PHY13L - E406 - PHOTOMETRY

DATA and OBSERVATION

 

TABLE 1. Inverse Square Law
	Transmittance	Experimental r2		%Error
r1 = 30cm	100% (I2 =  I1)	29 cm	30 cm	3.33%
	75% (0.75I2 =  I1)	25 cm	25.98 cm	3.77%
	50% (0.50I2 =  I1)	20.9 cm	21.21 cm	1.46%
	25% (0.25I2 =  I1)	14.7 cm	15 cm	2.00%
r1 = 45cm	100% (I2 =  I1)	44 cm	45 cm	2.22%
	75% (0.75I2 =  I1)	37 cm	38.97 cm	5.06%
	50% (0.50I2 =  I1)	31 cm	31.82 cm	2.58%
	25% (0.25I2 =  I1)	22 cm	22.5 cm	2.22%

 

 

TABLE 2. Polarization

 

Transmittance	Observation at 0°	Observation at 90°	Observation as polarizers are  rotated
100%	Not the same	dark	When 180°, it is same as the observation at 0°
	Experimental Φ		%Error
75% ()	30°	30°	0%
50% ()	45°	45°	0%
25% (0.25I2 =  I1)	50°	50°	0%

 

 

GUIDE QUESTIONS

How does varying the locations of the light sources with respect to the photometer affect the result of the experiment?

The intensity of the light as seen on the photometer depends on the distance of the light source. As the distance between the light source and the photometer decreases, the intensity of light increases and the other way around. 

How are the neutral density filter and polarizer different? In what way/s are they the same?

A neutral density filter or ND filter is a filter that reduces and/or modifies intensity of all wavelengths or colors of light equally, giving no changes in hue of color rendition. It can be a colorless (clear) or grey filter. A polarizer is an optical filter that passes light of a specific polarization and blocks waves of other polarizations. It can convert a beam of light of undefined or mixed polarization into a beam with well-defined polarization.

If you have two pairs of sunglasses, how would you test whether the sunglasses use polarized lenses or not without using polarizers?

Place the two pairs of sunglasses next to each other and rotate on of the pair of sunglasses. If the visibility of the light on the other side is changing, the sunglasses are polarized but if it's just the same, the sunglasses are not polarized.

 

SAMPLE COMPUTATION

TABLE 1

r_2act "=" √((I_1 〖r_1〗^2)/I_2 ) "=" √(〖r_1〗^2 )=30cm

"% Error=" |"30-29" /"30" |"×100=3.33%" 

TABLE 2

θ_act "="  cos^(-1)⁡√(I_1/I_2 ) "="  cos^(-1)⁡√(〖0.75I〗_2/I_2 )=30°

"% Error=" |"30-30" /"30" |"×100=0%" 

 

ANALYSIS

At 100% transmittance, are the distances of two sources from the photometer equal? Is it the expected result? Explain.

The distances, as expected, are equal. Since at 100% transmittance allows all of the light to pass through it and gives the same amount of light as.

At transmittance below 100%, how do the distances of the sources from the photometer compare?

The lower the percentage of the transmittance the closer the distance of the light source from the photometer and the other way around.

If the neutral density filter is set up to a transmittance below 100%, how should the orientation of the polarizers be change to produce the same illumination on both sides of the photometer?

If lower transmittance is used, we need to arrange the polarizer closer to zero degrees of for higher transmittance; it is moved closer to 90o.

CONCLUSION

Based on the experiment, it is found out that photometry can be used to determine various factors affecting the intensity of light and how can it be manipulated through a photometer. It is also concluded that the intensity of the light (luminosity) observed from a given view varies with respect to its distance from the light source. For shorter distances, light is more luminous.

Finally, light is multicomponent, and is polarized in all directions. When it is allowed to pass through a polarizer, it will be linearly polarized, having only one direction. When a second polarizer is used, the light that can be transmitted varies on the angle between the two polarizers. For angle is equal to 0o, the same amount of light from the first polarizer can passed through. On the other hand, as it approaches 90o, the intensity decreases for only component of the linearly polarized light is allowed to pass. At 90o, no light can be observed.

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. 

PHY13L - E404 - Index of Refraction

1.       If the transparent material has an index of refraction of 1.60, what is the angle of incident beyond which total internal reflection occurs?

Using the Snell's Law equation wherein the angle of incident when the angle of refraction becomes 90^o is called the critical angle if incident. Solving for it when the index of refraction is 1.60:

The angle of incident beyond which total internal reflection occurs is .

2.       Is it possible for the critical angle to exist when light goes from air (n=1.00) into water (n=1.33)? Explain your answer.

Critical angle will not be existing when light goes from air into water. Theoretically, n₂ which is index of refraction of the second medium wherein this case is the water must be less than n₁ which is the index of refraction of air as stated in Snell's Law. Therefore, it is not possible.

