Chapter 36

Refraction of Light:

When you are beside a swimming pool, you notice that the pool appears shallow.  If you look straight down into the pool, it still appears shallow, but not as shallow.  This phenomenon is a result of refraction of light.  Refraction is an abrupt change in the direction of light rays as they 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 in different media is different.  A detailed proof will be discussed at the end of the chapter when you have established a strong basic knowledge.

Refraction is also another phenomenon verifying the straight line motion of light   Refraction is the abrupt bending of light upon entering a new transparent medium.  In this connection, the index of refraction will be defined.  Light travels at different speeds in different media.  For example the speed of light in 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.  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 = 3.00x108 m/s is the speed of light in vacuum, and and v is speed of light in the medium of refraction index n.

Example 1:  Calculate the refraction indices of water, glass, and air (almost vacuum, because air around us is very dilute at one atmosphere of pressure).

Solution:   nwater = c / vwater     ;    nwater = (3.00x108 m/s) / (2.25x108 m/s) = 1.33

nglass = c / vglass     ;    nglass = (3.00x108 m/s) / (2.00x108 m/s) = 1.50

nvac.    = c / vvac.       ;    nvac.  = (3.00x108 m/s) / (3.00x108 m/s) = 1.00000...

nair = c / vair           ;    nair   =  (3.00x108 m/s) / (3.00x108 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, it bends and enters water through a different angle; 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. Fig. 1 When a ray enters a medium with a greater refraction index, it refracts and gets closer to the normal line, N (Left figure). When a ray enters a less dense medium, it refracts and goes away from the normal line, N (Right figure). Fig. 2 Light entering water from air gets closer to the normal line after refraction.  n1 and n2 are the refraction indices of medium 1 and medium 2, respectively θi = angle of incidence θr = angle of refraction D = angle of deviation Light entering air from water goes away from the normal line after refraction.

In reflection, θr = θi , but in refraction, θr  in not equal to θi .  It is possible to arrive at a relation between these two angles.  The relation that is not too difficult to derive is called the "Snell's Law of Refraction" as shown below:

n1sin θi = n2sinθr

θi is the angle of incidence, it means the angle that ray makes with normal line in medium 1 where its refraction index is labeled n1.    θr  is the angle of refraction, it means the angle that the refracted ray makes with the normal line in medium2 where its refraction index is labeled n2.

Example 2:  A ray of light makes a 40.0° angle with water before entering water.  What angle does it make with water surface after entering water?  nwater = 1.33.

 Solution:  As the figure shows, the angle of incidence is not 40.0° degrees.  It is 50.0° degrees, because the angle with normal at the point of incidence is important. θi = 90.0° - 40.0° = 50.0°  and n1 = 1.00 for air. n1sin θi = n2sinθr    ;   (1.00) sin (50.0°) = 1.33 sin (θr)  ;  sin θr  = sin 50.0° / 1.33    ;   θr  =  35.2°  ;  The angle with the water surface is 54.8° Fig. 3 Note that water has a greater index of 1.33 ; therefore, the refracted ray gets closer to the normal line, N.

Example 3:  A square slab of glass 1.00cm thick is 8.00cm 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° angle with that edge.  The ray enters the slab and exits the opposite edge.  (a) Show by calculation that the exit ray is parallel to the incident ray.  (b) Calculate the shift (s) between the incident and exit rays.  See the top view in the following diagram:

First answer these two questions:  1) If you find angle r1, knowing that I1K = 8.0cm, can you calculate I1I2 from triangle I1KI2?     2) If you find angle A, knowing I1I2, can you calculate I2H from triangle I1I2H?  If yes, that's a good sign.  First try to solve this problem without looking at the solution.

 Solution: (a) 1) Air to glass:  1.00sin(57°) = 1.5sin (r1)   ;   r1 = 34.0°  ;  It is clear that θ2  =  r1  =  34.0°.   2) Glass to air:  1.50sin( θ2) = 1.00sin (r2)   or,      1.50sin( 34.0° ) = 1.00sin (r2)   ;  r2 = 57° .   Since this exit angle is equal to the first incident angle, the two rays are parallel.  Solution: (b) cos r1 = I1K / I1I2   ;   I1I2 = 9.65cm  ;  A = 90 - 33 - 34 = 23°  sinA = S / I1I2    ;    S = I1I2 sinA = 3.77 cm.  ;  The shift between the incident and exit rays is 3.77cm. Fig. 4

Apparent Depth:

Lets us consider the case of looking straight down into a swimming pool and seeing the floor of the pool higher.  The apparent depth d' can be found from the formula where d is the actual depth and n2 and n1 are the refraction indices of the final and initial 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. Fig. 5

Example 4:  A coin is at the bottom of a swimming pool and appears to be at a depth of 5.5 ft when looking straight down onto it standing by the edge of the pool.  What is the actual depth at that point?

Solution: d' = d (n2 / n1)    ;    5.5 ft  =  d(1.00 / 1.33)    ;    d = 7.3 ft.

Example 5:  Knowing that it takes two rays to form the image of a point of an object, draw a ray diagram to come up with the apparent position of an object at the bottom of a swimming pool for the case when it is not being looked at straight down.  Choose the two rays as follows: a) Ray 1 emerges from point A on the object perpendicular to the water surface.  b) Ray 2 emerges from point A on the object and makes a 60° angle with the water surface as shown in the left diagram.

