physics week 12(sem. 2) - st. francis preparatory school · 2020. 10. 7. · physics week 12(sem....
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Ms. N. May ABC Math Student Copy
Physics Week 12(Sem. 2) Name____________________________
Light Chapter Summary Cont’d
Refraction
Light not only travels through a vacuum, but it also can travel through many materials such as air, water, and even glass. As a ray of light changes speed at the materials boundary it causes the ray the deviate from its incident direction. This change in direction is called refraction and is governed by Snell’s law of refraction. Snell’s law make use of the index of refraction, this is a ratio of the speed of light in a vacuum to the speed of light in the material. The equation to calculate the index of refraction is
In the above equation, index of refraction (n) is unit‐less. It uses the speed of light in a vacuum, 3.00x108 m/s as a reference speed. All refractive indices are greater than 1 because light travels slower through a medium than it does in a vacuum. For most gases (including air) the speed of light is very close to that in a vacuum so the index of refraction is taken to be 1. The index of refraction has a slight dependence on the wavelength of light, thus be sure to check the wavelength of light the indices were derived from.
Snell’s Law
When light strikes a boundary of two transparent materials the light usually divides into two parts. Part of the light is typically reflected at the boundary, while the other part is transmitted through the boundary. The angle the reflected light makes with the boundary is equal to the angle that the incident light makes with the normal line, this is called the law of reflection (θr= θi). If the incident light does not strike the boundary at a 90o angle than the refracted ray will continue through the material at a different angle/direction than the incident light. When light travels through a medium
with a smaller index of refraction to a material with a larger index of refraction, the light refracts towards the normal line. Also, if the light travels through a higher index material into a lower index material the light will bend away from the normal line. Note that the material where the incident light originates is labeled with a subscript of 1, while the refracted ray gets a 2 subscript. The relationship between the angle of refraction and the angle of incidence is given by Snell’s law
sin sin
Where n1 is the index of refraction where the incident ray originates, θ1 is the incident angle with respect to the normal line, n2 is the index of refraction of the material that contains the refracted ray, θ2 is the refraction angle with respect to the normal line.
Apparent Depth
With refraction changing the angle of travel through a boundary, objects under water appear closer to the surface than they are. As light rays travel through air into water they will be bent towards the normal.
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Ms. N. May ABC Math Student Copy
However, as light rays travel from the water into the air they will be bent away from the normal. So if the observer were to look from a top a water surface an object in the water will have an image at a different location than the object. If the observer were to be directly on top of the object looking down the objects true depth can be determined using the equation
Where d’ is the apparent depth, d is the actual depth, n2 is the index of refraction of the medium containing the refracted ray and n1 is the index of refraction containing the incident ray.
Total Internal Reflection
When light passes through a medium of larger refractive index into a material of a smaller index of refraction, the refracted ray bends away from the normal line. As the angle of incidence increases, the angle of refraction also increases. When the angle of incidence reaches a certain point, the critical angle (θc), the angle of refraction is 90o. This angle of refraction results in a ray that points along the surface, thus when the angle of incidence exceeds the critical angle there is no refracted light.
Thus all of the incident light to be reflected within the material that it came from resulting in total internal reflection. Total internal reflection only occurs when light travels from a higher refractive index into a medium of a lower refractive index, it does not occur in the reverse path. Using Snell’s law an equation for the critical angle can be obtained
sinsin 90
Since the sine of 90 degrees is 1 the equation reduces to a ratio of the index of refractions. Above which angle total internal reflection will occur so long as n1 is greater
than n2. This phenomenon is important in diamonds sparkling.
Dispersion of Light: Prisms & Rainbows
When light enters the prism on the left face, the refracted ray is bent toward the normal since the refractive index or glass is greater than air. When the light leaves the right face of the prism it is refracted away from the normal. Because the refractive index of glass depends on the wavelength, the rays corresponding to different colors are bent by different amounts by the prism and leave the prism in different directions. The greater index of refraction for a given color, the greater the bending. Violet, with the shortest wavelength bends the most and red light with the longest wavelength bands the least. Therefore when white light, that contains all colors, enters a prism it will separate into its colors and show a rainbow. The spreading of light into its color components is called dispersion.
