Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 24

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 δV White light 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 t To observer d Re et Viol ligh light t igh l ed R 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 −1 = sin 1.5 −1 = sin 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= h Image height = 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= h i =− 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 as “ ” across the front of ambulance vehicles People driving in front 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= i h =− 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 17.5 Formation of Images by Reflection 581 We can use Eqs 17.19 and 17.20 for either concave or convex spherical mirrors if we stick to the sign conventions presented in Table 17.2 This table gives the sign conventions for the quantities f, i, h , and M Table 17.2 Sign conventions for spherical mirrorsa Quantity Symbol Positive values when Negative values when Focal length f The mirror is concave The mirror is convex Image location i The image is in front of The image is in behind the mirror (real image) mirror (virtual image) Image height h The Image is upright The Image is inverted Magnification M The Image is upright The Image is inverted a The object location p and its height h are both positive Example 17.5 A concave mirror has a focal length of 10 cm Locate and describe the image formed by an object having distances: (a) p = 25 cm, (b) p = 15 cm, (c) p = 10 cm, and (d) p = cm Solution: Concave mirrors have a positive focal length, i.e f = +10 cm (a) To find the image distance, we use Eq 17.19 as follows: 1 1 1 1 + = ⇒ + = ⇒ = − p i f 25 cm i 10 cm i 10 cm 25 cm 25 cm − 10 cm 15 50 = = ⇒ i= cm ⇒ i 250 cm2 i 250 cm The positive sign of i indicates that the image is real and located on the front side of the mirror The magnification of the image can be determined using Eq 17.20 as: M=− i 50/3 cm =− p 25 cm ⇒ M=− The negative sign of M indicates that the image is inverted In addition, the image is reduced (66.7% of the size of the object) because the absolute value of M is less than unity, see Fig 17.16a (b) When p = 15 cm, the mirror and magnification equations give: 1 + = 15 cm i 10 cm ⇒ 1 = − i 10 cm 15 cm ⇒ i = 30 cm 582 17 Light Waves and Optics M=− 30 cm i =− p 15 cm ⇒ M = −2 The image is real when i is positive, inverted when M is negative, and enlarged when |M| > 1, see Fig 17.16b (c) When p = 10 cm, the mirror equation gives: 1 + = 10 cm i 10 cm ⇒ i=∞ This means that the reflected rays are parallel to one another and formed at an infinite distance from the mirror, see Fig 17.16c (d) When p = cm, the mirror and magnification equations give: 1 + = cm i 10 cm M=− ⇒ 1 = − i 10 cm cm (−10 cm) i =− p cm ⇒ ⇒ i = −10 cm M = +2 The image is virtual (or behind the mirror) because i is negative, upright because M is positive, and enlarged (twice as large) because M is greater than unity, see Fig 17.16d Example 17.6 An anti-shoplifter convex spherical mirror has a radius of curvature of 0.4 m Locate and describe the image formed by a man standing 3.8 m away from the mirror Solution: The focal length of a mirror is half of its radius of curvature, but for a convex mirror, the focal length that must be used in the mirror equation is: f = −R/2 = −0.2 m When f = −0.2 m and p = 3.8 m, the mirror and magnification equations give: 1 + = p i f ⇒ 1 + =− 3.8 m i 0.2 m 1 =− − i 0.2 m 3.8 m ⇒ i = −0.19 m 17.5 Formation of Images by Reflection M=− 583 (−0.19 m) i =− p 3.8 m ⇒ M = +0.05 The image is virtual (or behind the mirror) because i is negative, upright because M is positive, and reduced (5% of the man’s size) because M is less than unity 17.6 Formation of Images by Refraction Lenses gather and redirect light rays to form images of objects by refraction Again, we will use the ray-approximation model of geometric optics in which light travels in straight lines to form images 17.6.1 Spherical Refracting Surfaces Consider two transparent media having indices of refraction n1 and n2 and the boundary between them is a spherical surface of radius R, see Fig 17.18 Assume a point object O exists in the medium with an index of refraction n1 In addition, assume that all rays make small angles with the principal axis (paraxial rays) when they leave O and focus at point I after being refracted at the spherical surface Fig 17.18 Geometry used to derive Eq 17.26 for n1 < n2 n1 < n2 θ2 n1 Front q1 a A b g O R Back I C p n2 i Applying Snell’s law on the single ray of Fig 17.18 gives: n1 sin θ1 = n2 sin θ2 (17.21) Because θ1 and θ2 are assumed to be small angles, we use the small-angle approximation sin θ ≈ θ , where θ is measured in radians, to have: n1 θ1 = n2 θ2 (17.22) 584 17 Light Waves and Optics Next, we use the rule that an exterior angle of any triangle