Chapter 26
Refraction of Light:
Beside reflection, refraction is another phenomenon that verifies the straight line motion of light. When you are beside a swimming pool that is full, you notice that the pool appears shallower than it actually is. If you look straight down into the pool, it still appears shallow, but not as shallow as when you do not look straight down into it. This phenomenon is a result of refraction of light. Refraction is an abrupt change in the direction of light rays as they suddenly change medium, water to air, for example. The change in the direction of light from medium to medium is because of the change in speed it undergoes. Light speed is different in different media.
To incorporate the effect of speed change, the index of refraction is defined. Light travels at different speeds in different media. For example, the speed of light in clear glass is 200,000,000 m/s. Light speed in water is 225,000,000 m/s and in vacuum, 300,000,000 m/s. The "Refraction index (n) of a medium" is defined as the ratio of the speed of light in vacuum to the speed of light in that medium. We may write:

c = 3.00x10^{8} m/s is the speed of light in vacuum, and v is the speed of light in the medium of interest. 
Example 1: Calculate the refraction indices of water, glass, and air (Air acts almost like vacuum because the air around us is quite dilute at one atmosphere of pressure).
Solution: n_{water} = c/v_{wat.}; n_{water}= (3.00x10^{8}m/s)/(2.25x10^{8}m/s) = 1.33
n_{glass} = c/v_{glass}; n_{glass}= (3.00x10^{8}m/s)/(2.00x10^{8} m/s) = 1.50
n_{vac. }= c/v_{vac. }; n_{vac. }= (3.00x10^{8}m/s)/(3.00x10^{8}m/s) = 1
n_{air} = c/v_{air} ; n_{air} = (3.00x10^{8}m/s)/(2.997x10^{8}m/s) = 1.00
It is easy to see that when a ray of light (a laser beam, for example) is incident on water surface at a certain acute angle θ_{i} , it bends and enters water through a different angle θ_{r} ; in other words, it refracts. This refraction phenomenon is shown below for two cases: 1) when light enters water from air, and 2) when light enters air from water.
1) When a ray enters a medium with a
higher
refraction index (air to water): n_{2} > n_{1 ,} it refracts and bends toward the normal line N. θ_{r }< θ_{i }
Figure 1 
2) When a ray enters a medium with a
lower
refraction index (water to air): n_{2} < n_{1 ,} it refracts and bends away from the normal line N. θ_{r }> θ_{i }
Figure 2

Snell's Law of Refraction:
It can be shown that the relation between the incidence angle θ_{i} and the refraction angle θ_{r} is given by the "Snell's law of refraction" as
n_{1 }sin θ_{i} = n_{2 }sinθ_{r}
where n_{1} and n_{2} are the refraction indices of media 1 and 2, respectively. Note that θ_{i } and_{ }θ_{r} are with respect to the normal line at the point of incidence.
Example 2: A ray of light makes a 40.0^{o} angle with water surface before entering water. What angle does it make with the water surface after entering water? n_{water} = 1.33.
Solution: As
the figure shows, the angle of incidence is not 40.0^{o}.
θ_{i}
is the angle that incident ray makes with the normal line at the point
of incidence.
θ_{i} = 90.0^{o} 40.0^{o} = 50.0^{o}, and n_{1} = 1.00 for air. Substituting, we get: n_{1}sin θ_{i} = n_{2}sinθ_{r };_{ } 1.00sin50^{o} = 1.33sinθ_{r} sinθ_{r } = sin50.0^{o}/1.33 ; θ_{r }= 35.2^{o} ; The angle with the water surface is 90^{o}  35.2^{o} = 54.8^{o}.

Figure 3 Note that water has a greater index of 1.33 ; therefore, the refracted ray bends toward the normal line, N. 
Example 3: A ray of light is incident at right angle onto water surface. At what angle does it enter water?
Solution: Right angle to the interface means along the normal line or θ_{i } = 0. Plugging this into the Snell's formula, we get:
1.00sin0^{o} = 1.33sinθ_{r} or, 0 = 1.33sinθ_{r } or, 0 =sinθ_{r} that means θ_{r} = 0 as well.
This means that the ray enters water at 0 angle with the normal line N or the ray stays perpendicular to the water surface and goes straight through without bending or refraction.
Conclusion: A ray of light that is perpendicular to the interface between two transparent media goes through without refraction. Its speed in the new medium will however be different. θ_{i } = 0 makes θ_{r } = 0 as well. 
Example 4: A square slab of glass 1.00cm thick is 8.00cm long on each side. The slab is placed flat on a table and a laser ray parallel to the table is incident on one edge of it making a 33.0 ^{o} angle with that edge. The ray enters the slab and exits its opposite edge. (a) By calculation, show that the exit ray is parallel to the incident ray. (b) Calculate the shift (s) between the incident and exit rays. The top view is shown is Fig 4.
First answer these two questions: 1) If you find angle r_{1}, knowing that I_{1}K = 8.0cm, can you calculate I_{1}I_{2} from triangle I_{1}KI_{2}? 2) If you find angle A, knowing I_{1}I_{2}, can you calculate I_{2}H from triangle I_{1}I_{2}H? If yes, that's a good sign. First try to solve this problem without looking at the solution.
Solution: (a) θ_{i} = 90  33^{o} = 57^{o}. 1) Air to glass: 1.00sin57^{o} = 1.50sin r_{1} ; r_{1} = 34^{o} ; it is clear that θ_{2} = r_{1 }= 34^{o}. 2) Glass to air: 1.50sinθ_{2} =1.00sinr_{2} or, 1.50 sin 34^{o} = 1.00sin r_{2} ; r_{2} = 57^{o}. Since this exit angle is equal to the first incident angle, the two rays are parallel. (b) cos r_{1} = I_{1}K/I_{1}I_{2} ; I_{1}I_{2} = 9.65cm ; A = 90 33 34 = 23^{o}.

Figure 4 sinA = S/I_{1}I_{2 };_{ }S = I_{1}I_{2} sinA = 3.77cm. The shift between the 1st incident and last exit rays is 3.77cm. 
Apparent Depth:
Lets us consider the case of looking straight down into a swimming pool for which we see the pool floor higher. Suppose you are looking at a coin at the bottom of the pool. The apparent depth d' can be found from the formula:
where d is the actual depth and n_{2} & n_{1} are the refraction indices of the 2nd & 1st media for the ray of light. Note that in both figures, A is the object and A' is its virtual image. The image is virtual because it cannot be formed on a screen.
Figure 5
Example 5: A coin is at the bottom of a swimming pool and appears to be at a depth of 5.5 ft to a person who is standing by the edge of the pool looking straight down into it. What is the actual depth of the pool at where the coin is?
Solution: Light must come from the coin in water (medium 1) to reach the person who is in air (medium 2) for the coin to be seen; thus, n_{1} = 1.33 & n_{2} = 1.00.
d' = d (n_{2}/n_{1}) ; 5.5 ft = d(1.00/1.33) ; d = 7.3ft.
Example 6: In a ray diagram, show the method of obtaining the virtual image of a point chosen at the bottom of a swimming pool. The observer is of course assumed to be out of water.
Solution: In general: { The trajectory of 2 rays are needed to form an image. If the processed rays are convergent, they intersect and form a real image. If the processed rays are divergent, they do not intersect, but appear to be coming from a common point that means a virtual image}.Here, in this problem, let's find the image of point A, a point at the bottom of a pool. See Fig. 6a. We may select one of the rays (Ray 1) to be the one that leaves A and goes straight up perpendicular to the water surface or along the normal line. For this ray, θ_{i} = 0 that makes θ_{r} = 0 as well and the ray goes straight out of water. We may select Ray 2 to be incident at any desired and convenient angle θ_{i } . Let θ_{i} = 30^{o} for example, as shown. It refracts as it enters air through an angle greater than 30^{o} (entering a medium with a lower n, the ray bends away from the normal line as was discussed before). The refracted rays 1 and 2 are divergent in air and do not intersect to form a real image of A in air. We need to extend them in their opposite directions as shown in Fig. 6b. The extensions in opposite directions meet at A', the virtual image of A. That's why the pool floor appears to be higher.
