Physics 11 Unit 8 Geometric Optics Part 2
(c) Refraction (i) Introduction: Snell s law Like water waves, when light is traveling from one medium to another, not only does its wavelength, and in turn the speed, change, but also the direction of propagation is shifted. This is called the refraction. An example that demonstrates the idea of refraction is as follows: Unit 8 - Geometric Optics (Part 2) 2
The observation can be explained using the bent light from the coin immersed in water. The angle of the refracted light depends on the densities of the media and the incident angle of the incoming light. The optical density of a material is described in terms of the refractive index as introduced earlier. n = c v Unit 8 - Geometric Optics (Part 2) 3
The mathematical relationship between the angles of incidence and refraction was developed individually by many scientists such as I. Sahl (984), T. Harriot (1602), W. Snellius (1621), R. Descartes (1637), P. Fermat (1662), and C. Huygens (1687). This law, called the Snell s law, states that sin θ 1 sin θ 2 = v 1 v 2 = λ 1 λ 2 = n 2 n 1 The rule of thumb: the larger the refractive index, the smaller the refraction angle. Unit 8 - Geometric Optics (Part 2) 4
Example: A ray of light traveling in air has an angle of incidence of 30.0 on a block of quartz and an angle of refraction of 20.0. What is the index of refraction for this block of quartz? Since n 1 = 1.00, θ 1 = 30.0 and θ 2 = 20.0, by Snell s law, sin θ 1 sin θ 2 = n 2 n 1 n 2 = n 1 sin θ 1 sin θ 2 = 1.00 sin 30.0 sin 20.0 = 1.46 Unit 8 - Geometric Optics (Part 2) 5
Example: The speed of light in a clear plastic is 1.90 10 8 m/s. A ray of light traveling through enters the plastic with an angle of incidence of 22. At what angle is the ray refracted? We first have to determine the refractive index of the plastic. n = c 3.0 108 = v 1.90 10 8 = 1.579 Then by the Snell s law we have sin θ 2 = sin θ 1 n 1 n 2 = sin 22 1.00 1.579 = 13.7 Unit 8 - Geometric Optics (Part 2) 6
(ii) Total internal reflection (1) Critical angle According to the Snell s law, when a light is traveling from a low index medium (such as air) to a high index medium (such as water), it is partially reflected while the rest is refracted closer to the normal of the boundary. Conversely, when a light is traveling from a high index medium to a low index medium, it is partially reflected and the rest is refracted away from the normal line. Unit 8 - Geometric Optics (Part 2) 7
For a light that passes from a high index medium to a low index medium, when the angle of incidence increases, the angle of refraction also increases. There happens a moment in which the refracted angle is precisely 90 ; that is, the refracted ray travels parallel to the boundary of the two media. The corresponding angle of incidence is called the critical angle. At this moment, θ 2 = 90. Hence, by the Snell s law, sin θ c = sin 90 n 2 n 1 = n 2 n 1 or θ c = sin 1 n 2 n 1 8 Unit 8 - Geometric Optics (Part 2)
When the angle of incidence is larger than the critical angle, no light is refracted and the light is totally reflected. This phenomenon is called the total internal reflection (TIR). Total internal reflection only occurs when light is traveling from a high index medium to a low index medium. (Example: from water to air) For instance, from water to air, the critical angle is 1.00 θ c = sin 1 1.33 = 48.8 If the angle of incidence is greater than this, no light can pass through water and the water will behave like a mirror! Unit 8 - Geometric Optics (Part 2) 9
(2) Sparkle of diamonds An phenomenon that can be accounted for by the total internal reflection is the sparkle of diamond. When light enters into the diamond from the top, it gets refracted and hits the lower facet. If this second incident angle is greater than the critical angle of the diamond-air interface, it will be reflected totally back into the diamond and eventually exit the top face. That s why diamond glitters when it moves. It can be shown that the critical angle θ c of a diamond-air interface is about 24.4 provided the refractive index of diamond is 2.42. Unit 8 - Geometric Optics (Part 2) 10
