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Diffraction Grating Measurement of the Wavelength of Light

In: Science

Submitted By knight8e
Words 4741
Pages 19
1. Introduction

The aims of this experiment are to use diffraction gratings to measure the wavelength of light. Visible light is an electromagnetic radiation which is visible to human eye. Light source of different colors and wavelengths are often produced by a visible light source. A light source such as incandescent light bulb utilizing hot metal filaments is consisted of a continuous distribution of lights with a range of wavelengths, hence forming a white light. On the other hand, light produced from a mercury lamp and helium lamp contains only several discrete wavelength components.
A number of methods can be adopted to measure the separated wavelength components. However, for this experiment, we will be focusing on using diffraction grating in which light from the above three types of sources will be dispersed into its constituent colors through diffraction grating and then we are able determine the wavelength of light.

2. Objective

2.1. By comparing and analyzing the pattern of the three different light sources, we can understand the difference between a continuous and a discrete spectrum.

2.2. To be able to understand that the light produced has a spectrum that is a characteristic of the elements from the experiment results of mercury lamp and helium lamp.

2.3. To obtain or calculate the average spacing, given the characteristics wavelengths of a mercury lamp from our first experiment.

2.4. To determine the characteristic wavelengths of a helium lamp and an incandescent light bulb using the spacing obtained in experiment 1. (This is for the second and third experiment).

3. Theory

3.1. Continuous and discrete spectrum
When light is separated into its wavelength components, the resulting array of colors formed is called a “spectrum”. When a light source gives out the entire colors of visible light and one color fades with another, the spectrum is called a “continuous spectrum”. This spectrum can be obtained from heating solid metal filaments like incandescent light bulb.

In comparison to continuous spectrum, other light sources such as helium lamp and mercury lamp, they produce a “discrete spectrum” or “line spectrum”. A “discrete spectrum” appears mostly dark with a few discrete lines of colors. These light sources are produced by discharges in a gas of a distinct chemical element. The wavelength of the light emitted are characteristic of electronic structure of the element contained in the source. The term “line spectrum” is used as the images obtained are usually images produced from a narrow slit.

3.2. Diffraction grating method
A diffraction grating is a piece of transparent materials on which engraved with parallel lines through the accuracy machining technique. The distance between the lines is known as the “grating spacing” or d, which is a few times larger than the wavelength of the visible light. For instance, wavelength of visible light which ranges from 400nm to 700nm, their grating spacing d is estimated to be between 1000nm to 2000nm.

The wavelengths of light are associated with the colors of the light as seen by the human eye.
Starting from a shortest wavelength to a longest wavelength, the order of the colors is violet, blue, green, yellow, orange, and red. The actual range of the visible spectrum is slightly different for different individuals, and there may be a distinct difference in the ability of two experimental partners to see the wavelength colors at either end of the spectrum. It is often very difficult for some people to see the colors of very short wavelengths.

Light rays from the source will pass through the grating at all angles with respect to their original paths when the light rays strike the transparent portion of the diffracting grating. An interference image will occur when the output adjacent rays are in phase which means that the path length differs by an integer number of wavelengths of the light. The first image will be formed when the path difference between the adjacent rays is exactly equal to one wavelength. According to Figure 1, this condition will be true at all angle, θ1 such that the equation

λ = d sin θ1 (1)

is satisfied. At a larger angle θ2 , when the path difference between the adjacent rays from adjacent ruled lines is exactly equal to 2λ, then the equation will be

2λ = d sin θ2 (2)

must be satisfied. In general an image will be formed at any angle θn for which path difference between the adjacent rays from adjacent ruled lines is equal to nλ, where n is an integer called “order number”. Thus, the equation will be

nλ = d sin θn (3)

Figure 1 Ray diagram showing the condition required for the formation of a first order diffraction image.
Equation 3 is only applicable and valid only for integer values of n and for angle θn up to 90°. For given value λ and d, the maximum n value corresponds to the largest value of sinθ (where sinθ≤1). In this experiment, measurement will only be made on the first order even though we are able to see both the first and second order images from the experiment.

