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Chapter 27: Problem 13

A transparent film \((n=1.43)\) is deposited on a glass plate \((n=1.52)\) to form a nonreflecting coating. The film has a thickness that is \(1.07 \times 10^{-7} \mathrm{~m}\). What is the longest possible wavelength (in vacuum) of light for which this film has been designed?

Short Answer

Expert verified
The longest possible wavelength is 612 nm.

Step by step solution

01

Understand the Problem

We are tasked with finding the longest possible wavelength of light (in vacuum) that will experience destructive interference when passing through a thin film deposited on a glass plate. The film acts as a nonreflecting coating.
02

Recall Condition for Destructive Interference

For destructive interference in thin films to occur, the optical path difference should be an odd multiple of half the wavelength in the film. This condition can be expressed as: \( 2t = (m + \frac{1}{2}) \frac{\lambda}{n_f} \), where \(t\) is the thickness of the film, \(n_f\) is the refractive index of the film, \(\lambda\) is the wavelength in vacuum, and \(m\) is an integer.
03

Use Minimum Integer for Longest Wavelength

To find the longest wavelength, we use the smallest possible integer \(m\), which is zero. Therefore, the equation \( 2t = (0 + \frac{1}{2}) \frac{\lambda_{max}}{n_f} \) becomes \( \lambda_{max} = 4tn_f \).
04

Substitute Known Values

Plugging the known values into the formula, we get: \( \lambda_{max} = 4 \times 1.07 \times 10^{-7} \times 1.43 \).
05

Calculate the Longest Wavelength

Perform the calculation to find \( \lambda_{max} \):\[ \lambda_{max} = 4 \times 1.07 \times 10^{-7} \times 1.43 = 6.1156 \times 10^{-7} \text{ m} \]
06

Convert to Desired Units

Since wavelengths are typically expressed in nanometers (nm), we convert the result: \( 6.1156 \times 10^{-7} \text{ m} = 611.56 \text{ nm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nonreflecting Coating
A nonreflecting coating is a specially designed thin film that is applied to surfaces, such as glass, to reduce the reflection of light. This is often used in applications like eyeglasses, camera lenses, and solar panels to improve transparency and efficiency. The key idea behind nonreflective coatings is to utilize the principle of interference to minimize the unwanted reflection of light.

When light hits the surface of this coating, some of it reflects off the top layer, while the rest travels through the coating and reflects off the surface it covers. By carefully choosing the thickness of the coating and its material, these two reflected beams can interfere with each other destructively. This destructive interference leads to a significant reduction in reflected light, hence the term 'nonreflecting'.

The effectiveness of the coating depends on the thickness of the film, the refractive index of the film material, and the wavelength of the incoming light. When properly designed, it can cancel out reflections over a specified range of wavelengths.
Destructive Interference
Destructive interference is a phenomenon where two or more overlapping waves cancel each other out, reducing their amplitude or brightness. In the context of thin films, this happens when the path difference of two light waves leads to them being out of phase.

When light reflects off a thin film, part of it reflects off the top surface and another part off the lower surface of the film. For destructive interference to occur, the optical path difference must be an odd multiple of half the wavelength of the light in the film. This condition ensures the two light paths interfere destructively, canceling out the light and reducing reflections.

In the calculation of the longest possible wavelength for a nonreflecting coating, we set the path difference to the smallest such value that results in destructive interference. This usually involves starting with the smallest integer \(m\), which is zero, ensuring that the conditions are met for the largest possible destructive interference wavelength.
Optical Path Difference
Optical path difference is the key measurement in determining how waves will interact when they overlap. It's the difference in the paths taken by two waves as they travel through different media. This difference can lead to constructive interference (increased intensity) or destructive interference (decreased intensity).

In thin film interference, the optical path difference between the light reflected from the top and bottom surfaces of the film determines how the waves will interfere. It's mathematically calculated as twice the thickness of the film divided by the refractive index: \(d = \frac{2t}{n_f}\).

For the film to work as a nonreflecting coating, the optical path difference is manipulated so it results in destructive interference for a particular wavelength of light. By achieving an optical path difference that is an odd multiple of the film's half wavelength in the light, reflections are minimized, achieving the desired nonreflecting effect.
  • Thin film interference: Occurs due to the interaction of light waves reflected from surfaces of different thicknesses.

  • Compatibility with wavelength: The film is optimized for specific light wavelengths where the path difference matches interference conditions.

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Most popular questions from this chapter

Under these conditions the pupils of your eyes have diameters of about \(7.0 \mathrm{~mm}\). The taillights of this car are separated by a distance of \(1.2 \mathrm{~m}\) and emit red light (wavelength \(=660 \mathrm{~nm}\) in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00)\). A thin film of liquid floats on the gasoline and has a refractive index that is between \(1.00\) and \(1.40\). Light that has a wavelength of \(625 \mathrm{~nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is \(242 \mathrm{~nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

Light shines on a diffraction grating, and a diffraction pattern is produced on a viewing screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) From trigonometry, how is the distance \(y\) from the central bright fringe to the second-order bright fringe related to the diffraction angle \(\theta\) and the distance \(L\) between the grating and the screen? (b) From physics, how is \(\theta\) related to the order \(m\) of the bright fringe, the wavelength \(\lambda\) of the light, and the separation \(d\) between the slits? (c) In this problem, the angle \(\theta\) is small (less than a few degrees). When the angle is small, \(\tan \theta\) is approximately equal to \(\sin \theta\), or \(\tan \theta \approx \sin \theta\). Using this approximation, obtain an expression for \(y\) in terms of \(L, m, \lambda\), and \(d\). (d) If the entire apparatus in the drawing is submerged in water, would you expect the distance \(y\) to increase, decrease, or remain unchanged? Why? Light of wavelength \(480 \mathrm{~nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(5.0 \times 10^{-7} \mathrm{~m}\). The distance between the grating and the viewing screen is \(0.15 \mathrm{~m}\). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right)\), what is the distance \(y ?\) Be sure your answer is consistent with part (d) of the Concept Questions.

(a) In a single-slit diffraction pattern the width of the central bright fringe is defined by the location of the first dark fringe that lies on either side of it. For a given slit width, does the width of the central bright fringe increase, decrease, or remain the same as the wavelength of the light increases? (b) For a given wavelength, does the width of the central bright fringe increase, decrease, or remain the same as the slit width increases? (c) When both the wavelength and the slit width change, it is possible for the width of the central bright fringe to remain the same. What condition must be satisfied for this to happen? In each case, give your reasoning. A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{~m}\). When light with a wavelength of \(\lambda_{1}=510 \mathrm{~nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ) and a wavelength of \(\lambda_{2}=740 \mathrm{~nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\).

The wavelength of the laser beam used in a compact disc player is \(780 \mathrm{~nm}\). Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{~mm}\) apart at a distance of \(3.0 \mathrm{~mm}\) from the grating. Estimate the spacing between the slits of the grating.
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