Therefore, if a diamond is properly designed and cut, it would be able to reflect the incoming light totally and make the diamond sparkle. In general, diamonds can be graded according to its light performance in terms of brilliance, fire and sparkle: Unit 8 - Geometric Optics (Part 2) 11
The different light performance of diamonds depends on their cutting. It is one of the 4C used to evaluate the quality of a diamond (4C stands for cut, color, clarity and carat). Unit 8 - Geometric Optics (Part 2) 12
The flashes of rainbow colors, called fire, showed by a diamond are due to the breakdown of the incoming light into different colors. Note that different color components of light possess different wavelengths, and they are refracted differently in a medium because refractive index is dependent upon wavelength of the light. This is called dispersion, and is the working principle behind triangular prism. Unit 8 - Geometric Optics (Part 2) 13
(3) Applications of TIR Many optical instruments make use of the total internal reflection. Examples include binoculars, telescopes and periscopes. These species employ right-angled glass prisms to direct the incoming light to the eyepieces. Unit 8 - Geometric Optics (Part 2) 14
An important application of total internal reflection that has brought a great impact to our society is fiber optics which pipes light effectively from one place to another. An optical fiber is composed of a hair-like, flint-glass or plastic core wire wrapped by a concentric crown-glass shell called cladding. The cladding is covered by a thick layer of buffer/jacket in order to prevent physical damage. Unit 8 - Geometric Optics (Part 2) 15
The inner core has a higher refractive index (~1.6) than the outer shell (~1.5). When light enters into the optical fiber from one end, it strikes the core-cladding interface at an incident angle greater than the critical angle, and is therefore reflected back to the core. The light then travels along the fiber in a zigzag fashion. Unit 8 - Geometric Optics (Part 2) 16
Example: The figure shows an optical fiber that consists of a core made of flint glass (n = 1.667) surrounded by a cladding made of crown glass (n = 1.523). A ray of light in air enters the fiber at an angle θ 1 with respect to the normal. What is θ 1 if this light also strikes the core-cladding interface at an angle that just barely exceeds the critical angle? Unit 8 - Geometric Optics (Part 2) 17
The maximum angle of incidence with which the light is ensured to be reflected totally inside an optical fiber is called the numerical aperture (NA). Assume the refractive indices of air, core and cladding are n 1, n 2, and n 3 respectively. For a TIR occurring at the core-cladding interface, the condition is: n 2 sin θ c = n 3 sin 90 = n 3 Hence sin θ c = n 3 n 2 By the Snell s law, at the air-core interface n 1 sin θ 1 = n 2 sin θ 2 Unit 8 - Geometric Optics (Part 2) 18
Since θ 2 + θ c = 90, we have n 1 sin θ 1 = n 2 sin θ 2 = n 2 sin 90 θ c = n 2 cos θ c Using a trigonometric identity sin 2 x + cos 2 x = 1, we can write the last term as cos θ c = 1 sin 2 θ c Therefore, Recall that sin θ c = n 1 sin θ 1 = n 2 n 3 Τ n2, hence 1 sin 2 θ c n 1 sin θ 1 = n 2 1 n 3 n 2 2 = n 2 2 n 3 2 Unit 8 - Geometric Optics (Part 2) 19
Consequently, the numerical aperture is n 2 2 2 n 3 θ 1 = sin 1 n 1 For the previous example with n 1 = 1, n 2 = 1.667 and n 3 = 1.523, the NA is θ 1 = sin 1 1.667 2 1.523 2 = 42.7 1.000 which is what we have got before. Unit 8 - Geometric Optics (Part 2) 20
(ii) Lenses (1) Introduction The lenses used in optical instruments are made from transparent materials that refract light. They are carved in such a way that when light passes through them and gets refracted, an image of the light is formed. The simplest form of a lens is formed from two glass prisms. Due to the refraction, the light rays are bent toward the axis. Unfortunately, the rays do not cross at one point and thus no clear image of the object is formed. The quality of this lens is poor. Unit 8 - Geometric Optics (Part 2) 21