4. Experimental Setup and theory


1. Optical bench
2. Diffraction grating (600 lines/mm replica grating)
3. Spectrum-tube power suppliers
4. Ruler and slit arrangement
5. Mercury and helium lamps
6. Incandescent light bulb (10 W)

4.2 Method Explanation
The experiment set up is shown in Figure 2. Light coming from a light source passes through the tiny slit and reflected from the grating. This grating separates the light into different components of wavelength. When we look through the grating, the colorful images will appear on a white background ruler. Thus we can then locate the position of those bright color lines.
Firstly, the distance L from grating to the slit is kept fixed and constant at 50 cm throughout the whole experiment.
The ruler is perpendicular to the axis of the optical bench and level.
The slit is at 30.00-cm mark.
The first-order image will correspond to a particular wavelength component will correspond to a particular angle θ based on equation (1) and a particular distance D according to Figure 2. The angle can be determined using the formula tanθ=D/L. After obtaining the calculated angle, we can determine the grating spacing d from the formula λ = d sin θ, where n = 1 since we are doing only for the first order. Once we obtained the grating spacing d, we can use this calculated value for the grating (wavelengths which is valid because the same grating will be used throughout this experiment), of a helium light source and light bulb.

Figure 2 Arrangement of the diffraction grating, slit, light source and optical bench

Through equation tanθ=D/L (Figure 2), we can calculate angle θ. Given wavelength λ and this calculated θ, we can get d from equation “sin θ = λ/d”. However, given calculated d instead, we can get λ, using the same equation.

5. Experimental procedure
Please check the followings. The ruler (or meter stick) should be perpendicular to the axis of the optical bench and it should be level. The slit should be located at the center of the meter ruler (60cm) or at the 30cm mark. The distance, L between the grating and the slit should be approximately 50cm. Record the value of L (nearest to 0.1mm) in Data Table 1.
This experiment will be divided into 3 parts. They are Mercury Lamp, Helium Lamp and 10-W incandescent light bulb. This experimental setup will be used throughout the whole of the experiment.

5.1. Mercury Lamp
1). Mercury lamp should be placed behind the slit. Adjust its position such that the narrow output portion of the lamp is the most intense and is well aligned with the slit.

2). Turn on the power supply. While one of the partners is looking directly at the slit, the other partner will adjust the position of the lamp to place the bright narrow portion of the lamp in good alignment with the slit. Also, the light source should be positioned in such a way that that slit is a bright as possible and the images seen by the other partner is as brightest as possible.

3). Observe the grating from the left and the right of the slit. A series of images of different colors could be observed on the white background ruler.

4). Look through the grating. With the help of a pencil pointer, we can determine the position of PR the first order images (7 color lines) that are to the right of the slit. Record this values (nearest to 0.1mm) the position DR (DR = PR – 30cm) of each of the 7 wavelengths in the data table 1.

5). Similarly, we can read the position of PL of the first order images that are located on the left of the slit. Read the value of DL (DL = 30cm – PL) and record it (to the nearest 0.1mm) in the data table 1. Then calculate the average value of= (DL + DR) / 2, Calculate tanθ =/L, θ, sinθ and d from dsinθ = λ. Record all this values to four significant figures. Calculate the mean value and the standard error using the equation given in the appendix. .

5.2. Helium Lamp
1). Turn off the power supply to the mercury lamp. Remove the mercury lamp set and change it to the helium lamp and its power supply. Turn on the power supply and similarly repeat the experiment we did in 5.1.1), adjust the position for alignment.

2). Try to match the images we can see with the given colors of the corresponding wavelengths. Repeat the procedure 5.1.4 and 5.1.5. Record this value (to the nearest 0.1mm) the data for DR, DL and in Data Table 2. After obtaining the result, turn off the power supply of the helium lamp. You might like to perform the following calculations after you have completed the whole experiment. Calculate and record this values (to four significant values) in Data Table 2) tanθ =/L, θ, sinθ. Use the value obtained in calculation table 1 to help you determine the value of λ and record it. Compare the values of the given λ and the calculated λ values for the discussion.

5.3. 10-W incandescent light bulb
1). Use the same procedure, remove the helium lamp and change in incandescent light bulb. Turn on the power supply and adjust the position for alignment similar to what we did previously.

