Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 23

17.3 Total Internal Reflection and Optical Fibers Thus : θc = sin−1 571 n1 = sin−1 0.752 = 48.8◦ = sin−1 n3 1.33 (b) From the right-angle triangle at the glass-water interface we can find the refracted angle θ3 in water to be: θ3 = 90◦ − θc = 41.2◦ Using Snell’s law again at the glass-water interface, we have: n2 sin θ2 = n3 sin θ3 Thus: sin θ2 = n3 sin θ3 1.33 × sin 41.2◦ = = 0.584 n2 1.5 θ2 = sin−1 0.585 = 35.7◦ (c) Since the sides of the glass-walled fish tank are parallel, we can again apply Snell’s law at the air-glass interface to calculate θ1 as follows: n1 sin θ1 = n2 sin θ2 Thus: sin θ1 = n2 sin θ2 1.5 × sin 35.7◦ = 0.875 = n1 θ1 = sin−1 0.875 = 61◦ 17.4 Chromatic Dispersion and Prisms Except in vacuum, the index of refraction depends on the light’s wavelength, i.e its color, see Sect 27.7 Therefore if a beam of light consists of rays of different wavelengths (as in the case of white light), each ray will refract by a different angle from a surface This spread of light is called chromatic dispersion, or simply dispersion Generally, the index of refraction n decreases with increasing wavelengths This means that the violet light (with wavelength λ 425 nm and index n = 1.3435) bends more than the red light (with wavelength λ 700 nm and index n = 1.3318) when passing through the interface between two materials Figure 17.9a shows this for a glass block, and Fig 17.9b shows this for a glass prism The prism of Fig 17.9b is more commonly used to observe color separation of white light because the dispersion at the first surface is enhanced at the second 572 17 Light Waves and Optics interface Thus, the violet ray in the white light of Fig 17.9b will emerge from the right surface with an angle of deviation δV which is greater than the angle of deviation δR of the red ray The difference δV − δR is known as the angular dispersion, while δY is the mean deviation of the yellow rays White light A Air R δR V Glass R White light δV V Glass (a) (b) Fig 17.9 A schematic representation of the dispersion of white light The violet color is bent more than the red color (a) Dispersion in a glass block (b) Dispersion in a prism The general expression of δ for any color turns out to be rather complicated However, as the angle of incidence decreases from a large value, the angle of deviation δ is found to decrease at first and then increase The angle of minimum deviation δm is found when the ray passes through the prism symmetrically This angle is related to the angle of the prism A, and its index of refraction n by the relation: n= sin[(A + δm )/2] sin(A/2) −−−−−−−−−−→ When A is small n= A + δm A (17.12) The most charming example of color dispersion is that of a rainbow To understand the formation of a rainbow we consider a horizontal overhead white sunlight that is intercepted by spherical raindrops Figure 17.10 shows refractions and reflection in two raindrops that explain how light rays from the Sun reach an observer’s eye The first refraction separates the sunlight into its color components Each color is then reflected at the raindrop’s inner surface Finally, a second refraction increases the separation between colors, and these color rays finally make it to the observer’s eye Using Snell’s law and geometry, we find that the maximum deviation angles of red and violet are about 42 and 40◦ , respectively The rainbow that you can see is a personal one because different observers receive light from different raindrops 17.4 Chromatic Dispersion and Prisms 573 Raindrop White light Part of a rainbow 42° White light 40° ht t lig le Vio To observer d li Re et Viol ght light d Re ht lig Fig 17.10 A sketch of a rainbow formed by horizontal sunlight rays Only two enlarged raindrops are used to explain the rainbow’s formation for the case of the red and violet colors only Example 17.4 A monochromatic light ray is incident from air (with index n1 = 1) onto an equilateral glass prism (with index n2 = 1.5) and is refracted parallel to one of its faces (i.e we have a symmetric ray), see Fig 17.11 (a) What is the angle of incidence θ1 at the first face? (b) What is the subsequent angle of incidence at the second face? (c) Is the light ray totally reflected at the second face? If not, find the angle of minimum deviation of the light ray Then check that Eq 17.12 holds Fig 17.11 Air q1 60° q q 1′ 60° Glass dm q 2′ 60° Solution: (a) The path of a symmetric light ray going