Figure 6
Total Internal Reflection:
In general, a ray incident at the interface between two transparent media faces both reflection and refraction at the same time. In most cases a ray is partially refracted and partially reflected, but sometimes totally reflected. Reflection is governed by θ_{i} = θ_{rfl} , and refraction by n_{1}sinθ_{i} = n_{2}sinθ_{rfr} . As the angle of incidence θ_{i } increases, a greater portion of the ray gets reflected and a smaller portion refracted. The limiting value of θ_{i} when light enters a medium with a higher n is 90^{o} (Fig. 7a). This means that a θ_{i} of up to 90^{o} is possible when light enters a higher refraction index medium. In case of θ_{i} nearly 90^{o}, a small portion of light refracts and enters the second medium, but most of it reflects (the rightmost of Fig. 7a).
Figure 7
If light enters a medium with a lower n, as in Fig. 7b, since θ_{rfr } > θ_{i} , θ_{rfr} reaches its maximum of 90^{o} quickly before θ_{i} does. In other words, θ_{rfr} can reach its limit of 90^{o} before θ_{i} is even 50^{o} degrees for waterair interface, for example. If θ_{i} is large enough to want to make θ_{r } more than 90^{o}, then refraction does not occur at all. All of the incident light will be reflected back to the original medium. This is called " total internal reflection."
The limiting incident angle θ_{i}_{ }at which refracted rays travel parallel to the interface is called the "critical angle" and is shown as θ_{c }(Fig. 7b). Therefore, for θ_{i }= θ_{c }, θ_{r} = 90^{o}. Using these values in the Snell's formula, yields:
n_{1}sinθ_{c }= n_{2 }sin 90 ; n_{1}sinθ_{c }= n_{2 }( 1 ) ; sin θ_{c }= n_{2}/n_{1.}
Example 7: Find the critical angle for rays entering air from water?
Solution: n_{1} = 1.33 & n_{2} = 1.00. This results in sin θ_{c }= 1.00/1.33; θ_{c }= 48.8^{o}.
Chapter 26 Test Yourself 1:
1) Reflection and refraction are two phenomena that verify (a) the particlelike behavior of light (b) the wave nature of light (c) the straight line motion of light. click here.
2) The straight line motion of light is the main topic of (a) Wave Optics (b) Geometric Optics (c) Quantum Mechanics.
3) When a light source is on, every point of it emits (a) a large number of light rays (b) only one light ray (c) neither a nor b.
4) Rays of light that emerge from any point of a light source (a) travel in one direction only (b) travel in two directions only (c) travel in all directions, and a streak of light in any given direction forms a light ray in that particular direction.
5) We may think that each point of a light source (a) sends out rays in all directions (b) sends rays in one direction only (c) neither a nor b. click here.
6) Refraction is (a) the return of light back into its original medium upon striking on the interface between two transparent media (b) the abrupt bending of light upon a sudden change in medium (c) a & b.
7) Refraction index of a transparent medium is (a) the ratio of the speed of light in that medium to the speed of light in vacuum (b) the ratio of the speed of light in vacuum to the speed of light in that medium (c) neither a nor b. click here.
8) The speed of light in vacuum is (a)300,000 km/s (b)300,000,000 m/s (c)3.00x10^{8} m/s (d)186,000mi/s (e) a, b, c, & d.
9) The speed of light in water is 225,000km/s. Its refraction index, n_{water }is (a) 4/3 (b) 1.33 (c) both a & b.
10) The speed of light in glass is 200,000km/s. Its refraction index, n_{glass }is (a) 3/2 (b) 1.50 (c) both a & b.
11) The angle of incidence θ_{i} is the angle that an incident ray makes with (a) the interface between two media (b) the normal N to the interface between two media (c) neither a nor b. click here.
12) The angle of refraction θ_{r} is the angle that a refracted ray makes with (a) the normal N to the interface between two media (b) the interface between two media (c) neither a nor b.
13) For a nonzero θ_{i}, when light enters a medium with a greater refraction index, (a) it bends toward the normal line (b) it bends away from the normal line at the point of incidence (c) θ_{r }< θ_{i} (d) both a & c. click here.
14) When θ_{i} = 0, the incident ray is (a) parallel to the interface (b) perpendicular to the interface (c) neither a nor b.
15) For a nonzero θ_{i}, when light enters a medium with a smaller refraction index, (a) it bends toward the normal line (b) it bends away from the normal line at the point of incidence (c) θ_{r }> θ_{i} (d) both b & c. click here.
16) When θ_{i} = 0, the incident line is perpendicular to the interface and (a) light enters the new medium without refraction (b) light enters the new medium at a different speed (c) θ_{r} = 0 as well (d) a, b, & c.
17) The Snell's law is a) n_{1}sin θ_{i} = n_{2}sinθ_{r } b) n_{1}sin θ_{r} = n_{2}sinθ_{i } c) sin θ_{i} = sinθ_{r }.
18) A laser ray makes a 56.0^{o} angle with the interface between air and water. It enters water through an angle of (a) 42^{o} (b) 24.9^{o} (c) 15.8^{o}. click here.
19) A ray of light coming out of a liquid makes a 62.1^{o} angle with the liquid surface while in liquid. It enters air making an angle of 51.4^{o} with liquid's surface while in air. Is the liquid water? (a) Yes (b) No.
20) A coin is at the bottom of a swimming pool and is being observed by a kid that looks straight down onto it. He perceives it at a depth of 4.5 ft. The actual depth of the pool is (a) 6.0ft (b) 3.4ft (c) 4.5ft.
21) A swimming pool is 8.0ft deep. Looking straight down onto it, its floor appears to be at a depth of (a) 12.0ft (b) 10.0ft (c) 6.0ft. click here.
22) A kid under water holds a laser pointer such that it makes a 35^{o} angle with water surface. Does the laser ray exit water? (a) Yes (b) No.
23) A kid under water holds a laser pointer such that it makes a 40^{o} angle with water surface. Does the laser ray exit water? (a) Yes (b) No. click here.
(24) A kid under water holds a laser pointer such that it makes a 43^{o} angle with water surface. Does the laser ray exit water? (a) Yes (b) No.
25) The critical angle θ_{c} of a medium with respect to air is (a) sinθ_{c}= 2sinθ_{r} (b) sinθ_{c}= 2sinθ_{i} (c) sinθ_{c}= 1/n.
26) The critical angle for glass (n = 1.5) is (a) 45^{o} (b) 42^{o} (c) 38^{o}. click here.
Lenses:
Only thin lenses will be studied here. The equations we work with apply to thin lenses only. Lenses work on the basis of refraction. Like mirrors, lenses are two types: converging and diverging. If you have learned the spherical mirrors well, you will see that lenses have similar properties. One point that should be mentioned here is that a convex lens is a converging lens, and a concave lens is a diverging lens. These properties are opposite to those of mirrors.
A lens that is thicker in the middle is called "convex" and is converging. This means that it gathers parallel ray of light at its focal plane. A converging lens (like a converging mirror) creates real images in five cases and virtual image in one case only.
A lens that is thinner in the middle is called "concave" and is diverging. It makes parallel rays of light to spread apart and diverge. A diverging lens (like a diverging mirror) creates virtual images only. Again, like in mirrors, virtual images are upright with respect to their respective objects and real images are inverted.