2). Record (to the nearest 0.1mm) the positions PR and PL in Data Table 3 of the specific wavelength of the spectrum: a) the shortest wavelength visible; (b) the division between blue and green; (c) the division between green and yellow; (d) the division between yellow and orange; (e) the division between orange and red; and (f) the longest wavelength visible. Note, for example, tha the division between the blue and green images means the middle line dividing the blue and green images.. Use the same calculation procedure to determine average value of= (DL + DR) / 2, tanθ =/L, θ, sinθ and λ = sinθ (to four significant figures) for each wavelength and record it on the Calculation Table 3.

6. Experimental Results and Calculations

6.1 Data Table 1 (Mercury Lamp)

L=__________50.00_________cm | Colors | λ(nm) | PR(cm) | PL(cm) | Violet | 404.7 | 42.6 | 17.5 | Violet | 407.8 | 42.7 | 17.3 | Blue | 435.8 | 43.8 | 16.5 | Blue-green | 491.6 | 45.7 | 14.5 | Green | 546.1 | 47.5 | 12.6 | Yellow | 577.0 | 48.5 | 11.5 | Yellow | 579.0 | 48.6 | 11.4 |

Calculations Table 1

λ(nm) | DR(cm) | DL(cm) | (cm) | tanθ | θ | sinθ | d(nm) | 404.7 | 12.6 | 12.5 | 12.6 | 0.2520 | 14.14° | 0.2444 | 1655.9 | 407.8 | 12.7 | 12.7 | 12.7 | 0.2540 | 14.25° | 0.2462 | 1656.4 | 435.8 | 13.8 | 13.5 | 13.7 | 0.2740 | 15.32° | 0.2643 | 1648.9 | 491.6 | 15.7 | 15.5 | 15.6 | 0.3120 | 17.33° | 0.2978 | 1650.8 | 546.1 | 17.5 | 17.4 | 17.5 | 0.3500 | 19.29° | 0.3304 | 1652.8 | 577.0 | 18.5 | 18.5 | 18.5 | 0.3700 | 20.30° | 0.3470 | 1662.8 | 579.0 | 18.6 | 18.6 | 18.6 | 0.3720 | 20.41° | 0.3487 | 1660.5 | * = 1655.4 nm | αd= 1.90 nm |

You will need to show here the steps involved in obtaining d for a particular wavelength valueλ only. You will also need to show here the steps involved in obtaining the standard error αd.
DR= PR -30 cm DL= 30 cm – PL L = 50cm
= (DR+ DL)/2 tanθ =/L θ = tan-1(/L) dsinθ = nλ (n=1, d= ) =λ/sinθ

(APPENDIX – EQUATION (2)) = ((1/ (7-1)) * ((1655.9 – 1655.4)2 + (1656.4 – 1655.4)2 + (1648.9 – 1655.4)2 + (1650.8 – 1655.4)2 + (1652.8 – 1655.4)2 + (1662.8 – 1655.4)2 + (1660.5 – 1655.4)2) = ((1/6)*(25.365)) ½ = 5.04 (APPENDIX – EQUATION (1)) = (5.04)/ (71/2) = 1.90
6.2 Data Table 2 (Helium Lamp)

L=_________50.00_________cm | Wavelengths(nm) | Colors | PR(cm) | PL(cm) | 438.8 | Blue-violet | 42.0 | 16.0 | 447.1 | Deep Blue | 44.2 | 15.8 | 471.3 | Blue | 44.7 | 15.2 | 492.2 | Blue-green | 45.5 | 12.5 | 501.5 | Green | 45.8 | 14.0 | 587.6 | Yellow | 48.7 | 11.0 | 667.8 | Red | 52.0 | 8.0 | 706.5 | Red | 53.4 | 6.4 |

Calculations Table 2

DR(cm) | DL(cm) | (cm) | tanθ | θ | sinθ | λ(nm) | 14.0 | 14.0 | 14.0 | 0.2800 | 15.64° | 0.2696 | 446.3 | 14.2 | 14.2 | 14.2 | 0.2840 | 15.85° | 0.2732 | 452.3 | 14.7 | 14.8 | 14.8 | 0.2960 | 16.49° | 0.2838 | 469.8 | 15.5 | 15.5 | 15.5 | 0.3100 | 17.22° | 0.2961 | 490.2 | 15.8 | 16.0 | 15.9 | 0.3180 | 17.64° | 0.3030 | 501.6 | 18.7 | 19.0 | 18.9 | 0.3780 | 20.71° | 0.3536 | 585.3 | 22.0 | 22.0 | 22.0 | 0.4400 | 23.75° | 0.4027 | 666.6 | 23.4 | 23.6 | 23.5 | 0.4700 | 25.17° | 0.4254 | 704.2 |