through the prism (of apex angle 60◦ ) and back out again into the air is shown Using elementary geometry, this figure shows that the angle of refraction θ2 can be found as follows: θ2 + 60◦ = 90◦ Thus: θ2 = 30◦ 574 17 Light Waves and Optics Therefore, using Snell’s law: n1 sin θ1 = n2 sin θ2 We get: θ1 = sin−1 = sin−1 n2 sin θ2 n1 1.5 × sin 30◦ = sin−1 (0.75) = 48.59◦ (b) Again, by simple geometry the horizontal light ray inside the prism must be incident on the second face with an angle θ1 = θ2 = 30◦ (c) We know that if the incident angle is greater than the critical angle, then total internal reflection must occur Therefore, we first calculate the critical angle as follows: n1 n2 = sin−1 1.5 = sin−1 0.666 θc = sin−1 = 41.8◦ Since θ1 < θc , then the light ray refracts at the second face, and total internal reflection will not occur Using the geometry shown in the figure, we can find for this special case that the angle of minimum deviation is given by the following relation: δm = 2(θ1 − θ2 ) = 2(48.59◦ − 30◦ ) = 37.18◦ Substituting A = 60◦ and δm = 37.18◦ in Eq 17.12 gives: sin[(A + δm )/2] sin(A/2) sin[(60◦ + 37.18◦ )/2] = sin(60◦ /2) 0.75 = 1.5 = 0.5 n2 = 17.4 Chromatic Dispersion and Prisms 575 This value of n2 obtained from Eq 17.12 satisfies the given value of index of refraction of the prism 17.5 Formation of Images by Reflection Mirrors gather and redirect light rays to form images of objects by reflection To explain this, we will use the ray approximation model in terms of geometric optics, in which light travels in straight lines 17.5.1 Plane Mirrors A plane mirror is a plane surface that can reflect a beam of light in one direction instead of either scattering it in many directions or absorbing it Figure 17.12a shows how a plane mirror can form an image of a point object O located at a distance p from the mirror In this figure, we consider two diverging rays leaving O and strike the mirror and then are reflected to the eye of an observer The rays appear to diverge from point I behind the mirror Thus, point I is the image of point O The geometry of the figure indicates that the image I is opposite to object O and is located at a distance as far behind the mirror as the object is in front of the mirror p O i q q q I p O i q q h′ h q I q q Mirror Mirror Back side Front side (a) Front side Back side (b) Fig 17.12 A geometric sketch that is used to depict an image of an object placed in front of a plane mirror (a) An image formed for a point object (b) An image formed by an extended object, where the object is an upright arrow of height h 576 17 Light Waves and Optics Figure 17.12b shows how a plane mirror can form an image of an extended object O The object in this figure is an upright arrow of height h placed at a distance p from the mirror The full image can be inferred by locating the images of selected points on the object One of the two rays at the tip of the arrow follows a horizontal path to the mirror and reflects back on itself The second ray follows an oblique path and reflects according to the laws of reflection, as shown in the figure Using geometry we find that the image I is upright, opposite to the object, and located behind the mirror at a distance equal to the object’s distance in front of the mirror In addition, the height of the object and its image are equal Also, the geometry of Fig 17.12b indicates that h /h = i/p The image I in both parts of Fig 17.12 is called a virtual image because no light rays pass through it In addition, the value of i is considered to be negative since the image is behind the mirror and the value of h is considered to be positive since the image is upright We define the lateral magnification M of a horizontal overhead image as follows: M= Image height h = Object height h (17.13) We can use the relation h /h = i/p and the sign convention to write the lateral magnification M as follows: M= i h =− h p (17.14) For plane mirrors, M = 1, since h is positive and equal to h, or i is negative and has a magnitude equal to p The image formed by a plane mirror is upright but reversed The reversal of right and left is the reason why the word AMBULANCE is printed ” across the front of ambulance vehicles People driving in front as “ of such an ambulance can see the word “AMBULANCE” immediately evident when looking in their rear-view mirrors and make way 17.5.2 Spherical Mirrors A spherical mirror is simply a mirror in the shape of a small section of the