Example 8: Decide if the following lenses are convex (converging) or concave (diverging):
Figure 8
To check your answers, click here.
A converging (convex) lens is thicker in the middle and gathers parallel rays of light at a point on its focal plane. If the parallel rays are also parallel to the main axis of the lens, they gather at the main focal point, F on the main axis. Of course, the main axis of a lens is its axis of symmetry. See Fig. 9.
Figure 9
A diverging (concave) lens is thinner in the middle and diverges parallel rays of light in a manner as if they are coming from a point on its virtual focal plane. If parallel rays are also parallel to the main axis of the lens, they diverge as if they are coming from the main virtual focal point, F on the main axis. See Fig. 10.
Figure 10
Important Rays in Lenses:
We may use three important rays to draw ray diagrams for the image of objects in lenses. The way important rays are numbered may vary from text to text. As long as we keep it consistent, it results in common understanding. They are as follows:
Ray 1: Ray 1 may be chosen to be the one moving parallel to the main axis that passes through the focal point after refraction (going through the lens). Of course, for a diverging lens, the refracted ray appears to have come from the virtual focal point.
Ray 2: Ray 2 can be chosen as the one that has a tendency to passes through the focal point first and then travels parallel to the main axis after refraction.
Ray 3: Ray 3 is the one that passes through the center of the lens, and it goes straight through without refraction. Note that the center of a lens, O, is where its main axis crosses it. Also, note that we are talking about a thin lens for which thickness is negligible. The following diagrams show the important rays for both converging and diverging thin lenses:
Figure 11
Now that the basics for thin lenses are covered, we may proceed with the ray diagrams for the image of objects in thin lenses.
Image in Converging Lenses:
This type of lens is used in microscopes, telescopes, projectors, and more. Each human eye has a converging lens. Converging lenses are used in eyeglasses to correct farsightedness. There are six cases for image in a converging lenses. In each case we may use 2 of the 3 important rays to form the image. In drawing ray diagrams, we keep in mind the following assumptions:
a) The object is placed perpendicular to the main axis with its bottom on the main axis. b) An arrow is used to show the object with its tip indicating the top. c) Only the image of the top of the object is enough to find the image of the whole object. d)A real image is formed when real (refracted) rays of light intersect. e) A virtual image is formed when the refracted (real) rays are divergent and do not intersect. In such case, the extension of the divergent rays in the opposite direction do intersect to form a virtual image. A real image can be formed on a screen. A virtual image cannot be formed on a screen.
Case I : Object very far (d_{o} = ∞)
Object at infinity means very far from the lens. A distance of 100m may be considered infinity for a lens with a focal length of within half a meter. Rays coming from distant objects are essentially parallel or close to being parallel. If we hold a converging lens such that its main axis is parallel to the incoming parallel rays, the image forms at its focal point F on the main axis. This is the case when you hold a converging lens in front of the Sun and you see the accumulation of the refracted sunrays as a burning spot. You can change the lens orientation and form that spot on the main axis as shown. The spot on the main axis is the focal point, F. The distance from F to the center of the lens O is called the focal length of the lens.
d_{o} = ∞ Image Conditions: 1) Real 2) Inverted 3) A'B' <<AB 4) d_{i} = f " << " means much smaller. In this figure the assumptions made above are not completely observed. Although object AB is placed perpendicular to the main axis; however, 
the bottom of the object is not on the main axis. The center of the object is on the main axis. For the Sun, due to a nice symmetry, top A and bottom B are shown by arrowheads.
Figure 12 
Case II : Object beyond 2f ( d_{o} > 2f )
From every point of the object infinite rays of light emerge. All we need to do is to find A' the image of A that is the top of Object AB. Out of the infinite rays emerging from A, only two of them are of interest to us. Ray 1 is the ray that travels parallel to the main axis and after refraction passes through F, and Ray 2 is the one that goes through F first, and then travels parallel to the main axis after refraction. You need to carefully track these two rays. The refracted rays 1 and 2 intersect at a point that we call A', the image of A. From A', a line perpendicular to the main axis must then be drawn to intersect the main axis at B', the image of B, the bottom of the object. The ray diagram is shown below. You need to completely redraw it. As you will notice for an object placed beyond 2f, the image forms between f and 2f. Make sure that you practice it independently at least one time!
d_{o} >2f Image Conditions: 1) Real 2) Inverted 3) A'B' < AB 4) f < d_{i} < 2f Note: You could have used Ray 3 instead of 1 or 2. Ray 3 is the one that goes through the center O without refraction. 
Ray 3 would have also intersected 1 or 2 at A' as shown below.
Figure 13 
Case III: Object at 2f ( d_{o} = 2f )
As shown in Fig. 14, Ray 1 travels parallel to the main axis and passes through F on the right after refraction. Ray 2 passes through F on the left first, then travels parallel to the main axis after refraction. The two refracted rays intersect at A' that is the image of A. The line drawn from A' normal to the main axis intersects it at B' the image of B. B falls on 2F.
d_{o} = 2f Image Conditions: 1) Real 2) Inverted 3) A'B' = AB 4) d_{i} = 2f

Figure 14 
Case IV: Object between f and 2f ( f < d_{o} < 2f )
Ray 1 travels parallel to the main axis and goes through F after refraction. Ray 2 passes through F and travels parallel to the main axis after refraction. The two refracted rays intersect at A' that is the image of A. The line drawn from A' perpendicular to the main axis intersects it at B' that is the image of B.
f < d_{o} < 2f Image: 1) Real 2) Inverted 3) A'B' > AB 4) d_{i} > 2f

Figure 15 
Case V: Object at F ( d_{o} = f )
Ray 1 travels parallel to the main axis and passes through F after refraction. Ray 3 goes through the center O without refraction. The two rays are parallel and theoretically do not intersect. In practice, the object is placed very close to F so that d_{o} is slightly greater than f. The image forms far from the lens and much greater than the object. This case is used in projectors and movie theaters.
d_{o} = f Image: 1) Real 2) Inverted 3) A'B' >>AB 4) d_{i} → ∞ " >> " means much greater. 
Figure 16 
The following figure shows such ray diagram. When AB is slightly passed F, the image forms far from the lens and much greater than AB.
Figure 17
Case VI: Object within f ( d_{o} < f )
This is the only case that forms a virtual image. A virtual image cannot be formed on a screen. Ray 1 passes through F after refraction. Ray 3 goes through the lens' center without refraction. The refracted Rays 1 and 3 are divergent and do not intersect. Extending them in opposite directions, they intersect at A'. A' is the virtual image of A. It cannot be formed on a screen.
d_{o} < f Image: 1) Virtual 2) Upright 3) A'B' >AB 4) Forms behind the lens. 
Figure 18 
Real and Virtual Images:
In the first 5 cases, A' is called "real image" because it forms by real rays crossing each other. A real image can be formed on a screen.
In case 6, A' is called " virtual image" because we extend the trajectory of 2 or 3 real rays in their opposite directions, to form that image. A virtual image cannot be formed on a screen.
The Lens Formula:
Thin lenses use the same formula as mirrors. It is easy to show that the relation between the object distance d_{o}, the image distance, d_{i}, and the focal length f of a lens is:
This is true for all lenses, converging or diverging. Referring to the Fig. 19,
A'B' = HO & AB = KO. Also, FO = f , FB = d_{o}  f , & FB' = d_{ i}  f . With these in mind, writing the similarity of two pairs of triangles, the above formula can be proven.