You will need to show here the steps involved in obtaining a particular wavelength value λ only.
DR= PR -30 cm DL= 30 cm – PL L = 50cm
= (DR+ DL)/2 tanθ =/L θ = tan-1(/L) dsinθ = nλ λ = dsinθ (where n = 1, d = =1655.4)
1655.4 * 0.2696 = 446.3
1655.4 * 0.2732 = 452.3
1655.4 * 0.2838 = 469.8
1655.4 * 0.2961 = 490.2
1655.4 * 0.3030 = 501.6
1655.4 * 0.3536 = 585.3
1655.4 * 0.4027 = 666.6
1655.4 * 0.4254 = 704.2

6.3 Data Table 3 (10-W incandescent light bulb)

Portion of Spectrum | PR(cm) | PL(cm) | Shortest Wavelength | 42.8 | 17.4 | Division Blue and Green | 46.0 | 13.8 | Division Green and Yellow | 48.2 | 12.0 | Division Yellow and Orange | 49.0 | 11.0 | Division Orange and Red | 50.0 | 10.2 | Longest Wavelength | 54.5 | 5.6 |

Calculations Table 3

Portions of Spectrum | (cm) | tanθ | θ | sinθ | λ(nm) | Shortest Wavelength | 12.7 | 0.2540 | 14.25° | 0.2462 | 407.6 | Division Blue and Green | 16.1 | 0.3220 | 17.85° | 0.3065 | 507.4 | Division Green and Yellow | 18.1 | 0.3620 | 19.90° | 0.3404 | 563.5 | Division Yellow and Orange | 19.0 | 0.3800 | 20.81° | 0.3552 | 588.0 | Division Orange and Red | 19.9 | 0.3980 | 21.70° | 0.3698 | 612.2 | Longest Wavelength | 24.5 | 0.4900 | 26.10° | 0.4400 | 728.4 |

Involved Calculations:
You will need to show here the steps involved in obtaining a particular wavelength value λ only.
DR= PR -30 cm DL= 30 cm – PL L = 50cm
= (DR+ DL)/2 tanθ =/L θ = tan-1(/L) dsinθ = nλ λ = dsinθ (where n = 1, d = =1655.4)
1655.4 * 0.2462 = 407.6
1655.4 * 0.3065 = 507.4
1655.4 * 0.3404 = 563.5
1655.4 * 0.3552 = 588.0
1655.4 * 0.3698 = 612.2
1655.4 * 0.4400 = 728.4

7.1. Comment on the precision of your measurement of d.

Precision is the measure of degree to which further calculations show the similar result. In this experiment, the precision is measured by the Standard Deviation, of the measurements. For the precision to be reliable and accurate, the Standard Deviation should be small so that the precision will be high. For the mercury lamp, based on the data obtained in the calculation table 1, the measurements of the d result are 1655.9, 1656.4, 1648.9, 1650.8, 1652.8, 1662.8 and 1660.5 respectively. These values are quite near to the final mean value which is 1655.4 based on observation. After calculation (based on the calculation after calculation table 1), we are able to obtain our Standard Deviation, (5.04) which is relatively small. It is because it is approximately about 0.304% (5.04/1655.4*100%) of the mean value (1655.4). Hence the precision is very high and reliable.

In addition, we are also given the diffraction grating which is 600 lines/mm replica grating. From this, we are able to obtain the diffracting spacing by using the formula d = (1/N), where d is the diffracting grating and N is the number of lines. Therefore the value for d would be (0.001m/600) = 1666.67nm. Hence the actual value of d is 1666.67nm while the experimental result of which is obtained from the calculation table 1(Mercury Lamp experiment) is 1655.4. The magnitude of the percentage error is ((1655.4 – 1666.67)/1666.67)*100% = 0.676% which is very small. As a result, we could see that the precision of the result are high and satisfying enough for this experiment.

