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Light of wavelength620nmilluminates a diffraction grating. The second-order maximum is at angle 39.5°. How many lines per millimeter does this grating have?

Short Answer

Expert verified

There aren=512lines per millimeter does this grating have

Step by step solution

01

Step1:  definition of light of wavelength

As a result, the wavelength is defined as the distance between the crest or trough of one wave and the crest or trough of the next wave. The wavelength of light is defined as "the distance between the light wave's two successive crests or troughs."

02

Find how many lines per meter

In order to obtain constructive interference from a diffraction grating, the following conditions must be met.

dsinθm=mλ

m=0,1,2,3,

We can use the previous equation to calculate the distance between any two successive slits d if we know the angle of the second-order maximum and the wavelength used to illuminate the grating.

d=2λsinθ2=2×620×109msin39.5=1.95×106m

The number of lines per millimetre can now be calculated as

n=1mmd=1×103m1.95×106m=512

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Most popular questions from this chapter

Scientists shine a laser beam on a 35-μm-wide slit and produce a diffraction pattern on a screen 70cmbehind the slit. Careful measurements show that the intensity first falls to 25%of maximum at a distance of7.2mmfrom the center of the diffraction pattern. What is the wavelength of the laser light?

Hint: Use the trial-and-error technique demonstrated in Example 33.5to solve the transcendental equation.

To illustrate one of the ideas of holography in a simple way, consider a diffraction grating with slit spacing d. The small-angle approximation is usually not valid for diffraction gratings, because dis only slightly larger than λ, but assume that the λ/dratio of this grating is small enough to make the small-angle approximation valid.

a. Use the small-angle approximation to find an expression for the fringe spacing on a screen at distance Lbehind the grating.

b. Rather than a screen, suppose you place a piece of film at distance L behind the grating. The bright fringes will expose the film, but the dark spaces in between will leave the film unexposed. After being developed, the film will be a series of alternating light and dark stripes. What if you were to now “play” the film by using it as a diffraction grating? In other words, what happens if you shine the same laser through the film and look at the film’s diffraction pattern on a screen at the same distance L? Demonstrate that the film’s diffraction pattern is a reproduction of the original diffraction grating

Because sound is a wave, it's possible to make a diffraction grating for sound from a large board of sound-absorbing material with several parallel slits cut for sound to go through. When 10kHzsound waves pass through such a grating, listeners 10mfrom the grating report "loud spots" 1.4mon both sides of center. What is the spacing between the slits? Use 340msfor the speed of sound.

FIGURE shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating . As a practical matter, two peaks can just barely be resolved if their spacing yequals the width w of each peak, where wis measured at half of the peak’s height. Two peaks closer together than wwill merge into a single peak. We can use this idea to understand the resolution of a diffraction grating.

a. In the small-angle approximation, the position of the m=1peak of a diffraction grating falls at the same location as the m=1fringe of a double slit: y1=λL/d. Suppose two wavelengths differing by lpass through a grating at the same time. Find an expression for localid="1649086237242" y, the separation of their first-order peaks.

b. We noted that the widths of the bright fringes are proportional to localid="1649086301255" 1/N, where localid="1649086311478" Nis the number of slits in the grating. Let’s hypothesize that the fringe width is localid="1649086321711" w=y1/NShow that this is true for the double-slit pattern. We’ll then assume it to be true as localid="1649086339026" Nincreases.

c. Use your results from parts a and b together with the idea that localid="1649086329574" Δymin=wto find an expression for localid="1649086347645" Δλmin, the minimum wavelength separation (in first order) for which the diffraction fringes can barely be resolved.

d. Ordinary hydrogen atoms emit red light with a wavelength of localid="1649086355936" 656.45nm.In deuterium, which is a “heavy” isotope of hydrogen, the wavelength is localid="1649086363764" 656.27nm.What is the minimum number of slits in a diffraction grating that can barely resolve these two wavelengths in the first-order diffraction pattern?

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0cm. You have a helium neon laser (λ=633nm)and a0.12-mm-diameter pinhole. How far behind the pinhole should you place the screen that’s to be photographed?

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