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Chapter 36: Problem 34

Camera Lens A camera lens with index of refraction greater than \(1.30\) is coated with a thin transparent film of index of refraction \(1.25\) to eliminate by interference the reflection of light at wavelength \(\lambda\) that is incident perpendicularly on the lens. In terms of \(\lambda\), what minimum film thickness is needed?

Short Answer

Expert verified
t = \frac{\lambda}{5}

Step by step solution

01

- Understanding the problem

To eliminate the reflection of light at wavelength \(\lambda\) by interference, we need to determine the minimum thickness of a thin transparent film (with index of refraction 1.25) that is coated on a camera lens with a higher index of refraction (greater than 1.30).
02

- Formula for constructive interference

The condition for destructive interference (which eliminates reflection) at a certain wavelength for thin film interference is given by: \[2t = (m + 0.5) \frac{\lambda}{n_{film}}\] where \(t\) is the thickness of the film, \(m\) is an integer (0,1,2,...), \(\lambda\) is the wavelength of light, and \(n_{film} = 1.25\).
03

- Minimum film thickness

We are looking for the minimum thickness, so we set \(m = 0\). This simplifies to: \[2t = (0 + 0.5) \frac{\lambda}{1.25}\] \[ 2t = 0.5 \frac{\lambda}{1.25}\] Simplifying further, \[2t = \frac{\lambda}{2.5}\]
04

- Solving for thickness t

Finally, solve for \(t\): \[ t = \frac{\lambda}{5} \] Thus, the minimum film thickness needed is \( \frac{\lambda}{5} \).

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Key Concepts

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

Index of Refraction
The index of refraction, also known as the refractive index, is a measure of how much a light wave bends, or refracts, as it passes from one medium to another. The refractive index is defined by the ratio of the speed of light in a vacuum to the speed of light in the medium. Mathematically, it is expressed as:
\(n = \frac{c}{v} \), where
  • \(n\) is the index of refraction.
  • \(c\) is the speed of light in a vacuum.
  • \(v\) is the speed of light in the medium.
In the context of the exercise, the camera lens has a refractive index greater than 1.30, meaning that light slows down significantly when entering the lens compared to a vacuum. The thin transparent film, with a refractive index of 1.25, is used to manipulate how light interacts with the lens surface.
Destructive Interference
Destructive interference occurs when two waves combine to form a resultant wave of reduced amplitude. In the case of thin films, this happens when the path difference between the waves reflected from the film's top and bottom surfaces causes them to be out of phase. For light waves, this means the crests of one wave align with the troughs of another, canceling each other out.
For the thin film scenario, destructive interference can be used to minimize reflection. This requires satisfying the condition: \[2t = (m + 0.5) \frac{\lambda}{n_{film}}\]
where
  • \(t\) is the film thickness.
  • \(m\) is an integer (0,1,2,...).
  • \(\lambda\) is the wavelength of the light.
  • \(n_{film}\) is the refractive index of the film.
This formula ensures that the reflected waves interfere destructively, effectively reducing the reflection at the specific wavelength \(\lambda\).
Minimum Film Thickness
To achieve destructive interference, the thickness of the thin film must be carefully calculated. The exercise asks for the minimum film thickness that eliminates reflection for a specific wavelength of light. We start by setting \(m = 0\) in the destructive interference formula:
\[2t = (0 + 0.5) \frac{\lambda}{1.25}\]
Simplifying, we get:
\[2t = 0.5 \frac{\lambda}{1.25}\]
Dividing both sides by 2, we find:
\[t = \frac{\lambda}{5}\]
Thus, the minimum thickness for the film to eliminate reflection due to destructive interference is \(\frac{\lambda}{5}\).
This calculation ensures that any light of wavelength \(\lambda\) reflected from the top and bottom surfaces of the film will be out of phase and cancel each other out.

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Most popular questions from this chapter

Microwave Transmitter In Fig. 36-27, a microwave transmitter at height \(a\) above the water level of a wide lake transmits microwaves of wavelength \(\lambda\) toward a receiver on the opposite shore, a distance \(x\) above the water level. The microwaves reflecting from the water interfere with the microwaves arriving directly from the transmitter. Assuming that the lake width \(D\) is much greater than \(a\) and \(x\), and that \(\lambda \geq a\), at what values of \(x\) is the signal at the receiver maximum? (Hint: Does the reflection cause a phase change?)

Yellow Sodium Light The wavelength of yellow sodium light in air is \(589 \mathrm{~nm} .\) (a) What is its frequency? (b) What is its wavelength in glass whose index of refraction is \(1.52 ?\) (c) From the results of (a) and (b) find its speed in this glass.

Two Glass Plates Two glass plates are held together at one end to form a wedge of air that acts as a thin film. A broad beam of light of wavelength \(480 \mathrm{~nm}\) is directed through the plates, perpendicular to the first plate. An observer intercepting light reflected from the plates sees on the plates an interference pattern that is due to the wedge of air. How much thicker is the wedge at the sixteenth bright fringe than it is at the sixth bright fringe, counting from where the plates touch?

Directly Downward In Fig. \(36-25\), white light is sent directly downward through the top plate of a pair of glass plates. The plates touch at the left end and are separated by a wire of diameter \(0.048\) \(\mathrm{mm}\) at the right end; the air between the plates acts as a thin film. An observer looking down through the top plate sees bright and dark fringes due to that film. (a) Is a dark fringe or a bright fringe seen at the left end? (b) To the right of that end, fully destructive interference occurs at different locations for different wavelengths of the light. Does it occur first for the red end or the blue end of the visible spectrum?

Find Sum Find the sum \(y\) of the following quantities: $$ y_{1}=0 \sin \omega t \text { and } y_{2}=8.0 \sin \left(\omega t+30^{\circ}\right) \text { . } $$
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