physics week 13(sem. 2) - st. francis preparatory …...information and diagram below. the diagram...

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Physics Week 13(Sem. 2) Name____________________________

Light Chapter Summary Cont’d 2

Lens Abberation

Lenses can have two types of abberation, spherical and chromic. Abberation occurs when the rays forming an image (from an object) aren’t focused at exactly the same point causing a blurred image. Spherical abberation occurs with converging and diverging lenses with spherical surfaces. Ideally rays close and parallel to the principal axis are refracted and cross at the same point. However, some rays are far enough from the principal axis that the ray is refracted more and does not cross at the exact same point as other rays. Despite the lack of an exact position of rays crossing, the rays close the principal axis with a small cross sectional area and this is called the circle of least confusion. The circle of least confusion is where the most satisfactory image can be formed by the lens. Spherical abberation can be greatly reduced by only allowing incident rays that are close to the principal axis to pass through the lens. Also, parabolic shaped lenses are used to reduce this type of abberation, but they are expensive and hard to make.

Chromic abberation also contributes to blurred images. This arises because the index of refraction of a material is dependent on wavelength thus, refracting different colors of light to different degrees. This phenomenon leads to dispersion, where white light is separated into its component colors of light. Considering only the shortest and longest wavelength of visible light will be adequate to explain this abberation. For violet light, it is refracted more than red, so it crosses the principal axis closer to the lens than the red light. Thus the focal length of the lens is shorter for the violet light and longer for the red light, with intermediate points for the colors between red and violet. As a result of chromic abberation, an undesirable color fringe surrounds the image. Chromic abberation can be greatly reduced by using a compound lens. For example, a converging lens

followed by a diverging lens if the lenses are made of different types of glass then the different colors will be focused at a common point. This lens combination designed to reduce chromic abberation is called an achromatic lens, it is used in cameras.

Young’s Double‐Slit Experiment

This experiment demonstrated the wave nature of light and allowed for the determination of the wavelength of light for the first time. The experiment involved light passing through a single narrow slit falling on two closely spaced narrow slits (S1 &S2). The reason for the first narrow slit is provide a coherent light source that will be incident on the double slit and ultimately interact constructively and destructively at different points on the screen (beyond it all). Analyze the figure below to understand the presence of bright and dark fringes.

In the figure a and b demonstrate a bright fringe, the result of constructive interference, while figure c demonstrates destructive interference (a dark fringe). The brightness of the bright fringes varies in this experiment. The central fringe is labeled with a zero, while the other bright fringes are numbered in ascending order on either side of the bright fringe. It can be seen that the central fringe has the greatest intensity and the intensity of the other fringes decrease symmetrically in a way that depends on how small the slit widths are relative to the wavelength of the light. Using the following picture the position of the fringes can be calculated.

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If the screen is located far away from the slits compared to distance between the slits (d), then the lines l1 and l 2 in part a are parallel. Being nearly parallel, these lines make approximately equal angles θ with the horizontal. Thus l1 and l 2 differ by ∆l which is the length of the short side of the colored triangle in part b (bottom of picture). Since it is a right triangle it follows that ∆l =d sin θ. Constructive interference occurs when the distances differ by a multiple of wavelengths, or ∆l =d sin θ= mλ. Therefore the angle, θ, for the interference maxima (bright lines) can be determined using the equation

sin

Where m is equal to 0,1,2,3,… The value m specifies the order of the fringe. For example, m=2 identifies the second order bright fringe (indicated as #2 in part c). Similarly the dark fringes can be located using the equation

sin 1/2

Where m is equal to 0,1,2,3,… Example 2 will demonstrate how to apply these equations to a problem.

Single Slit Diffraction

For the single slit diffraction the equation is slightly different, the slit is split into parts and the light from each part can be analyzed separately. When analyzed there is still the production of bright and dark fringes on a screen, however the equation for single slit diffraction is

sin

Where m is equal to 1,2,3,… and where w is the slit width. Between each pair of dark fringes there is a

bright fringe due to constructive interference. Also, important to note that the brightness of the fringes is related to the light intensity just as loudness is related to sound intensity. The intensity of the light at any location on the screen is the amount of light energy per second per unit area that strikes the screen there. The central bright fringe (for single slit) is twice as wide as the other bright fringes and has by far the greatest light intensity.

Diffraction Gratings

Diffraction patterns of bright and dark fringes appear when monochromatic light passes through double and single slits. Fringe patterns also appear when light passes through more than two slits, an arrangement containing a large number of parallel, closely spaced slits is called a diffraction grating. Diffraction gratings can contain as many as 40,000 slits per centimeter can be made. Also the number of slits per centimeter is counted as the number of lines per centimeter.

