the wavelength: (2 * 480 * 1.33) = (2n + 1) * (640 / (2 * 1.33)) We can simplify this equation to: 640 = 2n + 1 Solving for n, we find that n = 319.5. Since n is not an integer, the path difference will not result in destructive interference, and this thickness will not fulfill the condition of no reflection. Since none of the thicknesses resulted in a path difference equivalent to an odd multiple of half the wavelength, none of the given thicknesses will not reflect the normally incident light with a wavelength of 640 nm in air.

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