Proof: From the figure, ABF is similar to HOF. It results in: AB/A'B' = FB/FO. (1) FKO is similar to A'FB'. It results in: AB/A'B' = FO/FB'. (2) Comparing (1) and (2) results in: FB/FO = FO/FB' or, (FB)(FB') = (FO)^{2} or, 
Figure 19 (d_{o} f)(d_{i}  f ) = f ^{2} or, d_{o}d_{i}  d_{o}f  d_{i}f + f^{ 2} = f^{ 2} or, d_{o}d_{i} = d_{i }f + d_{o}f ; dividing through by d_{o}d_{i }f , 1/f = 1/d_{o} + 1/d_{i} 
Magnification:
Magnification M is defined as the ratio of image size to the object size. It is easy to show that magnification is also the ratio of the image distance to the object distance. A negative sign is usually assigned to these ratios to account for the virtual or real and/or upright or inverted inclusions. It is always easier to use the absolute value of magnification M and not be bothered by the negative sign (s) that some texts include in these formulas (if seen anywhere).
Important:
In the lens formula: 1/f = 1/d_{o} + 1/d_{i} , d_{o} is always positive, because the object itself is real. If the image is real, d_{i} is also positive. If the image is virtual, d_{i} is negative. Even f, the focal length can be positive or negative. For a converging lens, f is real and positive. For a diverging lens, f is virtual and negative.
In general, anything real is positive, and anything virtual is negative. 
Example 9: A 10.0cm object is placed at 18.0cm from a converging lens perpendicular to its main axis. The focal distance of the lens is 12.0cm. Find the image distance, magnification, size, and determine if real or virtual. Also, provide an appropriate ray diagram.
Solution: 1/d_{o}+1/d_{i} = 1/f 1/18 +1/d_{i} = 1/12 ; d_{i} = 36.0cm. Since d_{i} is positive, the image is real. The magnification is: M = d_{i} /d_{o}= 36/18 M = 2.00. 
Figure 20 
Also, M = A'B'/AB ; 2.00 = A'B'/10.0 A'B' = 20.0cm. Image being real is also inverted. 
Example 10: A 5.0cm object is placed at 10.5cm from a converging lens perpendicular to its main axis. The focal length of the lens is 21.0cm. Find the image distance, magnification, size, and determine if real or virtual. Also, provide an appropriate ray diagram.
Solution: 1/d_{o}+1/d_{i} = 1/f 1/10.5+1/d_{i} = 1/21; d_{i} = 21.0cm. Since d_{i} is negative, the image is virtual. The magnification is: M = d_{i} /d_{o} = 21.0/10.5 = 2.00 Also, M = A'B'/AB 
Figure 21 2.00 = A'B'/5.0 ; A'B' = 10.0cm. Image being virtual is also upright. 
Example 11: A converging lens forms a real image that is 4.00 times greater than the object at 60.0cm from it. Find the focal length of the lens and the object size if the image size is 20.0cm.
Solution: Since the image is real, it is also inverted.
M = 4.00 ; M = d_{i} /d_{o} ; 4.00 = 60/d_{o} ; d_{o }= 15.0cm.
1/d_{o}+1/d_{i} = 1/f ; 1/15+1/60 = 1/f ; f =12.0cm ; AB = 20.0/4.00 = 5.00cm.
Figure 22
Image in Diverging Lenses:
There is only one case for image in a diverging lens. No matter where the object is placed on the main axis, the image is always upright, virtual, smaller than the object, and forms on the same side of the lens that the object is placed. This type of lens is used to correct nearsightedness. A ray diagram is shown below:
The only case: Object anywhere on the main axis
Only half of the lens is drawn. Ray 1 traveling parallel to the main axis refracts through the lens as if it is coming from the virtual F. Ray 3 goes through center O without refraction. The refracted rays 1 and 3 are divergent and do not intersect on the right side of the lens. Extending each in its own opposite direction brings them together at A'. A' is the virtual image that is also upright and smaller than the object.
Image is always: 1) Virtual 2) Upright 3) A'B' <AB 4) d_{i} forms on the same side that the object is. Figure 23 → . 

Note: When solving problems, for a diverging lens, f should be treated as a negative quantity.
Example 12: A 14cm object is placed at 8.0cm from a diverging lens perpendicular to its main axis. The focal length of the lens is 12.0cm. Find the image distance, magnification, size, and determine if it is real or virtual. Also, provide an appropriate ray diagram.
Solution: 1/d_{o}+1/d_{i} =1/f 1/8 +1/d_{i} =1/(12) ; d_{i} = 4.8cm. Since d_{i} is negative, the image is virtual and also upright. M = d_{i} /d_{o}= 4.8/8 = 0.60 Also, M = A'B'/AB 0.60 = A'B'/14.0 ; A'B' = 8.4 cm. 
Figure 24 
Example 13: Show that the lens formula (1/f = 1/d_{o}+1/d_{i} ) is valid for a thin slab of glass as well. A thin slab of glass may be viewed as a lens for which f →∞. Let f →∞ in the formula and see how d_{o} and d_{i} are related. In the results you get, what does the negative sign mean? What magnification do you obtain?
Solution: To be done by students.
Example 14: Using the lens formula, show that when an object is very far from a lens (d_{o}→∞), the image forms at the focal point F.
Solution: To be done by students.
Example 15: Using the lens formula, show that when object is at F, the image forms at infinity.
Solution: To be done by students.
Example 16: The distance between an object and its real image in a converging lens is 125cm. The lens has a focal length of 30.0cm. Knowing that A'B' > AB, find d_{o} and d_{i} .
Figure 25
Solution: From the figure: d_{ i} + d_{o} = 125.0 ; d_{ i} = 125.0  d_{o}.Substituting in the lens formula and solving for d_{o} yields:
1/d_{o} + 1/(125  d_{o}) = 1/30. Multiplying through by 30d_{o}(125  d_{o}), we get:
3750 30d_{o }+ 30d_{o }= 125d_{o}  d_{o}^{2} or, d_{o}^{2} 125d_{o} +3750 = 0.
Using the quadratic formula yields: d_{o}=50cm & d_{o}= 75cm.
d_{o}= 75cm places the object beyond 2F for which A'B' <AB (not acceptable).
d_{o} = 50cm is acceptable from which d_{i} = 75cm.
Example 17: The distance between an object and its virtual image in a converging lens is 32.0cm. The lens has a focal length of 24.0cm. Find d_{o}, d_{i}, and the image size, if the objects height is AB = 4.5cm.
Solution:
From the figure: d_{ i}  d_{o} = 32.0 and hence d_{ i} = 32 + d_{o}. Substituting in the lens formula and solving ford_{o} yields: 1/d_{o} 1/( 32 + d_{o}) = 1/24 768 + 24d_{o}  24d_{o }= 32d_{o} +d_{o}^{2} d_{o}^{2} + 32d_{o}  768 = 0, Using the quadratic formula yields: d_{o} = 48cm and d_{o} = 16cm are the possible answers. d_{o} = 16.0cm is acceptable (object position being positive). 
Figure 26 d_{i} becomes: d_{ i} = 32.0cm + 16.0cm = 48.0cm. M = d_{i} /d_{o} = 48/16 = 3.00. Also, M = A'B'/AB 3.00 = A'B'/4.5 ; A'B' = 13.5 cm. 
Example 18: The distance between an object and its virtual image in a diverging lens is 24.0cm. The lens has a focal length of 48.0cm. Find d_{o}, d_{i} , and the image size if the objects height is AB = 12.0cm. Draw a ray diagram before solving.
Solution: Left for students. Answer: 48.0cm, 24.0cm , and 6.0cm.