7.2. The values for the eight (8) wavelengths of the helium lamp are 438.8, 447.1, 471.3, 492.2, 501.5, 587.6, 667.8, and 706.5 nm. Calculate the percentage errors (including the signs as well) of your measured wavelength values using these given true wavelength values. Comment on the accuracy of your measurement results.

λ = 438.8 nm % Error = (446.3 - 438.8) /438.8*100% = +1.71% λ = 447.1 nm % Error = (452.3 - 447.1)/447.1*100% = +1.16% λ = 471.3 nm % Error = (469.8 - 471.3)/471.3*100% = -0.32% λ = 492.2 nm % Error = (490.2 - 492.2)/492.2*100% = -0.41% λ = 501.5 nm % Error = (501.6 - 501.5)/501.5*100% = +0.02% λ = 587.6 nm % Error = (585.3 - 587.6)/587.6*100% = -0.39% λ = 667.8 nm % Error = (666.6 - 667.8)/667.8*100% = -0.18% λ = 706.5 nm % Error = (704.2 - 706.5)/706.5*100% = -0.33%

From figure above, we can see that all the percentage error of the eight wavelengths is small. The absolute value of the percentage errors is less than 2% which means that the accuracy of the measurement is high and reliable. In conclusion, the accuracy of the measurement is high and satisfying.

7.3. If the grating had 600 lines/mm, the actual value of d would be 1667 nm. If you used that value of d (instead of the value) and values of sinθ in Calculations Table 2 in the calculations, would the resulting wavelengths for the helium lamp be better or worse than the wavelength values in that table? Justify your answer by showing the working out.

If we use the value of d which is 1667nm (given in the question), which is the actual value for the grating, and the value of sinθ (obtained in calculation table 2) we are able to calculate the value of λ. The values of λ will be affected because sinθ still remain unchanged. Based on this equation, we can calculate the value of λ. Thus, the only factors that change will be λ because the applied equation will be instead of.

Below is the diagram to show that sinθ will remain unchanged but λ will change.

Part of Changed Calculations Table 2

θ (unchanged) | sinθ(unchanged) | (nm) (changed) | 15.64° (unchanged) | 0.2696 (unchanged) | 1667*0.2696 = 449.4 | 15.85° (unchanged) | 0.2732 (unchanged) | 1667*0.2732 = 455.4 | 16.49° (unchanged) | 0.2838 (unchanged) | 1667*0.2838 = 473.1 | 17.22° (unchanged) | 0.2961 (unchanged) | 1667*0.2961 = 493.6 | 17.64° (unchanged) | 0.3030 (unchanged) | 1667*0.3030 = 505.1 | 20.71° (unchanged) | 0.3536 (unchanged) | 1667*0.3536 = 589.5 | 23.75° (unchanged) | 0.4027 (unchanged) | 1667*0.4027 = 671.3 | 25.17° (unchanged) | 0.4254 (unchanged) | 1667*0.4254 = 709.1 |

So the λ values change to 449.4, 455.4, 473.1, 493.6, 505.1, 589.5, 671.3 and 709.1 respectively. Thus the % error will change accordingly as shown below.

λ = 438.8 nm % Error = (449.4 – 438.8)/438.8*100% = +2.42% λ = 447.1 nm % Error = (455.4 – 447.1)/447.1*100% = +1.86% λ = 471.3 nm % Error = (473.1 – 471.3)/471.3*100% = +0.38% λ = 492.2 nm % Error = (493.6 – 492.2)/492.2*100% = +0.28% λ = 501.5 nm % Error = (505.1 – 501.5)/501.5*100% = -0.08% λ = 587.6 nm % Error = (579.5 – 587.6)/587.6*100% = -1.38 % λ = 667.8 nm % Error = (671.3 – 667.8)/667.8*100% = +0.52% λ = 706.5 nm % Error = (709.1 – 706.5)/706.5*100% = +0.37%