A better lens can be constructed from a single piece of transparent materials whose surfaces are properly curved. Every curved lens is associated with a focal point, center of curvature, principal axis, vertex (or optical center) and focal length. Their definitions are the same as those for mirrors. Unit 8 - Geometric Optics (Part 2) 22
Depending on whether the light rays exiting the lens are converged or diverged, lenses can be classified into two types: converging lens or diverging lens. They may come with many shapes but they possess the following unique features. Converging lens The lens is thicker at the middle than at the edges. It is also called convex lens. Diverging lens The lens is thinner at the middle than at the edges. It is also called concave lens. Unit 8 - Geometric Optics (Part 2) 23
Ray diagram is a method of analyzing how an image of an object is formed by a lens. The construction of a correct ray diagram for a converging lens follows the rules below: (1) An incoming light ray parallel to the principal axis is refracted by the lens and travels through the focal point at the other side of the lens. (2) An incoming light ray passing through the focal point is refracted by the lens, and travels in the direction parallel to the principal axis at the other side of the lens. (3) An incoming light ray passing through the vertex (or the optical center) of the lens is not refracted. As seen, these rules are similar to the rules used to draw the ray diagrams for a concave mirror. Unit 8 - Geometric Optics (Part 2) 24
The rules mentioned above are illustrated below. It is observed that the image formed by a converging lens depends on the location of the object. When an object is moved from far away toward the lens, the real image gets enlarged. But if the object is placed within the focal length, a virtual image will be formed instead. Unit 8 - Geometric Optics (Part 2) 25
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Similar rules are used to construct the ray diagrams for diverging lenses. Different from the converging lens, diverging lens always form virtual images which cannot be projected on a screen though visible. Unit 8 - Geometric Optics (Part 2) 27
Example: Complete the ray diagram below. Unit 8 - Geometric Optics (Part 2) 28
Example: Draw the ray diagram and find the location of the image. Unit 8 - Geometric Optics (Part 2) 29
Scenarios that we may encounter in daily life. Unit 8 - Geometric Optics (Part 2) 30
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(2) Thin-lens equation Using the Snell s law and the principles of refraction, one can derive the equation that describes the formation of images by lenses. The resulting equation, called the thin-lens equation, adopts the identical form of the mirror-equation which has been studied earlier. 1 f = 1 d 0 + 1 d i An equation relating the sizes of the object and the image can be derived in the way similar to that for mirrors. M = h i h o = d i d 0 Unit 8 - Geometric Optics (Part 2) 32
The meanings of the terms in the thin-lens equation can be shown by the following diagram: The sign conventions for lenses are summarized below. f: positive for converging lens; negative for diverging lens d 0 : positive for real object; negative for virtual object d i : positive if object and image are on different sides of the lens; negative if both object and image are on the same side. Unit 8 - Geometric Optics (Part 2) 33
Example: A person 1.70 m tall is standing 2.50 m in front of a digital camera. The camera uses a converging lens whose focal length is 0.0500 m. (a) Find the image distance and the determine whether the image is real or virtual. (b) Find the magnification and the height of the image on the image sensor. (a) Using d 0 = 2.50 and f = 0.05, we can find that 1 = 1 d i f 1 = 1 d 0 0.05 1 = 19.6 m 1 2.50 which gives d i = 0.0510 m. Since d i is positive, the image is real. Unit 8 - Geometric Optics (Part 2) 34
(b) The magnification is M = d i = 0.0510 d 0 2.50 = 0.0204 It means that the image is smaller and inverted. The height of the image is given by M = h i h o Hence, h i = Mh o = 0.0204 1.70 = 0.0347 m. Unit 8 - Geometric Optics (Part 2) 35
