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Chapter 36: Problem 105

Show that a grating made up of alternately transparent and opaque strips of equal width eliminates all the even orders of maxima (except \(m=0\) ).

Short Answer

Expert verified
Even orders are eliminated because they result in destructive interference for alternating transparent and opaque strips.

Step by step solution

01

Understanding the Grating Equation

The grating equation is given by \( d \sin\theta = m\lambda \), where \( d \) is the distance between adjacent slits, \( \theta \) is the angle of diffraction, \( m \) is the diffraction order, and \( \lambda \) is the wavelength of light. For maxima to occur, this condition must be satisfied.
02

Constructive Interference for Even Orders

A transmission grating with equal transparent and opaque regions will produce constructive interference when the path difference between light from adjacent slits is an integer multiple of the wavelength. For even orders (except \( m=0 \)), the phase difference between light from adjacent slits is \( (2m+1)\pi \), resulting in destructive rather than constructive interference.
03

Concept of Phase Difference

For light waves passing through a grating, each order \( m \) corresponds to a specific angular displacement where the phase difference is a multiple of \( 2\pi \). Since the strips are of equal width, for even \( m \), the phase difference leads to cancellation because half of one cycle occurs in the transparent part and half in the opaque part.
04

Conclusion for Even Orders

Because the interference for even orders results in destructive interference, alternating strips cause the light waves originating from these parts to cancel each other. Hence, only odd orders are observed, meaning even orders vanish except for \( m=0 \), where there is no path difference.

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Key Concepts

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

Constructive Interference
Constructive interference occurs when two or more light waves intersect and combine to form a wave with a larger amplitude than any of the individual waves. This happens with a diffraction grating when the path difference between waves from adjacent slits is an integer multiple of the wavelength (\( m\lambda \)). This results in bright fringes or maxima.
  • The waves meet in phase, which means that their peaks and troughs align perfectly.
  • The result is a bright spot on the screen because the amplitudes are added together.
  • In a grating, for odd orders, the waves combine constructively.
Destructive Interference
Destructive interference is the opposite of constructive interference and occurs when waves combine in such a way that they cancel each other out. This results in darker regions or minima. In a diffraction grating with equal transparent and opaque strips,
destructive interference is crucial in understanding why even orders are diminished.
  • The key condition is that the path difference is an odd multiple of \((\pi)\).
  • For even orders, destructive interference occurs as \((2m+1)\pi\) leads to waves being out of phase.
  • This results in cancellation of the light and hence no bright fringes for even orders.
Grating Equation
The grating equation is fundamental to the study of diffraction. It's an expression that relates the distance between grating lines \(d\), the wavelength \(\lambda\), the angle of diffraction \(\theta\), and the order of the maximum \(m\).
The formula is \(d\sin\theta = m\lambda\). This equation is vital in determining the angles at which the bright fringes occur.
  • It accounts for the conditions necessary for constructive interference to occur.
  • With equal transparent and opaque areas, even \(m\) doesn't satisfy these conditions due to destructive interference resulting from phase shifts.
Phase Difference
Phase difference refers to the difference in phase between two waves. It's an important concept in understanding interference. When light passes through the grating, each wave might be slightly out of sync with the others.
  • A full cycle in a wave is \(2\pi\), and phase difference measures how far 'off' they are from that full cycle.
  • For waves to be constructive, the phase difference must be a multiple of \(2\pi\).
  • In grating scenarios with equal strip widths, an odd phase difference (like \((2m+1)\pi\) for even orders) leads to destructive interference.
Thus, understanding phase difference is crucial for predicting where interference patterns will form.

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Most popular questions from this chapter

An \(x\) -ray beam of wavelength \(A\) undergoes first-order reflection (Bragg law diffraction) from a crystal when its angle of incidence to a crystal face is \(23^{\circ}\), and an x-ray beam of wavelength 97 pm undergoes third-order reflection when its angle of incidence to that face is \(60^{\circ} .\) Assuming that the two beams reflect from the same family of reflecting planes, find (a) the interplanar spacing and (b) the wavelength \(A\).

A grating has 400 lines/mm. How many orders of the entire visible spectrum \((400-700 \mathrm{nm})\) can it produce in a diffraction experiment, in addition to the \(m=0\) order?

The pupil of a person's eye has a diameter of \(5.00 \mathrm{~mm}\). According to Rayleigh's criterion, what distance apart must two small objects be if their images are just barely resolved when they are \(250 \mathrm{~mm}\) from the eye? Assume they are illuminated with light of wavelength \(500 \mathrm{nm}\).

In the two-slit interference experiment of Fig. \(35-10\), the slit widths are each \(12.0 \mu \mathrm{m},\) their separation is \(24.0 \mu \mathrm{m},\) the wavelength is \(600 \mathrm{nm},\) and the viewing screen is at a distance of \(4.00 \mathrm{~m} .\) Let \(I_{P}\) represent the intensity at point \(\bar{P}\) on the screen, at height \(y=70.0 \mathrm{~cm}\). (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where \(P\) is in the two-slit interference pattern by giving the maximum or minimum on which it lies or the maximum and minimum between which it lies. (c) In the same way, for the diffraction that occurs, determine where point \(P\) is in the diffraction pattern.

Assume that the limits of the visible spectrum are arbitrarily chosen as 430 and \(680 \mathrm{nm}\). Calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of \(20.0^{\circ}\).
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