Below is the table to compare the calculated value of λ and % error of these two conditions (using value of and using value of d). (nm)(=1655.4nm) | (nm)(d=1667nm) | % Error(=1655.4nm) | % Error(d=1667nm) | 446.3 | 449.4 | +1.71% | +2.42% | 452.3 | 455.4 | +1.16% | +1.86% | 469.8 | 473.1 | -0.32% | +0.38% | 490.2 | 493.6 | -0.41% | +0.28% | 501.6 | 505.1 | +0.02% | -0.08% | 585.3 | 579.5 | -0.39% | -1.38% | 666.6 | 671.3 | -0.18% | +0.52% | 704.2 | 667.8 | -0.33% | +0.37% |

Comparing the data obtained above, we can easily deduce that if we use the actual value of d (1667nm) and the values of sinθ (Calculation Table 2), the % error is much greater than using the value of (1655.4nm). From the table above, we observed that the absolute values of % errors using d = 1667nm is approximately 1% greater than using = 1655.4nm. This shows that the degree of accuracy has reduced. Thus the resulting wavelengths for the helium lamp will be worse off than the wavelength values in the table.

7.4. In the continuous spectrum of the incandescent light bulb, what is the wavelength range of the yellow image? What is the wavelength range of the orange image? What is the center/middle wavelength of the visible spectrum using your calculated wavelength range values of the visible spectrum in Calculation Table 3?

We can use the Calculations Table 3 to solve this question:
Calculations Table 3

Portions of Spectrum | (cm) | tanθ | θ | sinθ | λ(nm) | Shortest Wavelength | 12.7 | 0.2540 | 14.25° | 0.2462 | 407.6 | Division Blue and Green | 16.1 | 0.3220 | 17.85° | 0.3065 | 507.4 | Division Green and Yellow | 18.1 | 0.3620 | 19.90° | 0.3404 | 563.5 | Division Yellow and Orange | 19.0 | 0.3800 | 20.81° | 0.3552 | 588.0 | Division Orange and Red | 19.9 | 0.3980 | 21.70° | 0.3698 | 612.2 | Longest Wavelength | 24.5 | 0.4900 | 26.10° | 0.4400 | 728.4 |

Using the Calculation Table 3, we are able to determine the wavelength range.
The wavelength range for the yellow image is between 563.5nm to 588.0nm.
The wavelength range for the orange image is between 588.0nm to 612.2nm.

To determine the middle wavelength, we have to take the average of the addition of the longest and the shortest wavelength. It can be calculated by:

Middle wavelength = Shortest wavelength + longest wavelength = 407.6+728.4 = 568nm 2
8. Experimental Errors

1) The error that we encountered is that it is difficult to exactly locate and read off the position of the first images. Even when we use a pencil to point at the image, it is still hard to point at the right position as some of the regions are not clearly distinct. As a result, estimation errors will incur.

2) When we are doing the experiment, different people may observe the image at different location due to our eyesight. As such, it is difficult for us to determine the position of the images exactly as both partners will perceive them differently. Therefore, this may led to estimation reading from the observer and hence led to the rise in the percentage error.

3) Some of the images of the discrete line spectrum were produced with wavelengths that were extremely closed to each other. For instance, two visible colors produced by the mercury lamp and helium lamp are too closed to each other. This makes it very difficult for us to distinguish the color of the images and hence estimation is required. This will led to the rise in percentage error.

4) Another error is due to parallax error. When the observer has already located the image, the partner will try to read off from the ruler. It is difficult to look at eye level given the structure of the equipment. This will affect the value of DR, DL and. This will then further affect the value of tanθ, θ and sinθ. Then the value of calculated will be affected, hence affecting the wavelength values calculated in calculation table 2 and calculation table 3. Thus, this will give rise to the percentage error.

5) Another error could occur due to the accuracy of the ruler. Maybe the accuracy of the ruler is not accurate enough and this will affect the value DR, DL and. Hence the value such as θ, and wavelength will be affected. Hence this will led to the rise in a percentage error.