The Lens Makers Formula:
The following formula provides a relation between the radii of curvature of a thin lens, its refraction index, and its focal length. It is called the "lens makers formula."
where R_{1} and R_{2} are the radii of curvature of the sides of the lens, n its refraction index, and f its focal length. In this formula, if R_{1} or R_{2 } is positive, the corresponding side is convex (bulged out) and it adds to the converging property of the lens. If R_{1} or R_{2} is negative, the corresponding side is concave (caved in) and it adds to the diverging property of the lens. n, the refraction index, is usually 1.5 for clear glass.
Example 19: The radii of curvature of a convexconvex lens are 24cm and 36cm. The refraction index of glass is 1.50, Find its focal length.
Solution: The
diagram shows that the more convex a side is, the shorter the radius
of that side. Using the lens makers formula, we get: 1/f = (n1)(1/R_{1} + 1/R_{2}) 1/f = (1.51)(1/24 + 1/36) 1/f = 5/144 = 0.03472 f = 29cm. 
Figure 27 
Example 20: The radius of curvature of the convex side of a convexflat lens is 30.0cm. The refraction index of its glass is 1.50. Find its focal length.
Solution: The
diagram shows that the flat side has a radius of curvature of ∞.
Note that 1/∞
= 0. Using
the lens makers formula yields: 1/f = (n1)( 1/R_{1} + 1/R_{2}) 1/f = (1.51)(1/30 + 1/∞) = 1/60 f = 60.0cm 
Figure 28 
Example 21: The radius of curvature of the concave side of a concaveflat lens is 25.0cm. The refraction index of glass is1.50, Find its focal length.
Solution: Note
that the radius of
a
concave side is always treated as negative. The diagram shows
that the concave side has a radius of curvature of
R_{1} =
25.0cm,
and that of the flat side is R_{2} = ∞.
Note that 1/∞
equals zero. Using
the lens makers formula, we get: 1/f = (n1)( 1/R_{1} + 1/R_{2}) 1/f = (1.51)[1/25 + 1/∞] = 1/50 f = 50.0cm. 
Figure 29 
Example 22: The radius of curvature of the concave side of a concaveconvex lens is 25.0cm and that of its convex side is 40.0cm. The refraction index of glass is 1.50, Find its focal length.
Solution: Note
that the R of a
concave side is always treated as negative. The diagram shows
that the concave side has an R of curvature of
R_{1} = 25.0cm, and that of the convex side is
R_{2} = 40.0cm. Using
the lens makers formula, we get: 1/f = (n1)( 1/R_{1} + 1/R_{2}) 1/f =(1.51)(1/25 +1/40) = 15/2000 f = 133cm. 
Figure 30 
The Convergence Power ( C ) of a Lens:
The power of a lens is defined as the "reciprocal of its focal length." The shorter the focal length of a lens, the greater its converging or diverging power is. The formula is

In SI, f is in meters, then C is expressed in m^{1} called " diopter." 
Example 23: Calculate the convergence power of a lens that has a focal length of 50.0cm.
Solution: C = 1/f ; C = 1/0.500m = +2.00 diopters.
Example 24: A student experimenting with a lens notices that when sunlight is perpendicular to the lens surface, a bright spot forms at a distance of 80.0cm from the lens and on its main axis. If the spot is at its minimum size, determine (a) the focal length of the lens, and (b) its converging power.
Solution: When a lens can bring the parallel rays of sunlight that are also parallel to its main axis to the smallest bright spot on its main axis and on an object, the distance from that bright spot to the lens is the focal length of the lens. We may write:
(a) f = 80.0cm ; (b) C = 1/f ; C = 1/0.800m = +1.25 diopters.
Example 25: To measure the focal length and the power of a diverging lens, a student places it on an optical bench in a vertical position as shown. He sends two symmetric parallel laser beams with respect to its main axis to the lens and traces the refracted rays at points A, B, A', and B' as shown. The measurements are also shown. Find the virtual focal length f and the power C of the lens.
Solution: Points A, B, A', and B' are first made on a flat sheet of paper to trace of the refracted laser rays. The dotted lines BF and B'F are drawn afterward to locate the virtual focal point. Distances AH, HK, and BK are measured in (cm) and the values are shown. For the two similar triangles AHF and BKF, we may write: 25/10 = (45+f)/f ; 25f = 450 +10f f = 30.cm (negative) → 
Figure 31 C = 1/f ; C = 1/0.30m C = 3.33 diopters. 
Example 26: A farsighted person uses a pair of glasses that is rated as +1.50 diopters. Calculate the focal length of each lens of it in cm.
Solution: C = 1/f ; f = 1/C ; f = 1/(1.50m^{1}) = 0.667m ; f = 66.7cm.
The Convergence Theorem:
If a number of thin lenses are placed beside each other in touch with their main axes aligned, the total lens power is equal to the sum of individual powers. We may write:
C = C_{1} + C_{2} + C_{3} + . . .
Example 27: A concave lens with a focal length of 40.0cm is combined (placed in touch) with a convex lens with a focal length of 25.0cm with their main axes aligned. What is the focal length of the combined lenses?
Solution: C = C_{1} + C_{2} ; 1/f = 1/f_{1} + 1/f_{2} ; 1/f = 1/(40) + 1/25
1/f = 15/1000 ; f = 66.7cm.
Note that for the concave lens f = 40cm. Also, conversion to meter was not necessary because units would cancel from both sides anyway.
Example 28: A student combines a concave lens and a 5.00diopter convex lens, and the resulting lens has a focal length of 50.0cm. What is the power and focal length of the diverging lens?
Solution: C_{1} = ? ; C_{2} = 5.00m^{1} ; C = 1/(0.500m) = 2.00m^{1}. C_{1}+C_{2 } = C._{ }
2 = C_{1 }+ 5 ; C_{1} = 3.00m^{1} ; f_{1} = 1/(3.00m^{1}) ; f_{1} = 0.333m ; f_{1} = 33.3cm.
Optical Instruments:
A few optical instruments of interest are discussed below.
Telescope:
There are two types of telescopes: reflectors and refractors. A refractor uses two converging lenses and works completely on the basis of refraction.
A reflector uses a converging mirror and a converging lens.
Here, we will only discuss the twolens telescope or simply the refractor.
Refractor Telescopes:
A refractor telescope has a lens that is exposed to the light coming in from distant objects. It is called the "Objective Lens." The other lens through which the final image is viewed is called the "ocular" or the "eyepiece." The focal length, f_{o}, of the objective lens is a few meters and that of the eyepiece, f_{e }, is a few centimeters. The magnification of a refractor may be found from the formula M = f_{o} / f_{e }. A ray diagram showing the formation of the first and second images is shown below:
Figure 32
A'B', the first real and inverted image formed at F_{o}_{ }is not labeled for clarity and lack of space in the figure. The observer adjusts the eyepiece such that the first image is not formed exactly at F_{e} , the focal point of the eyepiece, but it is within the focal length of the eyepiece f_{e } . The Eyepiece forms a virtual image A''B'' from that first real image. The observer simply sees A''B''.
A'B' that forms at F_{o } (shown but not labeled) is a real image that acts as a virtual object for the eyepiece. This is the first time, we have mentioned the term "virtual object."
Human Eye:
For the anatomy of human eye, you may refer to an anatomy text. The main point here is that human eye has a converging lens the focal length of which can vary by the eye muscles. This causes a clear image to be formed on the retina where photosensitive nerves are located. The optic nerves transmit the image to brain for processing and recognition. As an object gets closer to one's eye (s), the eye muscles push in from all sides to make the eye lens more converging helping image to form on the retina. If for any reason the image is not formed on the retina, it will not be clear. We may discuss a few of the eye problems here, in brief.