Example: An object is placed 7.10 cm to the let of a diverging lens whose focal length is 5.08 cm. (a) Find the image distance and determine whether the image is real or virtual. (b) Obtain the magnification. (a) Recall a diverging lens has a negative focal length. Hence, f = 5.08 cm. By the thin-lens equation, 1 = 1 d i f 1 = 1 d o 5.08 1 7.10 and the image distance is d i = 2.96 cm. Since d i is negative, the image is virtual. = 0.338 cm 1 Unit 8 - Geometric Optics (Part 2) 36
(b) The magnification of the lens is M = d i d 0 = 2.96 7.10 = 0.417 Unit 8 - Geometric Optics (Part 2) 37
(3) Lenses in combination The magnification of a lens depends on the curvature, and in turns the focal length of the lens. There is therefore an intrinsic limitation of its magnification power. This problem can be resolved by adding one more piece of lens. The first piece of lens, called the objective, creates an image which serves as the object of the second piece of lens, called the eyepiece or ocular. Usually the two lenses are placed in such a way that the first image produced by the objective is located within the focal length of the eyepiece so that the resulting image is virtual (so that it is visible behind the eyepiece) and enlarged. Unit 8 - Geometric Optics (Part 2) 38
The following is the schematic of a two-lens magnifying instrument (e.g. compound microscope) Unit 8 - Geometric Optics (Part 2) 39
To see how the position of the final image can be determined, let s look at the following example. Example: The objective and eyepiece of a compound microscope are both converging lenses and have focal lengths of 15.0 mm and 25.5 mm respectively. A distance of 61.0 mm separates the lenses. The microscope is being used to examine an object placed 24.1 mm in front of the objective. Find the final image distance. The following diagram depicts the situation: Unit 8 - Geometric Optics (Part 2) 40
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First of all, we have to find the location of the first image. Since f 0 = 15.0 mm, and d o1 = 24.1 mm, by the lens equation, 1 = 1 1 = 1 d i1 f o d o1 15.0 1 24.1 = 0.0252 mm 1 which gives d i1 = 39.7 mm. It means that the first image is 61.0 39.7 = 21.3 mm from the eyepiece. Therefore d o2 = 21.3 mm. Using f e = 25.5 mm, 1 = 1 1 = 1 d i2 f e d o2 25.5 1 21.3 = 0.0077 mm 1 yielding d i2 = 130 mm. The image formed is virtual. Unit 8 - Geometric Optics (Part 2) 42
The overall magnification of the compound microscope is the product of the magnifications of the two lenses. That is, M = M o M e For the abovementioned example, M o = 39.7 24.1 = 1.647 M e = 130 21.3 = 6.103 Hence, the overall magnification is M = 1.647 6.103 = 10.05 The image is about 10 times larger and inverted. Unit 8 - Geometric Optics (Part 2) 43
Practice: A two-lens system consists of a converging lens (f 1 = 20.0 cm) and a diverging lens (f 2 = 15.0 cm) separated by a distance of 10.0 cm. An object with a height of 5.00 mm is placed at a distance of 45.0 cm to the left of the first (converging) lens. What is the height of the final image? [5.46 mm] Unit 8 - Geometric Optics (Part 2) 44
(4) The human eye An eye is a very sophisticated optical device that allows a person to see things in the world. The physical principle behind the operation of eyes can be explained in terms of lens. The structure of a human eye is illustrated below. 1. Cornea a transparent membrane through which light enters into the eye ball 2. Iris the circular diaphragm that controls the amount of light coming in 3. Lens responsible for focussing of incoming light onto the retina 4. Retina the light-sensitive part (with rods and cones) that create nerve signals Unit 8 - Geometric Optics (Part 2) 45
When light comes into an eye through the cornea, it is refracted and strikes on the lens. Since the distance between the lens and the retina is fixed, in order to make sure a sharp image can be formed on the retina, the curvature of the lens is adjusted by the ciliary muscle. To see a distant object, the ciliary muscle is not tensed, and the lens is fully relaxed, having the maximum focal length possible. This focal length coincides with the size of the eyeball, therefore producing a sharp image right on the retina. Unit 8 - Geometric Optics (Part 2) 46