9. Conclusions

1) Through this experiment, we are able to differentiate and observe the differences between a continuous and discrete spectrum. Continuous spectrum is a continuous array of colors which fade into one another which can be seen when we are doing the experiment using a 10-W incandescent light bulb whereas a discrete spectrum consists of mostly dark portions with a few discrete lines of colors when we are doing the experiment using a helium lamp or mercury lamp. 2) By comparing the mercury line spectrum and helium line spectrum, it is can be seen that both spectrums exhibit different sets of visible colors in its spectrum. It is because for helium lamp, it exhibits a visible red image over the mercury lamp. This reasserts the fact that light produced by the discharge in a particular chemical element has a spectrum that is the characteristics of the element. 3) The average grating spacing between lines of the diffraction grating has already been determined through the Mercury Lamp experiment. The diffraction grating spacing calculated was 1655.4nm which are consistent with the theory for the diffraction grating method as the value of the diffraction grating spacing should be between 1000nm to 2000nm. 4) From the value of the diffraction grating calculated from the experiment using helium lamp, the characteristics wavelength of a helium lamp and a 10-W incandescent light bulb were able to obtain. The results obtained were closed with the true values of the wavelength as the greatest percentage error of the wavelength is +1.71%. Hence we can conclude that this diffraction grating method is effective and reliable in determining the characteristic wavelength due to the small percentage error of the wavelength calculated based on the experiment.

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...laboratory 3, we will be exploring what light is and what gives an object its color. In addition, we will be using alternate light sources which will help us define the precise and characteristic wavelengths of light that certain elements will show through the colored bands seen when looking through the diffraction grating. The materials that are required for this lab will be the following: 3 meter sticks, 1 incandescent light source, 1 diffraction grating (600 lines per mm), 1 pencil, 1 element discharge tube, and 1 Helium, Hydrogen, Mercury or Neon light bulb. In the first section of lab 3, four lab partners will be divided into two pairing two students on the right side of the table and the other two on the left. The groups will then be assigned either an incandescent or an...

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Diffraction Grating

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Atomic Physics

...the Bohr Magneton 9/21/2015 Abstract In this lab, the Rydberg constant is found by observing the Balmer Series. With the experiment below, a Rydberg Constant was found to be 10116000 meters-1 with a 7.82% error. The Bohr Magneton is also found, and a value of 6.03*10-24 JT was obtained with a large error. The error can arise from each optical equipment having some fundamental error in its creation. Introduction With the help of atomic physics, quantum mechanics, and optics, the Rydberg constant and the Bohr magneton will be calculated in this experiment. The Rydberg constant is one of the most important constants in atomic physics because of its relation to other fundamental constants in atomic physics, such as the speed of light or Planck’s constant [1]. The Bohr Magneton tells us the magnetic moment of an electron by its angular momentum [2]. Attempting to calculate the Rydberg constant and the Bohr Magneton will inadvertently teach the basis of quantum mechanics, optics, and atomic physics. Atomic spectra of hydrogen, mercury, and helium will be studied in detail along with the Zeeman Effect. Theory In quantum mechanics, labeling often times helps discern descriptions of certain events. To describe the movement and trajectories of an electron in an atom, scientists use quantum numbers to label what is going on. The principal quantum number n, tells the energy level of the electron and the distance from the nucleus. The angular momentum quantum number l,......

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Blood Alcohol Measurement

...A BLOOD ALCOHOL MEASUREMENT DEVICE USING A FIXED FREQUENCY PHOTOSPECTROMETER BY LUCILLE J. DURFEE ABSTRACT There are many people who go out for a night on the town without thinking about the potential dangers of drinking alcohol and then driving. A simple, non-invasive system is needed to test these people before they leave the bar to determine if they are “okay” to drive. The purpose of this project is to study, design ,build, and test a fixed frequency photospectrometer that will test the blood alcohol content of a person non-invasively. BACKGROUND Over the past decades a new method for measurement has been rapidly envancing. This method is a photonic technique that envolves electromagnetic radiation. Electromagnetic radiation has several forms including visible and infrared light radiation. Visible light radiation ranges from 370nm to 760nm, while infrared ranges from 760nm to approximately 1OE-4nm. Mark A. Arnold at the university of Iowa in Iowa City and Gary W.Small at Ohio University in Athens have been focusing on near-infrared (800nm to 1000nm) absorption spectroscopy as a solution to non-invasive glucose monitoring. Problems that have occurred revolve around the broad absorption signals of near-infrared radiation making it more difficult to identi@ a specific analyte. Arnold estimates that a working device is still five years away. A Biophotonic success is the Pulse Oximeter. It is a ten year old deviqe used to monitor patient’s conditions under anesthesia. It...