surface of a sphere that has a center C and radius R When light is reflected from the concave 17.5 Formation of Images by Reflection 577 surface of the mirror, the mirror is called a concave mirror However, when light is reflected from the convex surface of the mirror, the mirror is called a convex mirror Focal Point of a Spherical Mirror The principal axis (or the symmetry axis) of a spherical mirror is defined as the axis that passes through its center of curvature C and the center of the mirror c, see Fig 17.13 We consider the reflection of light coming from an infinitely far object O located on the principal axis of a concave or convex spherical mirror Because of the great distance between the object and the mirror, the light rays reach the mirror parallel to its principal axis Convex mirror Concave mirror Principal axis C F c Principal axis F C c Virtual focal point Real focal point Front side f R Back side f Front side R Back side (b) (a) Fig 17.13 (a) Two parallel light rays will meet at a real focal point after reflecting from a concave mirror (b) The same rays will diverge from a convex mirror and appear to come from a virtual focal point When parallel rays reach the surface of the concave mirror of Fig 17.13a, they will reflect and pass through a common point F If we place a card at F, a point image would appear at F Therefore, this point is called the real focal point However, in the case of the convex mirror of Fig 17.13b, the parallel rays reflect from the mirror and appear to diverge from a common point F behind the mirror If we could place a card at F, no image would appear on the card Therefore, this point is called the virtual focal point The distance f from the center of the mirror to the focal point (real or virtual) is called the focal length of the mirror For concave and convex mirrors, the following relation relates the focal length f to the radius of curvature R: f = R (Spherical mirror) (17.15) 578 17 Light Waves and Optics 17.5.2.1 Concave Mirrors Sharp and Blurred Images Rays that diverge from any point on an object and make small angles with the principal axis (called paraxial rays) will reflect from the spherical concave mirror and intersect at one image point See Fig 17.14a for a point object on the principal axis On the other hand, rays that diverge from the same point and make large angles with the principal axis will reflect and intersect at different image points, see Fig 17.14b This condition is called spherical aberration Small angles incidence Large angles incidence c c O O I I1 I2 Sharp image (a) blurred image (b) Fig 17.14 (a) When rays diverge from point object O at small angles with the principal axis, they all reflect from the spherical concave mirror and meet at the same point image I (b) When rays diverge from O at large angles with the principal axis, they reflect from the spherical concave mirror and meet at different points I1 , I2 , The Mirror Equation The relationship between an object’s distance p, its image distance i, and the focal length f of a concave mirror can be found when light rays make small angles with the principal axis (paraxial rays) Figure 17.15a shows two rays (leaving an object O of height h) reflected to form an image I of height h The first ray strikes the mirror at its center c and is reflected The second ray passes through the focal point F and reflects parallel to the principal axis From the purple triangles of Fig 17.15a, we see that: tan θ = h h = p i ⇒ h i = h p (17.16) From the yellow triangles of Fig 17.15b, we see that: tan α = h h = p−f f ⇒ h f = h p−f (17.17) 17.5 Formation of Images by Reflection O 579 O Front h h′ q F q h c Front a h′ c F a h′ h′ I I f p i p (a) i f (b) Fig 17.15 (a) Intersection of two rays produced by a spherical concave mirror to form an image of the tip of an arrow (b) Demonstration of the geometry produced by only the second ray By comparing Eqs 17.16 and 17.17, we find that: f i = p p−f ⇒ ip − if = pf (17.18) Dividing both sides of this equation by pif, we get: 1 + = p i f (17.19) Equation 17.19 is known as the mirror equation for spherical mirrors, and this expression holds when we interchange p and i, i.e when we can replace the object O by the image I and vice versa For a given value of f, we notice the following for concave mirrors: • When p > f , the image distance i is positive A positive value of i means that the image is real and inverted See Fig 17.16a,b for images smaller or larger