On the other hand, to see a close object, the ciliary muscle is tensed to increase the curvature of the lens, reducing its focal length so that the sharp image can be formed on the retina instead of behind it. When a sharp image is formed on the retina, we say that the eye is focused on the object. The process in which the lens changes its focal length so that a sharp image can be formed on the retina is called accommodation. Unit 8 - Geometric Optics (Part 2) 47
There is a closest point to the eye at which our eyes can focus. This point is called the near point of the eyes. The ciliary muscle is fully tensed when an object is placed at this point. The near point for a normal eye is about 25 cm, and this point increases with age: 50 cm for 40-year-old adults and 500 cm for seniors. Eyes cannot focus on any object placed within the near point, and the image will look blurred. Similarly, there is a far point corresponding to the farthest position at which an object can be seen clearly when the eye is fully relaxed. A person with normal eyesight can see stars and planets in the sky. Therefore, his far point is essentially close to infinity. (Of course, it is not possible to be truly infinity!) Unit 8 - Geometric Optics (Part 2) 48
If the ability to adjust the curvature of the lens is weakened, a person may suffer from either of the following problems. Nearsightedness (myopia) A person with myopia can focus on nearby objects but cannot see clearly the things far away. When an eye with myopia tries to see a distant object, it is fully relaxed, but the maximum focal length is shorter than it should be, therefore making the image formed in front of the retina. Either the lens being too curved or the eyeball being too long causes myopia. Unit 8 - Geometric Optics (Part 2) 49
To correct the problem of nearsightedness, the person can wear a pair of glasses or contacts that use diverging lenses. The lenses diverge the rays from the object so that the final image can be formed right on the retina. The diverging lens serves to create a virtual image at the far point of the myopic eye so that the person can focus. Unit 8 - Geometric Optics (Part 2) 50
Farsightedness (hyperopia) Different from myopia, a person suffering from hyperopia can see distant objects clearly yet he cannot focus on nearby objects. The near point of a hyperopic eye is much larger than the normal value (e.g. 300 cm versus 25 cm). It is caused by either the eyeball being too short, the lens sitting farther back in the eye, or the lens being too stiff to be tensed. Unit 8 - Geometric Optics (Part 2) 51
The farsightedness can be corrected when the person wears a pair of glasses made of converging lenses. The lenses converge the incoming rays more before they enter into the eyes, thereby allowing the lens, which is already too flat, to focus the rays on the retina instead of somewhere behind it. Unit 8 - Geometric Optics (Part 2) 52
Example: A nearsighted person has a far point located only 521 cm from the eye. Assuming that eye-glasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lenses of the glasses so that the person can see distant objects. Referring to the diagram on p.50, the position of the image (i.e., the far point) is 521 2 = 519 cm from the glasses. For correcting nearsightedness, diverging lenses are used. Therefore, d i = 519 cm. Using the lens equation: 1 f = 1 + 1 = 1 d i d o 519 + 1 Hence, f = 519 cm. Unit 8 - Geometric Optics (Part 2) 53
Example: A farsighted person has a near point located 210 cm from the eyes. Obtain the focal length of the converging lenses in a pair of contacts that can be used to read a book held 25.0 cm from the eyes. A contact lens is placed right against the eyes. Therefore, the object distance is exactly 25.0 cm. If it is desired for the image to be formed at the near point, the image distance will be d i = 210 cm. (Refer to the diagram on p.52.) By the lens equation, 1 f = 1 + 1 = 1 d i d o 25 + 1 210 Hence, f = 28.4 cm. Unit 8 - Geometric Optics (Part 2) 54