PROBLEMS

 

1.       A ray of light from air, incident at an angle of 60^o on top of a transparent surface, is refracted and reflected partly. It is observed that the refracted and reflected rays are perpendicular to each other. Find the index of reflection and velocity of light in the transparent surface.

 

 

DATA and OBSERVATIONS

PARTS A and B. Determination of Index of Refraction of Glass

	PART A Light bends from GLASS to AIR	PART B Light bends from AIR to GLASS
Length of line GH	3.35cm	4.90cm
Length of line FL	5.00cm	3.25cm
Index of Refraction of Air, na	1.0	1.0
Index of Refraction of Glass, ng (experimental value)	1.49	1.51
Index of Refraction of Glass, ng (accepted value)	1.5	1.5
Percentage error	0.66%	0.66%

 

PART C. Determination of Critical Angle

 

Length of line MU	5.20cm
Length of line EU	6.40cm
Critical Angle, θC (experimental value)	39.09°
Critical Angle, θC (accepted value)	41.81°
Percent Error	6.51%

 

 

 

SAMPLE COMPUTATION

PART A

PART B

PART C

 

 

ANALYSIS

1.       Why can't a material have an index of refraction less than 1.00?

Since nothing can travel faster than the speed of light according to the theory of relativity, the speed of light in vacuum is always faster the speed of light in a medium so the index of refraction cannot have a less than 1.00.

2.       Does the angle of incidence affect the index of refraction of a materiel? Explain your answer.

The index of refraction is not affected by the angle of incidence of a material since the index of refraction has a constant value depending on a material or medium.

3.       What happens to the wavelength of light when it is refracted? Consider a case where light enters a denser medium.

The velocity of light wave decreases then light enters a denser medium while the frequency is stayed constant, the wavelength of refracted light increases.

CONCLUSION

Therefore I conclude that the index of refraction of a medium do not change even after seeing  it in 2 ways, one from less dense to more dense and second from more dense to less dense. I also found out that the angle of refraction is near the normal line if the condition was from less dense to more dense.

Also, we can only get critical angle when the first medium has higher index of refraction than the second medium. If the index of refraction of the first medium is less than the second medium, we cannot get the critical angle because the less dense will reach its critical angle first than the second medium in which obeys the Snell's Law.

PHY13L - E403 - Refraction from a Spherical Surface - Thin Lens

ANALYSIS OF DATA / REMARKS

For the entire experiment, we experimentally deter-mined the focal lengths of two lenses with an actual focal lengths of 10cm and 20cm respectively.

For the Part A of the experiment, focal length is determined using an infinity object. In this case, our best infinity object is the view of the sky from the laboratory room's window. Values shown are exact from the actual value which gives us no percentage error for both lenses. Focal length is computed using the lens equation where the image distance is approaching infinity which leads is to an equation where focal length is equal to the image distance.

Part B of the experiment uses a finite object where in this case is an image projected by the light source provided. In this part of the experiment, the screen and the light source is a meter apart from each other and the lens is somewhere in between. Lens is moved back and forth until the image formed on the screen is sharp. There would be two position of the lens where image formed is sharp where our data shows that these positions are interchangeable and are conjugate. This applies for both lenses with 10cm and 20cm focal lengths respectively. Focal length in this case is also obtained using the lens equation.

Using graphical method, we can also determine the focal length of the lens like what we did on the Part C of the experiment. In this part of the experiment, we used the 10cm-focal length lens. Distance between the screen and the light source is changed for every trial from 110cm to 90cm to 85cm apart respectively. Percentage difference is also computed where it shows that percentage difference is greater on the second position where lens is closer to the screen this is because the image height is very small and is hard to measure accurate using a ruler measuring millimeters.

Common possible source of errors for this experiment is measuring. Small difference on distance may show huge effect on the data.

 

CONCLUSION

We were able to determine the focal length of the lens using different location of the object including an infinity object and we were able to understand, with the help or out instructor, the graphical method in getting the focal length of a lens.

Refraction is the blending of light ray as it hits a material of different optical density. Transparent materials like lenses can refract parallel ray of light and can produce an image. Convex lens is used on our experiment which has middle-part thicker. Focal length's relation to the object and the image is given by the thin lens equation where focal length is the difference on product and sum of the object and image distance from the lens. . There are two positions of the lens where image formed is sharp, these positions are interchangeable and are conjugate. The magnification of the lens is the ratio of the image height and the object height, and is also related to the distances of the object and image from the lens.

Lenses play a very important role on the discovery of various things we know today.  Telescope and microscope uses lens to magnify things from afar or a very small particle. Without these lenses, such things would not be possible. This experiment helped me understand how light rays behave on thin lenses and focal points differences plays its part.