1) Farsightedness:
A farsighted person can see far objects clearly. This means that the image of far objects form on the retina without any problem. As an object gets closer to a farsighted eye, the image moves a way from the lens to the back of retina. A normal eye makes the eye lens more convex (converging) to form the image back on the retina; however, a farsighted eye is not capable of doing that and the image forms behind retina. To make the eye more convex (converging), a convex lens must be added to the eye lens. Therefore, the remedy for farsightedness is a converging lens.
Figure 33 (a) 
Figure 33 (b) 
The eyeball is shown with the eye lens on its left
side. In Fig. (a), the farsighted eye is comfortable with a far away object. In Fig. (b), the image is not formed on the retina for a nearby object. The eye lens is not convergent enough and forms the image behind retina. The remedy is to make the eye lens more convergent by using a converging lens as shown in Fig. (c). 
Figure 33 (c) 
2) Nearsightedness:
A nearsighted person can see near objects clearly. This means that the image of near objects form on the retina without any problem. As an object moves away from a nearsighted eye, the image moves toward the lens and forms in front of retina. A normal eye makes the eye lens less convex (converging) to form the image back on the retina; however, a nearsighted eye is not capable of doing that and the image forms in front of retina. To make the eye less convex, a concave (diverging) lens must be added to the eye lens. Therefore, the remedy for nearsightedness is a diverging lens.
Figure 34 (a) 
Figure 34 (b) 
The eyeball is shown with the eye lens on its left
side. In Fig. (a), the nearsighted eye is comfortable with a nearby object. In Fig. (b), the image is not formed on the retina for a far away object. The eye lens is very convergent and forms the image in front of retina. The remedy is to make the eye lens less convergent by using a diverging lens as shown in Fig. (c). 
Figure 34 (c) 
3) Astigmatism:
When eye is not capable of forming the image of a single point as a single point, the problem is called astigmatism. The reason is the loss of symmetry of the eye lens. The lens is more convergent or divergent in certain sectors. To solve this problem, the correcting lens must be custommade to account for the asymmetry of the eye lens.
Dispersion:
Dispersion is the separation of light into its constituent colors. This phenomenon can be treated as a ray optics subject in this chapter or as a wave optics topic in the next chapter. When a light ray is incident on the interface between two transparent media, it refracts. The refraction angle depends on the wavelength or color of the refracted ray. See Fig. 35. The shorter the wavelength, the greater the angle of refraction. Since white light is a mixture of different colors and consequently a mixture of different wavelengths, and different wavelength refract through different angles, when a ray of white light goes through refraction, it separates into its constituent colors and forms a rainbow. The best way to show this phenomenon is by a prism. A prism refracts a light ray twice, once at the entrance and once at the exit. A prism is a triangular slab of glass.
Fig. 35
Chapter 26 Test Yourself 2:
1) In the ray diagrams for image in lenses, the object (an upward arrow) is meant to be (a) perpendicular to the main axis (b) parallel to the main axis (c) neither a nor b. click here.
2) Placing an object perpendicular to the main axis of a lens (a) helps the consistency of all ray diagrams (b) allow us draw the image perpendicular to the main axis as well (c) Both a & b. click here.
3) The center of a thin lens O is (a) at 2F (b) at 3F (c) where the thin lens crosses the main axis.
4) A ray parallel to the main axis of a converging lens passes through (a) 2F (b) F (c) the lens center after refraction.
5) A ray parallel to the main axis of a diverging lens passes through as if it is coming from (a) virtual 2F (b) virtual F (c) O.
6) A ray going through O, the center of a thin converging lens, passes (a) through without refraction (b) refracts and continues along the main axis (c) refracts and always bend through a 30^{o} angle click here.
7) A ray going through O, the center of a diverging thin lens, (a) refracts and continues along the main axis (b) passes through without refraction (c) refracts and always bend through a 45^{o} angle.
8) A ray going through F of a convex lens refracts and (a) passes through 2F (b) passes through F (c) travels parallel to the main axis. click here.
9) A ray approaching a concave lens from its left side along a line that appears to pass through F on its right side, refracts and (a) passes through F (b) passes through 2F (c) travels parallel to the main axis on the right.
10) When an object is placed on the left beyond 2F of a converging lens, its image forms on the right (a) between F and 2F (b) at F (c) at 2F.
11) When an object is placed on the left at 2F of a converging lens, its image forms on the right (a) at F (b) at 2F (c) between F and 2F.
12) When an object is placed on the left at very far from a converging lens, its image forms on the right at (a) F (b) 2F (c) between F and 2F. click here.
13) When an object is placed on the left between F and 2F of a converging lens, its image forms on the right at (a) F (b) 2F (c) beyond 2F.
14) When an object is placed on the left within f of a converging lens, its image forms on the left (a) at ∞ (b) at 2F (c) upright, greater than the object, and virtual. click here.
15) The only case a converging lens forms a virtual image is when the object is placed (a) at F (b) at 2F (c) within f.
16) A diverging lens always forms (a) a real image that can be formed on a screen (b) a virtual image that can be formed on a screen (c) a virtual image that cannot be formed on a screen. click here.
17) It is correct to say that the image in a lens is (a) always inverted with respect to the object if the image is real (b) always upright with respect to the object if the image is virtual (c) both a & b.
18) When a set of parallel light rays (not necessarily parallel to the main axis) are incident on a converging lens, Fig. 9, the refracted rays are convergent and could initially come to a point (a) at F (b) on the focal plane (c) both a & b. The focal plane is a plane that passes through F and is _{┴} to the main axis. click here.
19) When a set of parallel light rays (not necessarily parallel to the main axis) are incident on a diverging lens from the left side, Fig. 10, the refracted rays are divergent on the right side. The extension of the refracted rays in the opposite direction could come to a point (a) at the virtual F on the left (b) on the virtual focal plane on the left (c) both a & b. click here.
20) A lens that shows your finger always smaller than its actual size is (a) diverging (b) converging (c) flat.
21) The farther an object is from a diverging lens, the closer its image to the (a) virtual 2F (b) virtual F (c) center O.
22) The formula that relates the object distance d_{o} to the image distance d_{i} and the focal length f of a thin lens is (a) 1/d_{o}+ 1/d_{i} = 1/f (b) 1/d_{o} 1/d_{i} = 1/f (c) 1/d_{o} 1/d_{i} = 1/f (d) a, b, & c.
23) In the lens formula, (a) object is always real and is treated as positive (b) a real image is given a positive sign (c) a virtual image is given a negative sign (d) a, b, & c. click here.
24) The absolute value of magnification M is defined as (a) the ratio of object size to that of the image size (b) the ratio of image size to that of the object size (c) both a & b.
25) If M >1, (a) the image is greater than the object (b) the image is smaller than the object (c) the image is equal to the object click here.
26) We may also write: (a) M = (do/di) = A'B'/AB (b) M = (di/do) = A'B'/AB both a & b.
In the following problems assume thin lenses:
Problem I: A 6.00cm high object is placed at 25.0cm on the left of a converging lens with a focal length of 10.0cm. Draw an appropriate raydiagram and answer the following questions:
27) The image is (a) virtual (b) real (c) greater than the object (d) both b & c.
28) The image forms on the right at (a) 16.7cm (b) 7.14cm (c) 15.0cm from the lens.
29) M is (a) 1.50 (b) 2.50 (c) 0.667. click here.
30) The object is placed (a) beyond 2F (b) at 2F (c) between F and 2F.
31) The image is formed (a) beyond 2F (b) at 2F (c) between F and 2F.
32) The image size is (a) 9.00cm (b) 4.00cm (c) neither a nor b. click here.
Problem II: A 9.00cm high object is placed at 30.0cm from a converging lens with a focal length of 15.0cm. Draw an appropriate raydiagram and answer the following questions:
33) The image is (a) virtual (b) real (c) equal to the object (d) both b & c.
