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FIGUREP33.49shows the interference pattern on a screen 1.0mbehind an 800line/mmdiffraction grating. What is the wavelength (in mm) of the light?

Short Answer

Expert verified

The wavelength's light is=500nm.

Step by step solution

01

Step: 1 Interference pattern:

When parallel plane waves of the same frequency meet at an angle, a simple sort of interference pattern results. Interference is fundamentally a process of energy redistribution. The energy wasted during destructive interference is recovered during positive interference.

02

Step: 2 Equating equation:

The angle of fringes as

Y1=Ltan11=tan1Y1L

Equation of grating wavelength is

localid="1649148192850" dsin1==1103800sintan1Y1L

03

Step: 3 Obtaining wavelength:

The wavelength to be

=1103800sintan10.4361=1.25106sin23.56=5.00107m=500nm.

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Most popular questions from this chapter

3. FIGURE Q33.3 shows the viewing screen in a double-slit experiment. FringeCis the central maximum. What will happen to the fringe spacing if

a. The wavelength of the light is decreased?

b. The spacing between the slits is decreased?

c. The distance to the screen is decreased?

d. Suppose the wavelength of the light islocalid="1649170567955" 500nm. How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit?

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鈥檚 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=L/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鈥檚 hypothesize that the fringe width is localid="1649086321711" w=y1/NShow that this is true for the double-slit pattern. We鈥檒l 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" ymin=wto find an expression for localid="1649086347645" min, 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 鈥渉eavy鈥 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?

Light of wavelength550nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of the m=2 bright fringe, how much farther is it to the more distant slit than to the nearer slit?

a. Green light shines through a 100-mm-diameter hole and is observed on a screen. If the hole diameter is increased by 20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.

b. Green light shines through a 100m-diameter hole and is observed on a screen. If the hole diameter is increased by20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.

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

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