than the object • When p < f , the mirror equation is satisfied by a negative value of the image distance i The negative image distance means that the image is virtual When we extend two rays from the object we find that the virtual image is upright and enlarged, see Fig 17.16d If we use this sign convention in the lateral magnification Eq 17.13, then we can also write M as follows: M= h i =− h p (17.20) We get an upright image for positive values of M and an inverted image for negative values of M as shown in Fig 17.16 580 17 Light Waves and Optics O Front Front O F C F Real image c I c C I Real image (a) (b) Front O O c C (i = + ∞) I Front c F C Virtual image F (c) (d) Fig 17.16 (a) An object O outside the center of curvature C (b) The object between the focal point F and C (c) The object at F (d) The object inside the focal point F and its virtual upright image I 17.5.2.2 Convex Mirrors Convex mirrors like those shown in Fig 17.17 are called diverging mirrors The images formed by these types of mirrors are virtual because the reflected rays appear to originate from an image behind the mirror Furthermore, the images are always upright and smaller than the object Because of this feature, these types of mirrors are often used in stores to prevent shoplifting Convex mirror Virtual image O Principal axis c I F Virtual focal point f Front side C R Back side Fig 17.17 When the object O is in front of a convex mirror, the image is virtual, upright, and smaller than the object 586 17 Light Waves and Optics Solution: For flat refracting surfaces, we use Eq 17.28 to find the location of the image Thus: i=− n2 p=− p = −0.752 p n1 1.33 The image of the fish is virtual because i is negative (both the object and image are in front of the flat surface in water) The apparent depth of the fish is approximately 3/4 of the actual depth 17.6.3 Thin Lenses A lens is a transparent object with two refracting surfaces of different radii of curvature R1 and R2 but with a common principal axis, and when light rays bend across these surfaces we get the image of an object When a lens converges light rays parallel to the principal axis, we call it a converging lens, see Fig 17.21a If instead it causes such rays to diverge, we call it a diverging lens, see Fig 17.21b Fig 17.21 (a) An (a) enlargement of the top part of a converging lens (b) An (b) R2 R1 enlargement of the top part of R1 R2 n n Converging lens Diverging lens a diverging lens The Thin Lens Equation First, we consider a thick glass lens bounded by two spherical surfaces, air-to-glass and glass-to-air This lens is defined by the radii R1 and R2 of the two surfaces, its thickness , and its index of refraction n, see Fig 17.22 Let us begin with an object O placed at a distance p in front of surface of radius R1 Using Eq 17.26 with n1 = and n2 = n, the position i1 of image I1 formed by surface satisfies the equation: n n−1 + = p i1 R1 (17.29) 17.6 Formation of Images by Refraction R1 n R2 R1 R2 n1 = O p 587 Real image I1 Virtual image n1 = O I1 C1 n p C1 p1 Δ (a) i1 i1 Δ p1 (b) Fig 17.22 When we ignore the existence of surface (of radius R2 ): (a) the first possibility is that an object O produces a real image I1 by surface (of radius R1 ), and (b) The other possibility is that the image I1 is virtual Point C1 is the center of curvature of surface The position i1 is positive in Fig 17.18a when the image I1 is real and negative in Fig 17.18b when the image I1 is virtual In both cases, it seems as if I1 is formed in the lens material with index n Next, we consider the image I1 as a virtual object placed at a distance p1 in front of surface of radius R2 Again, applying Eq 17.26 with n1 = n and n2 = 1, the position i of the final image I formed by surface satisfies the equation: n 1−n + = p1 i R2 (17.30) We note from Fig 17.22a, b that p1 = −i1 + , where i1 is positive for real images and negative for virtual objects For thin lenses, is very small and therefore p1 − i1 Thus, the last equation becomes: − 1−n n + = (For thin lenses) i1 i R2 (17.31) Adding Eqs 17.29 and 17.31, we get: 1 1 − + = (n − 1) p i R1 R2 (17.32) The focal length f of a thin lens is obtained when p → ∞ and i → f in this equation Thus, the inverse of the focal length for a thin lens is: 1 = (n − 1) − f R1 R2 (17.33) which is called the lens-makers’ equation because it can be used to determine R1 and R2 for the desired values of n and f 588 17 Light Waves and Optics In conclusion, a thin lens of index