34) The image forms at (a) 45.0cm (b) 9.50cm (c) 30.0cm from the lens.
35) M is (a) 4.50 (b) 1.00 (c) 2.33. click here.
36) The object is placed (a) beyond 2F (b) at 2F (c) between F and 2F.
37) The image is formed (a) beyond 2F (b) at 2F (c) between F and 2F.
38) The image size is (a) 9.00cm (b) 4.00cm (c) neither a nor b. click here.
Problem III: A 2.00cm high object is placed at 12.0cm from a converging lens with a focal length of 10.0cm. Draw an appropriate raydiagram and answer the following questions:
39) The image is (a)virtual (b)real (c)smaller than the object (d)both b & c. click here.
40) The image forms at (a) 60.0cm (b) 17.5cm (c) 30.0cm from the lens.
41) M is (a) 10.0 (b) 0.20 (c) 5.00. click here.
42) The object is placed (a) beyond 2F (b) at 2F (c) between F and 2F.
43) The image is formed (a) beyond 2F (b) at 2F (c) between F and 2F.
44) The image size is (a) 9.00cm (b) 4.00cm (c) 10.0cm. click here.
Problem IV: A 1.50cm high object is placed at 10.1cm from a converging lens with a focal length of 10.0cm. Draw an appropriate raydiagram and answer the following questions:
45) The image is (a )real (b) virtual (c) much greater than object (d) a & c. click here.
46) The image forms at (a) 343cm (b) 575cm (c) 1010cm from the lens.
47) M is (a) 10.0 (b) 100 (c) 1000 click here.
48) The object is placed (a) beyond 2F (b) at 2F (c) almost at F.
49) The image is formed (a) way beyond 2F (b) at 2F (c) between F and 2F.
50) The image size is (a) 150.cm (b) 90.00cm (c) 60.0cm. click here.
Problem V: A 3.50cm high object is placed at 8.00cm on the left of a converging lens with a focal length of 12.0cm. Draw an appropriate raydiagram and answer the following questions:
51) The image is a) upright b) virtual c) greater than the object d) a, b,&c. click here.
52) The image forms at (a) 24.0cm (b) 24.0cm (c) 12.0cm from the lens.
53) M is (a) 1.00 (b) 3.00 (c) 2.00. click here.
54) The object is placed (a) within f (b) at 2F (c) almost at F.
55) The image is formed (a) on the left of the lens, upright and virtual (b) at 2F (c) between F and 2F.
56) The image size is (a) 15.cm. (b) 10.5cm. (c) 24.0cm. click here.
Problem VI: A 3.50cm high object is placed on the left at 18.0cm from a diverging lens with a focal length of 12.0cm. Draw an appropriate raydiagram and answer the following questions:
57) The virtual focal length of the lens is (a)12.0cm (b)24.0cm (c) neither a nor b. click here.
58) The image forms at (a) 36.0cm (b) 7.20cm (c) 10.8cm from the lens on the left.
59) M is (a) 2.00 (b) 3.00 (c) 0.400 click here.
60) The object is placed (a) within f (b) at 2F (c) neither a nor b. In diverging lenses, F is virtual and on the same side of the lens where the object is; therefore, there are no bench marks as to where an object is placed from a diverging lens. Even 2F does not make any sense for a diverging lens. The closer the object to a diverging lens, the closer its image to the lens as well. The farther an object from a diverging lens, the closer its image to the virtual focal point of it. You may verify this by drawing a few ray diagrams and placing the object at different positions.
61) The image is formed (a) on the left, smaller, upright and virtual (b) within the virtual focal length (c) both a & b.
62) The image size is (a) 1.40cm (b) 10.5cm (c) 7.00cm. click here.
Problem VII: An object is placed at 8.00cm from a converging lens. The image formed on a screen is 5.00 times greater than the object. Draw an appropriate raydiagram and answer the following questions:
63) The absolute value of M is a) 5.00 b) 0.200 c) neither a nor b.
64) The image is forms (a) 20.0cm (b) 40.0cm (c) 60.0cm from the lens.
65) The f of the lens is (a) 10.0cm (b) 6.67cm (c) 21.5cm. click here.
Problem VIII: An object is 56.0cm from its real image in a converging lens. The image is 6.00 times greater than the object. Draw an appropriate raydiagram and answer the following questions:
66) The image is real; therefore, the object and its image (a) are both on the left of the lens (b) one is on the left and the other is on the right of the lens. click here.
67) The 56.0cm that is the distance from the object to its image is (a) d_{i }+ d_{o} (b) d_{i}  f (c) d_{i } d_{o}. To understand this clearly, a ray diagram must be drawn.
68) (a) d_{i }+ d_{o} = 56.0cm (b) d_{i } f = 56.0cm (c) d_{i } d_{o} = 56.0cm.
69) To solve for the unknowns d_{i }and d_{o} , (a) another relation between d_{i }and d_{o} must exist (b) these two unknowns can be solved with only one equation (c) neither a nor b. click here.
70) from the M formula, we get (a) d_{i }= d_{o }(b) d_{i }= 5d_{o} _{ }(c) d_{i }= 6d_{o}.
71) The value of d_{o} is (a) 4.00cm (b) 9.00cm (c) 8.00cm.
72) The value of d_{i }is (a) 48.0cm (b) 28.0cm (c) 18.0cm. click here.
73) The focal length f of the lens is (a) 6.0cm (b) 8.0cm (c) 6.86cm.
Problem IX: An object is 16.0cm from its virtual image in a converging lens. The image is 3.00 times greater than the object. Draw an appropriate raydiagram and answer the following questions:
74) The image is virtual; therefore, the object and its image (a) are both on the same side of the lens (b) one is in front and the other is behind the lens. click here.
75) The 16.0cm that is the distance from the object to its image is (a) d_{i }+ d_{o} (b) d_{i}  f (c) d_{i } d_{o}. To understand this clearly, a ray diagram must be drawn. click here.
76) (a) d_{i }+ d_{o} =16.0cm (b) d_{i } f = 16.0cm (c) d_{i } d_{o} = 16.0cm.
77) From magnification formula (a) d_{i }= 3d_{o}_{ }(b) d_{i }= 5 d_{o}_{ }(c) d_{i }= d_{o}.
78) Solving for d_{o} we get (a) 4.00cm (b) 9.00cm (c) 8.00cm.
79) The value of d_{i }is (a) 48.0cm (b) 24.0cm (c) 18.0cm click here.
80) The focal length f is (a) 6.0cm (b) 12.0cm (c) 15.0cm.
81) The lens makers formula is a) 1/f = (n1)[1/R_{1}+1/R_{2}] b) 1/f = (n+1)[1/R_{1}+1/R_{2}]. (c) 1/f = (n)[1/R_{1}+1/R_{2} ].
82) The radii of curvature of a convexconcave lens are 20.0cm and 40.0cm with n = 1.50. Its focal length f is (a) 40.0cm (b) 60.0cm (c) 80.0cm.
83) The radii of curvature of a convexconvex lens are 20.0cm and 40.0cm, with n = 1.50. Its focal length f is (a) 30.0cm (b) 26.7cm (c) 80.0cm.
84) The radii of curvature of a concaveconcave lens are 80.0cm and 80.0cm with n = 1.50. Its f is (a) 80.0cm (b)60.0cm (c)90.0cm. click here.
85) The radius of curvature of a convexflat lens is 20.0cm on the convex side with n = 1.50. Its focal length is (a) 40.0cm (b) 60.0cm (c) 80.0cm.
86) The radius of curvature of a flatconcave lens is 30.0cm on the concave side with n = 1.50. Its focal length is (a) 40.0cm (b) 60.0cm (c) 80.0cm click here.