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Led Composition

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...unique frequency The frequency of a vibration depends on: The strength of the chemical bond between atoms The mass of each atom So we can learn a lot about a molecule when we study its vibrations! © ABB Inc. - 4 - Infrared spectroscopy is a method for the analysis of molecular vibrations IR Spectroscopy Analysis based on the absorption at different wavelengths of an infrared beam by a sample Source radiation Transmitted radiation Sample 100 80 100 Relative intensity of IR beam 80 60 60 40 40 20 20 0 © ABB Inc. - 5 - transmittance spectrum of the sample 0 4000 3000 2000 1000 4000 3000 2000 1000 Principles of Infrared Spectroscopy When a frequency of light corresponds to a molecular vibration it is absorbed by the sample © ABB Inc. - 6 - The fraction of light transmitted by the sample compared with the light incident as a function of frequency gives the infrared spectrum of the sample Infrared Spectrum of a Polystyrene Film Wavelength is expressed in number of waves per cm. (e.g. 1000 /cm is the same as wavelength λ=1/1000th cm) Vertical scale is percent remaining intensity transmittance spectrum of the sample 100 80 60 % 40 20 0 © ABB Inc. - 7 -...

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...Polymer-stabilized blue phase liquid crystals: a tutorial [Invited] Jin Yan and Shin-Tson Wu* College of Optics and Photonics, University of Central Florida, Orlando, FL 32816, USA * Abstract: Blue phase liquid crystals exhibit several attractive features, such as self-assembled three-dimensional cubic structures, optically-isotropic in the voltage-off state, no need for alignment layers, and submillisecond response time. This tutorial gives step-by-step introduction on basic bluephase materials and properties, monomers and polymerization processes, and key device performance criteria for display and photonics applications. ©2011 Optical Society of America OCIS codes: (160.3710) Liquid crystals; (160.5470) Polymers. References and links 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. F. Reinitzer, “Beiträge zur Kenntniss des Cholestherins,” Monatsh. Chem. 9(1), 421–441 (1888). A. Saupe, “On molecular structure and physical properties of thermotropic liquid crystals,” Mol. Cryst. Liq. Cryst. (Phila. Pa.) 7(1), 59–74 (1969). S. A. Brazovskii and S. G. Dmitriev, “Phase transitions in cholesteric liquid crystals,” Zh. Eksp. Teor. Fiz. 69, 979–989 (1975). R. M. Hornreich and S. Shtrikman, Liquid Crystals of One- and Two- Dimensional Order (Springer-Verlag, Berlin, 1980). S. Meiboom, J. P. Sethna, W. P. Anderson, and W. F. Brinkman, “Theory of the blue phase cholesteric liquid crystals,” Phys. Rev. Lett. 46(18), 1216–1219......

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Why the Sky Is Blue

...oxygen, .7% argon, and a small amount of trace gases. White light from our sun is made up of all the colors in the rainbow from red to violet. When the light hits our atmosphere the molecules scatter blue light more than red or green etc. so we see a blue sky! Let's do an experiment to see the effect of scattering, and some other properties of light. Get as many of the following materials as possible: A 1 liter beaker or 1 qt glass jar (a 12 oz glass will do), some milk, a laser pointer or other low power laser, a prism or diffraction grating and some lenses, and a lab notebook to record your observations. Fill the beaker with cold tap water and shine your laser through it. Mix 2-3 drops of milk into the water. Take the mixture into a dimly lit, or dark, room. Shine the laser through the side of the beaker and look through the top of the beaker, jar, or glass. What do you see? The mixture you made is called a colloidal suspension. Now shine the laser up through the side of the beaker toward the center of surface of the suspension. [Physics FAQ] - [Copyright] Original by Philip Gibbs May 1997. Why is the sky blue? A clear cloudless day-time sky is blue because molecules in the air scatter blue light from the sun more than they scatter red light.  When we look towards the sun at sunset, we see red and orange colours because the blue light has been scattered out and away from the line of sight. The white light from the sun is a mixture of all colours of the rainbow. ......

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Garbage Sorter

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Nestle Report

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