n and two surfaces of radii R1 and R2 has an equation identical to the mirror equation, written as: 1 + = , where p i f 1 − = (n − 1) f R1 R2 (17.34) This is called the thin-lens equation The sign conventions for R1 and R2 are presented in Table 17.3 Just as with mirrors, the thin lens lateral magnification is: M= i h =− h p (17.35) Since light rays can travel in both directions of a lens, then each lens has two focal points F1 and F2 Both focal points are at the same distance f (the focal length) from a thin lens The focal length f is the same for light rays passing through a given lens in either direction This is illustrated in Fig.17.23 for a biconvex lens (converging lens) and a biconcave lens (diverging lens) (b) (a) f F1 f F2 F1 f f F2 F1 F2 F1 F2 Fig 17.23 Parallel rays passing through: (a) a converging lens, and (b) a diverging lens Ray Diagrams for Thin Lenses Ray diagrams are convenient tools that help us locate images formed by thin lenses They also clarify our sign conventions For the purpose of locating an image, we only use two special rays drawn from the top of the object to the top of the image as follows: • Ray starts parallel to the principal axis – For a converging lens, the ray is refracted by the lens and passes through the focal point F2 on the back side of the lens – For a diverging lens, the ray is refracted by the lens and appears to originate from the focal point F1 on the front side of the lens 17.6 Formation of Images by Refraction 589 • Ray passes through the center of the lens and continues in a straight line Figure 17.24 shows such ray diagrams for converging and diverging lenses F2 O I F1 I F1 O Back Front F2 Back Front (a) (b) O F1 F2 I Front Back (c) Fig 17.24 Ray diagrams for locating the image formed by a thin lens (a) An object in front of a converging lens (double convex lens) When the object is outside the focal point, the image is real, inverted, and on the back side of the lens (b) When the object is between the focal point and the converging lens (double convex lens), the image is virtual, upright, larger than the object, and on the front side of the lens (c) When an object is anywhere in front of a diverging lens (double concave lens), the image is virtual, upright, smaller than the object, and on the front side of the lens When using Eq 17.34, it is very important to use the proper sign conventions introduced in Table 17.4 Table 17.4 Sign conventions for thin lenses Quantity Symbol Positive values when Negative values when Radii R1 or R2 The center of curvature is in back of lens The center of curvature is in front of the lens Object location p The object is in front of lens (real object) The object is in back of lens (virtual object) Image location i The image is in back of lens (real image) The image is in front of lens (virtual image) Image height h The Image is upright The Image is inverted Magnification M The Image is upright The Image is inverted Table 17.5 shows a comparison of the image positions, magnifications, and types of images formed by convex and concave lenses when an object is placed at various 590 17 Light Waves and Optics positions, p, relative to the lens Notice that a converging (biconvex lens) can produce real images or virtual images, whereas a diverging (biconcave) lens only produces virtual images Table 17.5 Properties of a single spherical lens system Type of lens Converging lens (Biconvex lens) Diverging lens (Biconcave lens) ∗ f + − p i M Image p>2 f f >i>f Real f >p>f i>2 f f >p>0 |i| > p (Negative) Reduced inverted Enlarged inverted Enlarged upright p>0 |f | > |i| > (Negative) Reduced upright Virtual Real Virtual Combination of Thin Lenses To understand and locate the image produced by two lenses, we follow two steps The first image formed by the first lens is located as if the second lens were not present Then this first image is treated as a virtual object and we use the second lens to find the final image This procedure can be extended to three or more lenses Let us consider the case were two lenses of focal lengths f1 and f2 are in contact with each other If p is the object distance from the system and i1 the is image distance produced by the first lens, then: 1 + = p i1 f1 and M1 = − i1 p (17.36) (17.37) This image is the object for the second lens Since this image is behind the second lens, it serves as a virtual object and its distance for the second lens is negative, i.e its distance to the second lens is −i1 (see