Lenses
Lenses are similar to mirrors, but they have one main difference. Lenses refract light allowing it to travel through the transparent material compared to mirrors that reflect light.
Converging Lenses
A converging lens is one that is thicker in the center of the lens when compared to the edges. With a spherical lens, rays relatively close to the principal axis and parallel to it converge at a single point, this is called the focal point (F). Thus an object located infinitely car away far on the principal axis will produce an image at the focal point. Considering only thin lenses where making the focal length (f) measurement from either surface of the lens is acceptable and will provide an accurate answer. Such examples of converging lenses would be the double convex lens (protruding both sides), planoconvex(flat on back of lens) and convex meniscus (curved on back to match front).
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Ray Diagrams for Converging Lenses
Similar to mirrors, however the ray of light is able to pass through the lens (assuming left to right). For the first paraxial ray, it travels parallel to the principal axis and then is refracted toward the axis and travels through the focal point on the right side of the lens. The second ray passes through the focal point on the left and then once hitting the lens it is refracted parallel to the principal axis. The third ray passes through the center of the lens and is not refracted or bent at all, so long as the front and back of the lens are nearly parallel at the center.
Diverging Lenses
Another type of lens causes the parallel incident rays to diverge after exiting the lens, diverging lens. For this type of lens, paraxial rays that are parallel to the principal axis appear to originate from a single point on the axis after passing through the lens. This point is called the focal point (F), and f is the focal length to the lens (assuming it is thin). Examples of diverging lens are double concave, plano concave and concave meniscus.
Ray Diagrams for Diverging Lenses
For the first paraxial ray, it travels parallel to the principal axis and then is refracted away from the axis and appears to have originated from the focal point on the left of the lens. The second ray leaves the object and moves towards the focal point on the right of the lens, but before reaching the focal point it is refracted parallel to the axis. The third ray passes through the
center of the lens and is not refracted or bent, so long as the front and back of the lens are nearly parallel at the center.
Images for Converging Lens
If the object is placed at a distance greater than 2F, any of the three rays will confirm, the resulting image will be real, inverted and smaller. This is used in cameras. If the object is placed between 2F and F the image is still real and inverted, but now it is larger than the object. This is used in film projectors, the film must be placed upside down to get the proper image. When the object is placed between the focal point and the lens, the rays diverge after leaving the lens, producing a virtual image appearing to originate from behind the lens. This virtual image will be upright and larger than the object. This is the image produced when using a magnifying glass.
Images for Diverging Lens
Light rays diverge when leaving a diverging lens therefore the images formed are virtual, meaning on the same side as the object (or incident ray). Regardless of the position of the object (for a diverging lens), the image will be virtual, upright and reduced in size.
Equations for Lenses
These equations are the same that apply to mirrors. The first equation relates the focal length to both the image and object distance (to the lens).
1 1 1
Where f is the focal length, do is the distance of the object to the lens and di is the distance of the image to the lens. For converging lenses, the focal length (f) is positive because it is on the opposite side of the lens. For a diverging lens, the focal length (f) is negative. For the magnification equation, it is a ratio of the height of the object relative to the image height. If the object height is greater than the image height than the magnification (m) is greater than 1. If the object height is less than the image height than the magnification (m) is less than 1. Also, the height of the object and image
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Ms. N. May ABC Math Student Copy
can be related to the distance relative to the lens. Thus the equation for magnification can be
Where m is the magnification, hi is the height of the image, ho is the height of the object, di is the distance of the image to the lens and do is the distance of the object to the lens. The value of m will determine if the image is upright or inverted, a positive number indicating an upright image. Diverging lenses always produce virtual images on the same side of the lens as the object. See table below for sign conventions for lenses.
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Ms. N. May ABC Math Student Copy
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Ms. N. May ABC Math Student Copy
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Ms. N. May ABC Math Student Copy
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Ms. N. May ABC Math Student Copy
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1. Which diagram best represents the path taken by a ray of monochromatic light as it passes from air through the materials shown?