equals the sum of the two opposite interior angles Applying this rule to triangles OAC and AIC of Fig 17.18, we get: θ1 = α + β and β = θ2 + γ (17.23) where α, β, and γ are also small angles Eliminating θ1 and θ2 from the last two equations gives: n1 α + n2 γ = (n2 − n1 )β (17.24) From the figure we find that the following holds true for paraxial rays: tan α ≈ α ≈ p tan β ≈ β ≈ R tan γ ≈ γ ≈ i (17.25) When substituting these expressions into Eq 17.24 and eliminating h from the result, we get the following relation: n2 n2 − n1 n1 + = p i R (17.26) which is valid regardless of which index of refraction is greater We notice that for a fixed object distance p, the image distance i is independent of the small angle that the paraxial ray makes with the axis Therefore, we conclude that all paraxial rays from point O focus at the same point I The magnification is given by: M= n1 i h =− h n2 p (17.27) Again, we must use a sign convention if we want to apply Eq 17.26 to a variety of cases; see Table 17.3 We define the side of the surface in which light rays originate as the front side The other side is called the back side and is the side in which real images are formed 17.6.2 Flat Refracting Surfaces When the refracting surface is flat, its radius of curvature R is infinite (i.e R = ∞) and Eq 17.26 reduces to: i=− n2 p n1 (17.28) 17.6 Formation of Images by Refraction 585 where the sign of i is opposite that of p Thus, according to Table 17.3, the image formed by a flat refracting surface is on the same side of the surface as the object, see Fig 17.19 Table 17.3 Sign conventions for refracting surfacesa Quantity Positive values when Negative values when Radius R The center of curvature is behind the surface The center of curvature is in front of the surface Image location i The image is in behind the surface (real image) The image is in front of the surface (virtual image) Image height h The image is upright The image is inverted Magnification M The image is upright The image is inverted a Symbol When the object is in front of the surface, the object location p and its height h are positive Fig 17.19 A virtual image n1 formed by a flat refracting surface when n1 > n2 All rays n2 Front are assumed to be paraxial Back n1 > n I O i p Example 17.7 A small fish is at a distance p below the water surface, see Fig 17.20 The index of refraction of water and air are n1 = 1.33 and n2 = 1, respectively What is the apparent depth of the fish as viewed by an observer directly above the water? Fig 17.20 n2 =1 i p n1 = 1.33 586 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 = −0.752 p 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 n1 = O p R2 R1 R2 n 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 Real Virtual 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) [...]... 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 1 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 2 passes through the center of the lens and continues in a straight line Figure 17 .24 shows... measured in radians, to have: n1 θ1 = n2 2 (17 .22 ) 584 17 Light Waves and Optics Next, we use the rule that an exterior angle of any triangle equals the sum of the two opposite interior angles Applying this rule to triangles OAC and AIC of Fig 17.18, we get: θ1 = α + β and β = 2 + γ (17 .23 ) where α, β, and γ are also small angles Eliminating θ1 and 2 from the last two equations gives: n1 α + n2... 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... 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 1 of radius R1 Using Eq 17 .26 with n1 = 1 and n2 = n, the position i1 of image I1 formed... 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... 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... principal axis (paraxial rays) when they leave O and focus at point I after being refracted at the spherical surface Fig 17.18 Geometry used to derive Eq 17 .26 for n1 < n2 n1 < n2 2 n1 Front q1 a A b g O R Back I C p n2 i Applying Snell’s law on the single ray of Fig 17.18 gives: n1 sin θ1 = n2 sin 2 (17 .21 ) Because θ1 and 2 are assumed to be small angles, we use the small-angle approximation sin... image distance i is independent of the small angle that the paraxial ray makes with the axis Therefore, we conclude that all paraxial rays from point O focus at the same point I The magnification is given by: M= n1 i h =− h n2 p (17 .27 ) Again, we must use a sign convention if we want to apply Eq 17 .26 to a variety of cases; see Table 17.3 We define the side of the surface in which light rays originate... is at a distance p below the water surface, see Fig 17 .20 The index of refraction of water and air are n1 = 1.33 and n2 = 1, respectively What is the apparent depth of the fish as viewed by an observer directly above the water? Fig 17 .20 n2 =1 i p n1 = 1.33 586 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 1 p = −0.752