Table 17.4) Therefore, the distance i of the final image produced by the second lens satisfies: − and 1 + = i1 i f2 M2 = − i i = (−i1 ) i1 (17.38) (17.39) 17.6 Formation of Images by Refraction 591 Adding the two Eqs 17.36 and 17.38 gives: 1 + = , where p i f 1 = + f f1 f2 (17.40) Thus, two thin lenses in contact are equivalent to a single thin lens of focal length f given by f −1 = f1−1 + f2−1 The overall magnification of the two lenses is: i1 i i =− (Thin lenses in contact) p i1 p M = M1 M2 = − (17.41) Example 17.8 A converging lens of focal length 20 cm forms an image of an object of height 30 cm located at a distance 40 cm from the lens Locate and describe the image Draw two rays to locate the image Solution: A converging lens has a positive value for its focal length, i.e f = +20 cm To find the image distance when p = 40 cm and f = +20 cm, we use Eq 17.34 as follows: 1 + = p i f ⇒ Consequently we have : 1 + = ⇒ i = 40 cm 40 cm i 20 cm 40 cm i ⇒ M = −1 M=− =− p 40 cm The image is real and on the back side because i is positive, inverted because M is negative, and as large as the object, see Fig 17.25 Fig 17.25 O C1 F2 F1 I Back Front Example 17.9 Repeat Example 17.8 using a diverging lens Solution: The diverging lens would have f = −20 cm Thus: 1 + = p i f ⇒ 1 + =− 40 cm i 20 cm C2 ⇒ i = −40/3 cm 592 17 Light Waves and Optics Consequently, we have: M = − (−40/3 cm) i =− p 40 cm ⇒ M = +1/3 The image is virtual and on the front side because i is negative, upright because M is positive, and reduced because M is less than unity, see Fig 17.26 Fig 17.26 O C1 F1 F2 I Front Back Example 17.10 An object is placed 20 cm from a symmetrical lens that has an index of refraction n = 1.65 The lateral magnification of the object produced by the lens is M = −1/4 (a) Determine the type of the lens and describe the image (b) What is the magnitude of the two radii of curvature of the lens? Solution: (a) Using the lateral-magnification equation, we have: M=− i =− p ⇒ i= p 20 cm = 4 ⇒ i = +5 cm Because i is positive, the obtained image must be real The only type of lens that can produce a real image is a converging lens According to Fig 17.24a, the object must be outside the focal point and the image must be inverted and on the back side of the lens (b) To find the focal length f of the lens when p = 20 cm and i = +5 cm, we use Eq 17.34 as follows: 1 + = p i f ⇒ 1 + = 20 cm cm f ⇒ f = cm From the general lens-makers’ Eq 17.33, f is related to the radii of curvatures R1 and R2 of the two surfaces of the lens and its index of refraction n by the relation: 1 = (n − 1) − f R1 R2 17.6 Formation of Images by Refraction 593 For a symmetric lens, R1 and R2 have the same magnitude R If R1 is for the surface where the center of curvature is in the back of the lens, and R2 is for the surface where the center of curvature is in the back of the lens, then using the sign convention of Table 17.4, we have R1 = +R and R2 = −R Thus: 1 = (n − 1) − f R −R Hence, = 2(n − 1) R ⇒ R = 2(n − 1) f , R = 2(n − 1) f = 2(1.65 − 1) × (4 cm) = 5.2 cm Example 17.11 Two thin coaxial lenses and 2, with focal lengths f1 = +24 cm and f2 = +9 cm, respectively, are separated by a distance L = 10 cm; see part (a) of Fig 17.27 An object is placed cm in front of lens Locate and describe the image Draw the necessary sketches to show how you can reach to the answer Solution: We first ignore the presence of lens and find the image I1 produced by lens alone, see part (b) Fig 17.27 Equation 17.34 written for lens leads to the following steps: 1 + = p i1 f1 ⇒ 1 + = cm i1 24 cm i1 = −8 cm Consequently, we have the following lateral magnification: M1 = − (−8 cm) i1 =− p cm ⇒ M1 = +4/3 This tells us that image I1 is virtual (8 cm in front of lens 1), upright because M is positive, and enlarged because M is greater than unity, see part (b) of Fig 17.27 For the second step, we ignore lens and treat the image I1 as a virtual object O1 in front of the second lens The distance p1 between the virtual object O1 and lens is: p1 = L − i1 = 10 cm − (−8 cm) = 18 cm Equation 17.34 written for lens leads us to the following: 594 17 Light Waves and Optics Lens Lens O (a) p L Lens Front Back F1 I1 O F1 (b) p f1 L i1 Front O1 Lens Back F2 F2 (c) I f2 p1 = L+ |i1| i Fig 17.27 1 + = p1 i f2 ⇒ 1 + = 18 cm i cm i = +18 cm Consequently, we have the following lateral magnification: M2 = − i 18 cm =− p1 18 cm ⇒ M2 = −1 17.6 Formation of Images by Refraction 595 The final image is