(1) (3)
(2) (4)
2. A laser beam is directed at the surface of a smooth, calm pond as represented in the diagram below.
Which organisms could be illuminated by the laser light?(1) the bird and the fish (3) the crab and the seaweed(2) the bird and the seaweed (4) the crab and the fish
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3. A beam of monochromatic light (f = 5.09 × 1014 hertz) passes through parallel sections of glycerol, medium X, and medium Y as shown in the diagram below.
What could medium X and medium Y be?(1) X could be flint glass and Y could be corn oil.(2) X could be corn oil and Y could be flint glass.(3) X could be water and Y could be glycerol.(4) X could be glycerol and Y could be water.
4. Base your answer to the following question on the information and diagram below.
A converging lens has a focal length of 0.080 meter. A light ray travels from the object to the lens parallel to the principal axis.
Which phenomenon best explains the path of the light ray through the lens?(1) diffraction (3) reflection(2) dispersion (4) refraction
5. The diagram at the right represents the path of periodic waves passing from medium A into medium B. As the waves enter medium B, their speed(1) decreases (3) remains the same(2) increases
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6. The diagram below represents straight wave fronts passing from deep water into shallow water, with a change in speed and direction.
Which phenomenon is illustrated in the diagram?(1) reflection (2) refraction (3) diffraction (4) interference
7. Which diagram best represents the path of light rays passing through a glass prism?(1)
(2)
(3)
(4)
8. Total internal reflection can occur as light waves pass from(1) water to air(2) Lucite to crown glass(3) alcohol to glycerol(4) air to crown glass
9. The diagram below shows a ray of light passing from medium X into air.
What is the absolute index of refraction of medium X?(1) 0.500 (3) 1.73(2) 2.00 (4) 0.577
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10. The diagram below shows a ray of light (Ø = 5.9 × 10-7 meter) traveling from air into medium X.
If the angle of incidence is 30.° and the angle of refraction is 19°, medium X could be(1) air (3) Canada balsam(2) alcohol (4) glycerol
11. The speed of a ray of light traveling through a substance having an absolute index of refraction of 1.1 is(1) 1.1 × 108 m/s (3) 3.0 × 108 m/s(2) 2.7 × 108 m/s (4) 3.3 × 108 m/s
12. Which quantity is equivalent to the product of the absolute index of refraction of water and the speed of light in water?(1) wavelength of light in a vacuum(2) frequency of light in water(3) sine of the angle of incidence(4) speed of light in a vacuum
13. When a student 1.5 meters tall stands 5.0 meters in front of a lens, his image forms on a screen located 0.50 meter behind the lens. What is the height of the student's image?(1) 0.015 m (3) 1.5 m(2) 0.15 m (4) 15 m
14. The diagram below shows an arrow placed in front of a converging lens.
The lens forms an image of the arrow that is(1) real and inverted (3) virtual and inverted(2) real and erect (4) virtual and erect
15. Base your answer to the following question on the information and diagram below. A convex lens having optical center O and principal focus F is used to produce an image of a candle. Ray RF is shown.
As the candle is moved toward the left, the size of its image will(1) decrease (3) remain the same(2) increase
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Base your answers to questions 16 and 17 on the diagram below which represents an object placed at point O, located 1.0 meter from a lens of focal length 0.50 meter.
16. Compared to the image when the object is at point O, the image when the object is moved to point B will be(1) smaller(2) changed from inverted to erect(3) nearer the lens(4) farther from the lens
17. Compared to the size of the object, the size of the image is(1) smaller (3) the same(2) larger
18. A student uses a magnifying glass to examine the crystals in a mineral specimen. The magnifying glass contains a(1) convex (diverging) mirror(2) convex (converging) lens(3) concave (diverging) lens(4) plane mirror
19. Which optical device may form an enlarged image?(1) plane mirror (3) converging lens(2) glass plate (4) diverging lens
20. An object 0.16 meter tall is placed 0.20 meter in front of a concave (diverging) lens. What is the size of the image that is formed 0.10 meter from the lens?(1) 0.040 m (3) 0.16 m(2) 0.080 m (4) 0.32 m
21. Virtual images can be formed by(1) converging lenses, only(2) diverging lenses, only(3) neither converging nor diverging lenses(4) either converging or diverging lenses
Base your answers to questions 22 and 23 on the diagram below which represents an object placed 0.20 meter from a converging lens with a focal length of 0.15 meter.