Constructive interference creates the bright fringes, including the central fringe, first order (principal) fringes, second order fringes, and more. Assuming the rays are parallel to each other and taking into account the location of the first order bright fringe, the light from slit 2 has to travel a distance of one wavelength longer than the light from slit 1. Similarly, light from slit 3 has to travel one wavelength further than the light from slit 2, and so forth. Therefore the equation for a diffraction grating is the same as the double slit experiment

sin

The separation d between the slit can be calculated from the number of slits per centimeter of the grating; for instance, a grating with 2500 slits per centimeter has a slit separation of d=1/2500 cm = 4.0 x 10‐4 cm. Though the equation is the same as that for the double slit the resulting bright and dark fringes are narrower and sharper than those from the double slit. Also, there are second order maxima that are much less intense than the very bright (principal) fringes.

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Example 10 demonstrates the use of a diffraction grating applied to a mixture of violet and red light.

Thin‐Film Interference

The slit experiment is one example of interference between light waves, however interference happens in more common circumstances. For example, a thin film of gasoline floating on water. Assuming the film to have a constant thickness, consider what occurs when monochromatic light strikes the surface perpendicularly.

At the top surface reflection occurs, however refraction also occurs. Then parts of the refracted ray reflects from the bottom surface and passes back through the film and eventually reenters the air again. So ray 2, the refracted ray, travels a much further distance than ray 1 (the reflected ray). Because of this distance there could be interference of the two rays of light, if a uniform bright film is observed than the interference was constructive. If destructive interference occurred than a dark film would be observed. The controlling factor in the appearance would be whether the extra distance traveled (down and back up through the thin film) is a whole number of wavelengths or an odd number of half wavelengths. The difference between wave 1 & 2 occurs in the thin film, therefore the wavelength that is important for thin film interference is the wavelength

within the film (not the wavelength in a vacuum). Using the equation for the index of refraction and combining it with the speed of light in a vacuum the following equation is found to be true

Where n is the index of refraction of the film. To be thorough the reflection needs to be analyzed further. When light is reflected at a boundary it is possible for the phase to change. If one end is fixed than the wave will invert and be changed by ½ wavelength, as if the wave traveled an additional ½ wavelength. In contrast when a wave is reflected from a moveable boundary no such phase change occurs. So for light waves

1. When light travels through a material with a smaller index of refraction toward a medium with a larger index of refraction (e.g. air to gasoline), reflection at the boundary occurs along with a phase change (1/2 λ).

2. When light travels from a larger index of refraction to a smaller refractive index, there is no phase change upon reflection at the boundary.

Example 4 will explain how to account for this possible phase change in light at a boundary.

Another example of thin film interference would be the air wedge. An air wedge contains two pieces of glass at an angle of θ with respect to each other with air in between them. If a piece of paper were inserted between them to catch the light an arrangement of bright and dark fringes would appear. The fringes that appear are the result of both constructive and destructive interference that results from the distance between the glass plates and the angle of the top plate of glass.

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Base your answers to questions 1 through 3 on the information and diagram below.

The diagram below represents the interference pattern produced by a source of monochromatic light with a wavelength of 6.0 × 10–7 meter. The light is incident upon two slits that are 2.0 × 10–6 meter apart and 2.0 meters from the screen.

1. If the distance between the slits and the screen is increased then the distance x will(1) decrease (3) remains the same(2) increase

2. The color of the incident light is(1) green (3) red(2) yellow (4) orange

3. If the light source is made brighter, then distance x will(1) decrease (3) remain the same(2) increase

Base your answers to questions 4 and 5 on the diagram below which shows the top view of a double-slit barrier. The distance between the slits d is 1.0 × 10–3 meter. The barrier is illuminated with light whose wavelength is 6.0 × 10–7 meter. An interference pattern is observed on the screen.

4. If point M on the screen is the location of a bright line, the path difference between AM and BM may be(1) one wavelength(2) three-quarter wavelength(3) one-half wavelength(4) one-quarter wavelength

5. If point M is the location of the first maximum, distance x is(1) 1.2 × 10–3 m (3) 0.2 m(2) 2.4 × 10–3 m (4) 0.4 m

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6. Base your answer to the following question on the diagram below which represents red light incident upon a double-slit barrier, producing an interference pattern on a screen. The wavelength of the red light is 6.6 × 10–7meter, the distance (d) between the slits is 2.0 × 10–3 meter, and X is 3.3 × 10–4 meter.

The red light source is replaced with a blue light source. Compared to distance x when the red light source is used, what would distance x be when the blue light source is used?(1) less (3) the same(2) greater

7. Base your answer to the following question on the information below.

Light of wavelength 5.4 × 10–7 meter shines through two narrow slits 4.0 × 10–4 meter apart onto a screen 2.0 meters away from the slit.