87) The lens power formula is a) C = 1/f^{2} b) C = 1/f^{3} c) C = 1/f.
88) The Convergence power C is in diopter if the focal length is in (a) meters (b) cm (c) mm. click here.
89) The convergence power C of a convex lens with a focal length of 40.0cm is (a) 2.0 (b) 3.0 (c) 2.5 diopters.
90) The power of a concave lens with f = 25cm is a) 2.0 b) 4.0 c) 5.0 diopters.
91) The radii of curvature of a concaveconvex lens are 25.0cm and 50.0cm with n = 1.50. Its power is a) 2.0 b) 4.0 c) 1.0 diopters. click here.
92) The radii of curvature of a concaveconcave lens are 30.0cm and 60.0cm with n = 1.50. Its power is (a) 2.5 (b)4.0 (c) 1.0 diopters.
93) The radii of curvature of a convexconvex lens are 120.0cm and 60.0cm with n = 1.50. Its power is (a) 2.5 (b) 1.25 (c) 1.0 diopters. click here.
94) Three lenses with powers +1.25, 1.5, and +1.75 diopters are placed side by side and they share the same main axis. The focal length of the combined lens is (a) 50.0cm (b) 60.0cm (c) 66.7cm.
95) Two lenses of focal lengths 66.7cm and 80.0cm are placed side by side such that they share the same main axis. The focal length of the combination is (a) 146.7cm (b) 36.4cm (c) 13.3cm. click here.
96) In combining lenses, (a) it is correct to add the convergence powers in diopters (b) it is correct to add the focal lengths (c) neither a nor b.
Problem X: A flatconcave lens is placed on a horizontal table and on its flat side. The concave side is then filled with water. Knowing that the radius of curvature of the concave side is 25.0cm, and the refraction indices of glass and water are 1.50 and 1.33, respectively, find the focal length of the combined lens system and answer the following questions:
97) f_{glass} is (a) 25.0cm b) 50.0cm (c) 50.0cm. click here.
98) f_{water} is (a) 25.0cm (b) 75.8cm (c) 50.0cm.
99) The total convergence power is (a) 0.68 (b) 0.68 (c) 3.87diopters.
100) f_{combo} is (a) 147cm (b) 47cm (c) +147cm. click here.
101) A refractor is a telescope that uses (a) a converging mirror and a converging lens (b) two converging mirrors (c) two converging lenses.
102) A reflector is a telescope that uses (a) a converging mirror and a converging lens (b) two converging mirrors (c) two converging lenses. click here.
103) The magnification of a refractor telescope may be found by (a) M = f_{e}/f_{o} (b) M = f_{o}*f_{e} (c) M = f_{o}/f_{e}.
104) In refractors, f_{o} and f_{e} are respectively of the order of (a) cm & cm (b) m & cm (c) m & m. click here.
105) The magnification of a refractor with f_{o} = 2.20m and f_{e} = 1.10cm is (a) 100 (b) 150 (c) 200.
106) For an image to be seen clearly by a human eye, it must be formed (a) on the retina (b) in front of the retina (c) behind the retina. click here.
107) A farsighted eye forms the image of nearby object (a) on the retina (b) in front of the retina (c) behind the retina.
108) The reason why a farsighted eye forms the image of a nearby object behind the retina is that the eye lens is (a) very convergent (b) not convergent enough (c) normal. click here.
109) The remedy for a notconvergingenough eye (farsighted) is to use (a) a converging lens (b) diverging lens (c) a piece of flat glass.
110) A nearsighted eye forms the image of nearby object (a) on the retina (b) in front of the retina (c) behind the retina.
111) The reason why a nearsighted eye forms the image of a faraway object in front of the retina is that the eye lens is (a) very convergent (b) not convergent enough (c) normal. click here.
112) The remedy for a very converging eye (nearsighted) is to use (a) a converging lens (b) diverging lens (c) a piece of flat glass.
113) The reason for astigmatism is that the eye lens is (a) too convergent (b) too divergent (c) not symmetric and cannot form the image of a point as a single point. click here.
114) Dispersion is the (a) separation of light into its constituent colors through a prism (b) the return of light as it is incident on a rough surface (c) the bending of light as it passes through an opening.
115) when light passes through a prism, (a) it refracts once (b) it refracts twice and separates more (c) the shorter wavelength bends more (d) b & c. click here.
Problems:
1) (a) Speed of light in a type of glass is 183,000km/s. Find (a) it refraction index. b) The refraction index of gasoline is 1.40. How fast can light travel in it?
2) A ray of light makes a 24.0^{o} angle with gasoline surface before entering it from air. (a) What angle does it make with it after entering? n_{gasoline}=1.40. Draw a ray diagram that shows the incident ray, normal line, the refracted ray, and label all angles. (b) Show and calculate the deviation angle.
3) A coin is at the bottom of a swimming pool and appears to be at a depth of 3.0 ft when looking straight down onto it standing by the edge of the pool. What is the actual depth at that point? The refraction index of water is 1.33.
4) An 8.0cm object is placed at 22.0cm from a converging lens perpendicular to its main axis. The focal length of the lens is 15.0cm. Find the image distance, size, and type (real or virtual), and the absolute value of magnification. Also, provide an appropriate ray diagram.
5) The distance between an object and its real image in a converging lens is 150.0cm. The lens has a focal length of 24.0cm. Knowing that A'B' > AB, find d_{o} and d_{i} . Draw an appropriate ray diagram.
6) The distance between an object and its virtual image in a converging lens is 40.0cm. The lens has a focal length of 30.0cm. Find d_{o}, d_{i} , and the image size, if the objects height is AB = 3.5cm. Draw an appropriate ray diagram.
7) The distance between an object and its virtual image in a diverging lens is 20.0cm. The lens has a focal length of 15.0cm. Find d_{o}, d_{i}, and the image size if the objects height is AB = 18.0cm. Draw a ray diagram before solving.
8) The radii of curvature of a convexconcave lens are 25cm and 40cm, respectively. Use a glass refraction index of 1.50 and find its focal length and convergence power.
9) The radius of curvature of the concave side of a concaveflat lens is 40.0cm. Use a glass refraction index of 1.50 and find its focal length and power.
10) A student experimenting with a lens notices that when sunlight is perpendicular to the lens surface, a bright spot forms at a distance of 66.7cm from it. If the spot is at its minimum size, determine (a) the focal length of the lens, and (b) its converging power.
11) If a farsighted person uses a pair of glasses with a convergence power of +2.00 diopters. (a) Find the focal length of its lenses in cm? How many times larger will he see his thumb if he holds it under this lens at 40.0cm from the lens?
12) A concave lens with a focal length of 80.0cm is combined with a convex lens with a focal length of 40.0cm with their main axes aligned. (a) Find the focal length of the combo lens. (a) Is the combination lens converging or diverging?
13) A student combines a concave lens and a 2.50diopter convex lens, and the resulting lens has a focal length of 60.0cm. What is the power and focal length of the concave lens?
14) A refractor telescope is 5.000m long. The focal length of its eyepiece is 4.0cm. Find its magnification.
Answers:
1) 1.64, 214,000km/s 2) 49.3^{o}, 25.3^{o} 3) 4.0 ft
4) 47.1cm, 17cm, Real, 2.14 5) 30.0cm, 120.0cm
6) 20.0cm, 60.0cm, 10.5cm 7) 30.0cm, 10.0cm, 6.00cm
8) 133.cm, 0.751diopters 9) 80.0cm, 1.25diopters
10) 66.7cm, 1.50diopters 11) 50.0cm, 5.00 times
12) 80.0cm, Converging 13) 120.cm, 0.833diopters 14) 120