22. The image produced by the lens is(1) enlarged and real(2) enlarged and erect(3) diminished and virtual(4) diminished and inverted
23. If the object distance were increased, the image would become(1) larger and erect (3) smaller, only(2) smaller and virtual (4) larger, only
24. The diagram below shows an object placed in area A before a converging lens. Where is the image produced?
(1) at F (3) in area Y(2) at 2F (4) in area Z
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25. Which diagram below shows the path of light rays as they pass from an object at 2F through a converging lens to the image formed at 2F?(1)
(2)
(3)
(4)
26. Base your answer to the following question on the diagram below which represents a converging lens with a focal length of 0.20 meter. DE represents a light ray parallel to the principal axis. OD is an object perpendicular to the principal axis.
This type of lens can not be used to form images which are(1) real and enlarged (3) virtual and enlarged(2) real and reduced (4) virtual and reduced
27. Base your answer to the following question on the diagram below which represents a converging crown glass lens with a focal length of 8.0 centimeters. Ray A represents one of the light rays incident upon the lens.
Which ray best represents the path of ray A after it emerges from the lens?(1) 1 (3) 3(2) 2 (4) 4
28. The diagram below shows the refraction of the blue and red components of a white light beam.
Which phenomenon does the diagram illustrate?(1) total internal reflection(2) critical angle reflection(3) spherical aberration(4) chromatic aberration
29. An object is located 0.15 meter from a converging lens with focal length 0.10 meter. How far from the lens is the image formed?(1) 0.060 m (3) 0.15 m(2) 0.10 m (4) 0.30 m
30. An object 0.080 meter high is placed 0.20 meter from a converging (convex) lens. If the distance of the image from the lens is 0.40 meter, the height of the image is(1) 0.010 m (3) 0.080 m(2) 0.040 m (4) 0.16 m
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31. Base your answer to the following question on the formation below:
A crown glass converging lens has a focal length of 0.10 meter.
An object is placed 0.30 meter from the lens. How far from the lens will an image of the object be formed?(1) 0.30 m (3) 0.15 m(2) 0.20 m (4) 0.10 m
32. Base your answer to the following question on the diagram below which represents a convex lens being used to form the image of an object. The distance from the center of the lens to the object is 0.060 meter. The distance from the center of the lens to the image is 0.120 meter.
If the height of the object is 2.6 × 10–2 meter, the height of the image is(1) 1.3 × 10–2 m (3) 2.6 × 10–2 m(2) 2.0 × 10–2 m (4) 5.2 × 10–2 m
33. Images formed by diverging mirrors are always(1) real and inverted (3) virtual and inverted(2) real and erect (4) virtual and erect
34. When an object is placed 0.40 meter from a diverging lens with a focal length of –0.10 meter, the image produced will be(1) virtual and smaller than the object(2) virtual and larger than the object(3) real and smaller than the object(4) real and larger than the object
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35. The diagram below shows light ray R incident on a glass lens in air.
Which ray best represents the path of light ray R after it passes through the lens?(1) A (2) B (3) C (4) D
36. Which diagram below correctly represents rays of light passing through a glass lens?
(1) (3)
(2) (4)
37. The diagram below shows an object placed between 1 and 2 focal lengths from a converging lens.
The image of the object produced by the lens is(1) real and inverted (2) real and erect (3) virtual and inverted (4) virtual and erect
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38. Which graph best represents the relationship between image size (Si) and image distance (di) for real images formed by a converging lens?
(1) (3)
(2) (4)
39. A lens produces a real image by causing light rays from a common point to(1) converge and intersect at a point(2) disperse into component wavelengths(3) reflect constructively(4) diverge and appear to come from a point
40. Base your answer to the following question on the information below.
When a 0.020-meter-tall object is placed 0.15 meter in front of a converging mirror, the object's image appears 0.30 meter in front of the mirror.
How tall is the image?(1) 0.010 m (3) 0.030 m(2) 0.020 m (4) 0.040 m
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