The distance between the bright areas in the stationary interference pattern formed on the screen is(1) 1.1 × 10–10 m (3) 5.4 × 10–3 m(2) 2.7 × 10–3 m (4) 3.7 × 102 mN. M

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8. The diagram below represents monochromatic light incident on a pair of slits, S1 and S2, that are separated by a distance of 2.0 × 10–6 meter. A, B, and C are adjacent antinodal areas that appear on a screen 1.0 meter from the slits. The distance from A to B is 0.34 meter.

What is the wavelength of the incident light?(1) 6.8 × 10–7m (2) 5.9 × 10–6 m (3) 1.7 × 105m (4) 6.8 × 107 m

9. Which will most likely occur when light passes through a double slit?(1) refraction(2) diffraction, only(3) interference, only(4) diffraction and interference

10. An interference pattern is produced on a screen by a green monochromatic light beam that has passed through a single narrow slit. Which diagram best represents this pattern?(1)

(2)

(3)

(4)

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11. The diagram below represents shallow water waves of constant wavelength passing through two small openings, A and B, in a barrier.

Which statement best describes the interference at point P?(1) It is constructive, and causes a longer

wavelength.(2) It is constructive, and causes an increase in

amplitude.(3) It is destructive, and causes a shorter

wavelength.(4) It is destructive, and causes a decrease in

amplitude.

Base your answers to questions 12 and 13 on monochromatic light passes through a single narrow slit forming a diffraction pattern on a screen.

12. Which graph best represents the light intensity of the single-slit diffraction pattern for monochromatic light?

(1) (3)

(2) (4)

13. If the intensity of the light is increased, the width of the central maximum in the diffraction pattern will(1) decrease (3) remain the same(2) increase

14. Waves pass through a 10.-centimeter opening in a barrier without being diffracted. This observation provides evidence that the wavelength of the waves is(1) much shorter than 10. cm(2) equal to 10. cm(3) longer than 10. cm, but shorter than 20. cm(4) longer than 20. cm

15. A beam of monochromatic light approaches a barrier having four openings, A, B, C, and D, of different sizes as shown below.

Which opening will cause the greatest diffraction?(1) A (3) C(2) B (4) D

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16. Two point sources of monochromatic light radiate in phase with a constant wavelength of 500 nm. The first order interference maximum appears at 14º (use sin 14º = 0.25). The separation of the sources is most nearly(1) 125 nm (4) 1000 nm(2) 250 nm (5) 2000 nm(3) 500 nm

17. Young's double slit experiment did much to the confirm the(1) wave nature of light(2) structure of the atom(3) existence of the photon(4) invariance of the speed of light(5) law of universal gravitation

18. One way to decrease the distance between the bright and dark bands of the diffraction pattern seen on the screen in a double slit experiment would be to(1) use light of a longer wavelength(2) increase the intensity of the light(3) move the screen away from the slits(4) increase the separation of the slits(5) use light with a lower frequency

19. Two point sources of monochromatic light with equivalent wavelengths radiate in phase. The first order interference appears at 8.5o (use sin 8.5o = 0.15). The separation of the sources is 4,000 nm. The wavelength of the light is most nearly(1) 60 nm (4) 3,000 nm(2) 400 nm (5) 6,000 nm(3) 600 nm

20. Which of the following is NOT illustrated by Young's Double Slit Experiment?(1) the wave-particle duality of light(2) diffraction(3) constructive interference(4) destructive interference(5) the wave nature of light

21. The diffraction pattern produced by a double slit will show greatest separation of maxima when the color of the light source is(1) yellow (4) green(2) orange (5) red(3) blue

22. Compared to a double-slit diffraction pattern, a single-slit diffraction pattern would exhibit(1) a different color of the bright bands(2) a different intensity of the dark bands(3) evenly spaced bright and dark bands(4) relatively wider central bright band(5) relatively smaller central bright band

23. In a double slit experiment, monochromatic light is shined through slits with separation 0.015 mm. The first order maximum appears at an angle of 6o from the center (use sin 6° = 0.1). What is the wavelength of the light? (1) 1.5 × 10–5 m (4) 3 × 10–6

(2) 3 × 10–5 m (5) 1.5 × 10–7

(3) 1.5 × 10–6 m

24. A transparent film with an index of refraction of 1.25 is used to coat a sheet of glass with an index of refraction of 1.5. What is the minimum thickness of the film needed to minimize the reflection of light with a wavelength of 600 nm?(1) 120 nm (4) 240 nm(2) 150 nm (5) 300 nm(3) 200 nm

25. The critical angle for a transparent material in air is 37º. The index of refraction of the material is most nearly(1) 0.6 (4) 1.25(2) 0.8 (5) 1.67(3) 1.0

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physics week 13(sem. 2) - st. francis preparatory …...information and diagram below. the diagram...

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