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Michael Lustwerk

Wavelength of Light: The Grating Spectroscope

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Objective: To use a grating spectroscope to measure the wavelengths of spectrum lines from mercury,

and examine the continuous spectrum emitted by a hot source.

Introduction: When people see a source of light, they see a single color composed of all the colors

emitted by the source. A spectroscope enables the separation of these colors so they can be studied.

Here we will use a diffraction grating to do this. Since each color corresponds to a wavelength of light,we can describe the spectrum emitted by a light source based on the wavelengths that comprise it.

There are two types of emission spectra, continuous spectra and line spectra. When the light from a

glowing solid is broken down into its spectrum of colors, the spectrum is shown to be composed of all

wavelengths of light from violet to red. This is known as a continuous spectrum and the intensities of the

wavelength regions depend on the temperature of the light source. For a gas however, there are only a

limited number of wavelengths making up the light. Therefore, the spectroscope will show a number of

colored lines. Therefore the spectrum in this case is known as a line spectrum. The wavelengths and

relative intensities of the lines depend on the particular gas. Since different elements and compounds

emit different line spectra, this process also helps identify unknown materials. In this experiment the

spectral lines of mercury will be measured and the wavelengths compared with accepted values.

A transmission grating is made of a large number of fine, evenly spaced slits very close together. For rays

of light diffracted through an angle from the incident light direction, the rays can be focused on a point

I by means of a converging lens. If the separation between slits in the grating is d, then the path

difference between adjacent slits is dsin. If the path difference is equal to an integral number of

wavelengths, then the rays will all arrive at I in phase and therefore interfere constructively to produce a

bright fringe of light there. The condition for this fringe is dsin=n, where n=0,1,2, which is the same

relationship for two slits separated by a distance d. Between the maxima, there is destructive

interference and there will be no light. For a point O that is directly across from the grating, if we move

away from ), there will first be a dark area and then a bright region, where dsin1 =, meaning n=1. This

is the first order spectrum. Continuing away from O, there will be another bright region at n=2, the

second order spectrum, at an angle 2. Here dsin 2=2. This pattern of light and dark happens on both

sides of O. The locations of the maxima do not depend on the number of slits in the grating, but the

intensities of the maxima increase and the widths decrease as the number of slits increase, making the

bright regions sharper and brighter. If the light source has only one wavelength, then the image is only a

single colored image of the source, and the angle is unique for each order. However, if the source has

several different wavelengths, then at each order there will be several different colored regions, one for

each wavelength. A laser is an example of a monochromatic source.

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Procedure: There are three parts to the experiment. The first is a measurement of d, the distance

between slits in the diffraction grating. Knowing d, the wavelength of a diode laser will then be

measured using the grating formula, and finally, the continuous spectrum from a filament light will be

examined.

First, to determine the grating constant, use light of known wavelength and use the grating formula,dsin=n. Here we will use a diode laser with a wavelength of 650 nm. The laser is pointed at a screen

with the grating in between. Adjust the distance L between the grating and screen so that the first and

second order images are on the screen. Record the values of L, y, and n, where y is the distance between

the center of the spectrum, O, and the spectrum line. To find y1 for the first order image, n=1, and y2 for

the second order image, n=2, the positions of the images to the left and right of O are found which yield

the values 2y1 and 2y2. Finding these values means not having to find the position of O. Repeat this

process for a larger distance L so the first order images are at the ends of the screen. For each set of

measurements calculate the grating constant d (in centimeters) using the grating formula, and compute

the average value of the measurements. Given the length of the grating, determine the number of lines

on the grating.

Now go to the diode laser setup, where a 2 meter ruler serves as a screen and is mounted close to a slit,

with their planes perpendicular to a line joining them. The grating is mounted away from the ruler and

slit, and the diode laser is mounted behind the slit. Looking through the grating, see the images of the

slit on either side of O. The first and second order spectra should be seen above the ruler. Since the

lasers spectrum is composed of a number of wavelengths, in each order the wavelengths will be

diffracted through different angles and the image of the slit is a line spectrum composed of a series of

bright colored lines. Find each pair of lines, and find the distance between them to get 2y for each line

for both n=1 and n=2. Using tan=y/L, find the angle for each of the line spectra. Do this for the blue and

green light in the spectrum. Using the grating formula and the average d calculated, calculate thewavelength of each line in the first and second order spectra. From these values find the average

wavelength for each line.

Now go to the setup with a filament lamp. Record observations of the boundary wavelengths of each

colored region of the spectrum.

Analysis:

Experiment 1

n=1 n=2

y1(cm) tan (degrees) y2(cm) tan

L=70cm L=50cm

13.95 0.1993 11.27 21.4 0.428 23.17

L=50cm d1 d2 d3

10.1 0.202 11.42 3.3265E-6 3.28283E-6 3.304E-6

Average d: 3.3044E-6 Lines/cm:3026

Table 1

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Experiment 2

n=1 n=2 L=100cm

Y (cm) tan (degrees) Y tan

8.2 0.082 4.668 26.75 0.2675 14.976 Blue

17 0.17 9.65 34.3 0.343 18.932 green

270nm 427nm Blue

550nm 536nm green

Table 2

Experiment 3

Outer bound (cm) 18.5cm

Inner bound (cm) 9.75cm

tan in (degrees) 0.113

tan out (degrees) 0.214

in (degrees) 6.431

out(degrees) 12.072in(nm) 370

out(nm) 690

Table 3

For experiment 1, as the actual lines per centimeter was not known, there is no way to determine the

accuracy of the experiment. For experiment 2, the experiment was reasonably successful, as the final

calculated wavelength for n=2 for blue was 427 was reasonably close to the lower bound of the blue

spectrum, 450nm, and the green was within the actual green spectrum, as the obtained value was

550nm and the actual range is from 495-570nm. The third experiment was also relatively successful, as

the final calculated range for the wavelengths, 370nm to 690nm, is almost the same as the actual range

for the lower bound, 380nm, and is not very far from the upper bound, 750nm.

Conclusion: Overall, although there was some definite error in the calculations, most notably in

experiment 1 and part of experiment 3, they were close enough to declare the lab a success.

Questions:

1. dsin=n, for =4000 angstroms = 4E-7m and n=1, =6.902 degrees. For n=2, =13.61 degrees.For =7000 angstroms=7E-7m and n=1, =11.96 degrees. For n=2, =22.96 degrees. From this,

for n=1 the angles range from 6.902 t0 11.96 degrees and for n=2 from 13.61 t0 22.96 degrees.

In the second order, spectra are farther apart, making them easier to discern, and therefore

making it easier to find measurements of distance and angles compared with n=1, where they

may be close together and more difficult to tell one line from another.