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A diffraction grating having 500lines/mmdiffracts visible light at 30.What is the light's wavelength?

Short Answer

Expert verified

The wavelength of sunshine is =500nm.

Step by step solution

01

Step: 1 Finding the grating spacing:

The constant value of grating is 500lines/mm,the spacing by

d=1mm500d=2106m.

02

Step: 2 Equating the equation:

In equation of grating,the light diffraction light wavelength by

dsinm=m=dsinmm.

03

Step: 3 Obtaining the wavelength value:

From expression of wavelength

=2106sin30m=1000nmm.

As we're seeing, there should only be one value of mthat corresponds to a wavelength within the visible region and therfore the mwill be 2,The wavelength are500nm.

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Most popular questions from this chapter

Light from a helium-neon laser (=633nm)is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.0mbehind the slits. Twelve bright fringes are seen, spanning a distance of52mm. What is the spacing (in mm) between the slits?

In a double-slit interference experiment, which of the following actions (perhaps more than one) would cause the fringe spacing to increase? (a) Increasing the wavelength of the light. (b) Increasing the slit spacing. (c) Increasing the distance to the viewing screen. (d) Submerging the entire experiment in water.

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?

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 1cmacross, and you estimate that the distance from the window shade to the wall is about 3m.. Estimate (a) the average wavelength of the sunlight (in nm ) and (b) the diameter of the pinhole (in mm ).

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