real because i is positive and on the back side of lens 2, inverted because M is negative, and as large as the virtual object I1 , see part (c) of Fig 17.27 The overall magnification of the two lenses is: M = M M2 = i1 i (−8 cm) (18 cm) = p p1 (6 cm) (18 cm) M = −4/3 The final image is enlarged because |M| > Notice that when L = 0, we get M = −i/p as expected from Eq 17.41 17.7 Exercises Section 17.2 Reflection and Refraction of Light (1) A beam of light travels in vacuum and has a wavelength λ = 500 nm The beam passes through a piece of diamond (n = 2.4) What is the wave’s speed and wavelength in diamond? (2) Assume that the wavelength of a yellow beam of light in vacuum is λ = 600 nm, and that the index of refraction of water is 1.33 (a) What is the speed of this light when it travels in vacuum? (b) What is the speed of this light when it travels in water? (c) What is the frequency of this light when it travels in vacuum? (d) What is the wavelength of this light when it travels in water? (e) What is the frequency of this light when it travels in water? (3) A beam of light having a wavelength λ = 600 nm is incident perpendicular to a glass plate of thickness d = cm and index of refraction n = 1.5 (a) How long does it take a point on the beam to pass through the plate? (b) Calculate the number of wavelengths in the glass plate (4) At what angle must a ray of light traveling in air be incident on acetone (n = 1.38) in order to be refracted at 30◦ ? (5) The index of refraction of alcohol is n = 1.4 (a) What is the speed of light in alcohol? (b) Find the angle of refraction in alcohol assuming light meets the air-alcohol boundary at an angle of incidence of 60◦ ? (6) A beam of light in air falls on a liquid surface at an angle of incidence of 55◦ The liquid has an unknown index of refraction (a) If the beam is deviated by 20◦ , what is the value of n? (b) What is the speed of light in this liquid? 596 17 Light Waves and Optics (7) A beam of light in air strikes a glass plate at an angle of incidence of 53◦ If the thickness of the glass plate is cm and its index of refraction is 1.6, what will be the lateral displacement of the beam after it emerges from the glass? (8) A beam of light in air falls on water at an angle of incidence of 45◦ and then passes through a glass block before it emerges out to air again The surfaces of water and glass are parallel and their indexes of refraction are 1.33 and 1.63, respectively (a) What is the angle of refraction in water? (b) What is the angle of refraction in glass? (c) Show that the incoming and outgoing beams are parallel (d) At what distance does the beam shift from the original if the thickness of water and glass are both equal to cm? Section 17.3 Total Internal Reflection and Optical Fibers (9) Diamond has a high index of refraction n = 2.42 To some extent, Diamond’s “brilliance” is attributed to its total internal reflection Find the critical angle for the diamond-air surface (10) A beam of light passes from glass to water The index of refraction of glass and water are 1.52 and 1.333, respectively (a) What is the critical angle of incidence in glass? (b) If the angle of incidence in glass is 45◦ , what is the angle of refraction in water? (11) As it travels through ice, light has a speed of 2.307 × 108 m/s (a) What is the index of refraction of ice? (b) What is the critical angle of incidence for light going from ice to air? (c) If the angle of incidence in ice is 45◦, what is the angle of refraction in air? (12) As the sun sets, its rays are nearly tangent to the surface of water, see Fig 17.28 The index of refraction of water is 1.33 (a) At which angle from the normal would the fish in the figure see the sun? (b) Refraction at the water-air boundary changes the apparent position of the Sun What is the apparent direction of the Sun with respect to the fish (measured above the horizontal)? (13) Figure 17.29 shows a sketch of an Optical fiber cable that has a length L = 1.51 m, diameter of D = 251 µm, and index of refraction n = 1.3 A ray of light is incident on the left end of the cable at an angle of incidence θ1 = 45◦ (a) What is the critical angle of incidence for light going from inside the cable to air? (b) Find the angle of refraction θ2 and the length Does the angle θ 17.7 Exercises 597 fulfill the condition of total internal reflection? (c) How many reflections does the light ray make before emerging from the other end? Direction of the sun as