PHY13L - E405 - DIFFRACTION

GUIDE QUESTIONS

How does slit width affects the result of the experiment? How about slit screen separation? Are the experiment results consistent with the theory?

 

The size of the minima and maxima increases as the slit width decreases, and vice versa. Moreover, the number of minima and maxima increases when the slit width increases and vice versa. The theory in diffraction is consistent and correct; the theory tells us that slit is divided into 2 elements and that light from this area element strike the same point on the screen and it is divided into equal elements.

 

Compare the patterns formed in a single-slit diffraction and in a double-slit interference? How are the patterns similar? How are the patterns different?

If light is incident onto an obstacle which contains two very small slits a distance d apart, then the wavelets emanating from each slit will constructively interfere behind the obstacle.When light passes through a single slit whose width w is on the order of the wavelength of the light, then we observe a single slit diffraction pattern.  Huygen's principle tells us that each part of the slit can be thought of as an emitter of waves.  All these waves interfere to produce the diffraction pattern.

The principal difference between an interference pattern caused by two slits and that caused by a grating is that a grating has more intense bright fringes that are more widely-spaced. These two effects should make sense. The brightness would increase since more light can reach the screen (there are more openings, or slits, in the barrier). The greater spacing can be accounted for by the fact that since d is getting increasingly smaller, y would get increasing larger.

 

PROBLEMS

Monochromic light is incident upon a slit of width 0.55mm. A diffraction pattern is formed on a screen 1.5m away. If the distance from the central maximum to the minimum is measured to be 1.25mm, what is the wavelength of the light?

a=0.55m

m=1

x=1.50m=1500m

y_m=mxλ/a

λ=(y_m a)/mx=1.25mm(0.55mm)/1(1500mm) =458nm

What frequency of light produces a diffraction pattern on a screen such that the distance between the first minimum and the third minimum is approximately 4mm? The slit-to-screen distance is 1.00m while slit width is 0.80mm.

a=0.8mm

x=1m=1000mm

m_1  to m_2=4mm

y_m=2mm

λ=(y_m a)/mx=2mm(0.8mm)/1(1000mm) =1600nm

 

SAMPLE COMPUTATION

TABLE 1

"y" _"m"  "="  "mxλ" /"a" 

"λ="  (y_m a)/"mx"  "="  (4mm)(0.16mm)/(1)(900mm) =1.11〖x10〗^(-4) mm

"Average λ=" |"711nm+667nm+652nm" /"3" |"=677nm" 

TABLE 2

"y" _"m"  "="  "mxλ" /"a" 

"λ="  (y_m a)/"mx"  "="  (1mm)(0.5mm)/(1)(600mm) =8.33〖x10〗^(-3) mm

"Average λ=" |("833" (2)"+625" (2)+444(2))/"6" |"=634nm" 

 

ANALYSIS

Diffraction is a property of the motion of all waves which is a phenomenon by which a wave encounters a barrier or an obstacle. Interference is phenomenon by which two or more waves tend to overlap, thus, interfere with one another. When waves interfere with each other, the amplitude of the resulting wave depends on the frequencies, relative phases (relative positions of the crests and troughs), and amplitudes of the interfering waves.

The first part of the experiment measures the wave length of a single slit disk. Part II of the experiment measures the wavelength using two slit interference. Just like in the first part of the experiment, the light should be diffracted. Having the measured values, wavelength is now computable. By getting the wavelength at different position we attained its average and this value should conform in the accepted value of the wavelength. On both parts, diffraction is observed when the light rays tend to bend from a much smaller barrier or passageway to be projected on the screen with an image having a larger magnitude than the width of the slit. Interference of light waves is observed on the second part of the experiment when the light rays pass on two different slits and interfere with the wave of the other which is observed on the screen as much smaller divisions of bright and dark fringes. The slit may be divided into equal area elements with each are element acting as a source of secondary waves. These wave patterns are the one which could be observed by us in the screen. We had observed that the measurement of minima and maxima increases as the slit decreases.

CONCLUSION

I therefore conclude that the distance of every minima and maxima are equal to each other. The measurement of the minima and maxima increases as the slit decreases. The measurement of the minima and maxima decreases as the distance of the screen to the slit decreases. We must also consider several factor in performing the experiment, we must know what kind of color the laser produces in able to know the wavelength of that specific color because every color have different wavelength.

In real life application, for diffraction, if a radio is turned on in one room, the sound from the radio can be heard in an adjacent room even from around a doorway. Similarly, whenever water waves pass an object on the surface of the water, such as a jetty or boat dock, waves that pass the object's edge spread out into the region behind the object and directly blocked by it. An example of interference also radio waves which interfere with each other when they bounce off buildings in cities, distorting radio signals. Sound-wave interference must be taken into account when constructing concert halls, so that destructive interference does not result in areas in the hall where the sounds produced on stage cannot be heard.