seen by the fish Fig 17.28 See Exercise (12) Apperant position Air n =1 Sun Water n1 = 1.33 L Air n1= θ θ2 Optical fiber n = 1.3 D θ1 Fig 17.29 See Exercise (13) (14) Using the figure of Exercise 13, show that the largest angle of incidence θ1 for which total internal reflection occuring at the top surface is given by the relation sin θ1 = (n2 /n1 )2 − Now find the value of this angle using the data of Exercise 13 Section 17.4 Chromatic Dispersion and Prisms (15) Find the difference in time needed for two short pulses of light to travel 12 km through a fiber optics cable, assuming that the cable’s index of refraction for a pulse of wavelength 700 nm is 1.5 and 1.53 for a pulse of wavelength 400 nm (16) A monochromatic light ray is incident from air (n1 = 1) onto one of the faces of an equilateral prism that has an index of refraction n2 = 1.5, see Fig 17.30 If the angle of incidence θ1 is 40◦ , then at what angle from the normal would this ray leave the prism? 598 17 Light Waves and Optics Fig 17.30 See Exercise (16) Air n1 60° Glass n θ1 θ2 θ4 θ3 120° 60° 60° (17) A narrow beam of white light is incident from air onto a plate of fused quartz at an angle of incidence θ1 = 60◦ ; see Fig 17.31 The index of refraction of quartz for violet and red light is nV = 1.470 and nR = 1.458, respectively Find the angular width δV −δR between the violet and red light rays inside the quartz Fig 17.31 See Exercise (17) White light θ1 Air R δR Quartz V δV (18) A prism has an index of refraction n = 1.5 and an apex angle A = 30◦ The prism is set for minimum deviation by allowing a ray of monochromatic light to pass through it symmetrically, as shown in Fig 17.32 (a) Find the angle of minimum deviation δm (b) Find the value of the angle of incidence θ1 Fig 17.32 See Exercise (18) Air 30° δm θ1 Glass 17.7 Exercises 599 Section 17.5 Formation of Images by Reflection (19) The height h of a man is 200 cm The top of his hat t, his eyes e, and his feet f are marked by dots on Fig 17.33 In order for the man to be able to see his entire length in a vertical plane mirror, he needs a mirror of height H, as shown The figure also shows two paths, one for the light ray leaving his hat t and entering his eyes e, and another for the light ray leaving his feet f and again entering his eyes e (a) Find the height H of the mirror (b) Use two rays to make a geometric sketch for the location and the height of the man’s image Fig 17.33 See Exercise (19) t a b e H Mirror h c f p (20) A concave mirror has a radius of curvature of 1.5 m Where is the focal point of this mirror? (21) A concave mirror has a focal length f = +0.2 m An object of height cm is placed 0.1 m along its principal axis Locate and describe the image formed by the mirror (22) Repeat Exercise 21 using a convex mirror (23) Assume a spherical concave mirror has a positive focal length | f | Use the mirror equation 1/p + 1/i = 1/| f | to determine where an object must be placed if the image created has the same size as the object, i.e when M = | − i/p| = (24) Assume a spherical convex mirror has a negative focal length −| f | Use the mirror equation 1/p + 1/i = −1/| f | to show that the condition M = |−i/p| = cannot not be satisfied 600 17 Light Waves and Optics (25) Six objects are located at the following positions from a spherical mirror: (i) p = ∞, (ii) p = 15 cm, (iii) p = 10 cm, (iv) p = 7.5 cm, (v) p = cm, and (vi) p = 2.5 cm Locate and describe the image for each object when the spherical mirror is: (a) concave, with a focal length of cm, (b) convex, with a focal length of cm (26) Repeat Exercise 25, this time sketching the lateral magnification M for each object’s location p Section 17.6 Formation of Images by Refraction (27) (a) A cylindrical glass rod (n2 = 1.6) has a hemispherical end of radius R = cm An object of height h = 0.2 cm is placed in air (n1 = 1) on the axis of the rod at a distance p = cm from the spherical vertex, see Fig 17.34 (a) Locate and describe the image (b) Repeat part (a) when p = cm Fig 17.34 See Exercise (27) n =1 n1 < n2 n =1.5 Back h Front O I h' C p R i (28) A spherical fish bowl filled with water (n1 = 1.33) has a radius of 15 cm A small fish is located at a horizontal distance p = 20 cm from the left side of the bowl, see Fig 17.35 Neglecting the effect of the glass walls of the bowl, where does an observer see the fish’s image? What is the lateral magnification of the fish? Fig 17.35 See Exercise (28) n2 = i O Back Front I p n1 1.33