 Solution: Ray 1 is incident on water surface from underneath through θi = 0 with the normal line N.  From    n1sinθi = n2sinθr  , it turns out that θr = 0 as well, and Ray 1 goes out of water without refraction.  Ray 2 is incident on water surface from underneath through θi = 30° with the normal line N.  It refracts as it enters air through an angle greater than 30°.  However, the refracted rays 1 and 2 are divergent in air and do not intersect to form a real image of A in air.  We extend them in the opposite direction as shown in the right figure.  The extensions in the opposite direction meet at A' ,  the virtual image of A. Fig. 6  A', the virtual image of A appears to be higher as expected.

Total Internal Reflection:

In general, a ray incident at the interface between two transparent media faces reflection and refraction at the same time.  Any ray will be partially reflected and partially refracted.  Reflection is governed by θi = θr , and  refraction by  n1sinθi = n2sinθr .   As the angle of incidence, θi  increases, a greater portion of the ray gets reflected and a smaller portion refracted.  The limiting value of θi when light enters an optically denser medium is 90° (left figure) .  This means that a θi  very close to 90° is possible for when light enters an optically denser medium.  In such case, a small portion of light refracts and enters the second medium, and most of it reflects.  In the figures shown, except the very right one, only the refracted portions are shown. Fig. 7

Now if light is going from a medium into an optically less dense medium (middle figure), we know that it refracts though a greater angle than being incident.  θi is always less than θin this case.  In other words, θr can reach its limit of 90° before θi is even 50 degrees for water-air interface.  If θi is large enough to make θr  more than 90°, then no light leaves the denser medium and all will be reflected back to the denser medium.  This is called " total internal reflection".  The limiting incident angle θi at which refracted rays travel parallel to the interface is called the "critical angle" and is shown as θc (middle figure).  Therefore, for θi = θc θr = 90°.  Using these values in the Snell's formula, yields:

n1sinθc = n2 sin 90    ;    n1sinθc = n2 ( 1 )    ;        sin θc = n2 / n1.

Example 6:  What is the critical angle for rays entering air from water?

Solution:   n1 = 1.33, and n2 = 1.00.  This results in sin θc = 1.00 / 1.33    ;    θc =  48.8°.

Chapter 36 Test Yourself 1:

1) Reflection and refraction are two phenomena that verify (a) the particle-like 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 the 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)3.00x108 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, nwater  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, nglass  is (a) 3/2.  (b) 1.5  (c) both a & b.

11) The angle of incidence, θi, is the angle that an incident ray makes with (a) the interface between two media.  (b) the normal to the interface between two media.  (c) neither a nor b.    click here.

12) The angle of refraction, θr, is the angle that a refracted ray makes with (a) the normal to the interface between two media.  (b) the interface between two media.  (c) neither a nor b.

13) For a nonzero θi, when light enters a medium with a greater refraction index, (a) it bends and gets closer to the normal line at the point of incidence.  (b) it bends and moves away from the normal line at the point of incidence.  (c) θr < θi .   (d) both a & c.    click here.

14) When θi = 0, the incident line is (a) parallel to the interface.  (b)  perpendicular to the interface.  (c)  neither a nor b.

15)  For a nonzero θi, when light enters a medium with a smaller refraction index, (a) it bends and gets closer to the normal line at the point of incidence.  (b) it bends and moves away from the normal line at the point of incidence.  (c) θr > θi .   (d) both b& c.    click here.

16) When θi = 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) θr = 0 as well.   (d) a, b, & c.

17) The law of refraction is (a) n1sin θi = n2sinθr .    (b) n1sin θr = n2sinθi .    (c) sin θi = sinθr .

18) A laser ray makes a 56.0º-angle with the interface between air and water.  It enters water through an angle of (a) 42º.  (b) 24.9º.   (c) 15.8º.        (Draw an appropriate ray diagram).    click here.

19) A ray of light that makes a 62.1º angle with the surface of a liquid (inside that liquid) enters air through an angle of 51.4º with that liquid's surface.  Is the liquid water?   (a) Yes.    (b) No.        (Draw an appropriate ray diagram).

20) A coin is at the bottom of a swimming pool and is being observed by a kid that looks straight down onto it and 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 into 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 35º 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 40º 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 43º angle with water surface. Does the laser ray exit water?  (a) Yes.  (b) No.

25) The formula for critical angle,θc, of a medium is (a) sinθc = 2 sin θr .   (b) sinθc = 2 sin θi .    (c) sinθc = 1/n .

26) The critical angle for glass (n = 1.5)  is  (a) 45º.    (b) 42º.    (c) 38º.    click here.

Lenses:

Thin lenses are studied here.  The equations we work with apply to thin lenses only.  Lenses work on the basis of refraction.  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 converging, and a concave lens is diverging.  This is one difference between lenses and 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 miror) 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 virtual images are upright with respect to their respective objects and real images are inverted.

Example 7:  Decide if the following lenses are convex (converging) or concave (diverging):

 Lens Type A B C D E F

Fig. 8

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 focal point on the main axis.  Of course, the main axis of a lens is its axis of symmetry.  See Figure 9, below: Fig. 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 virtual focal point, F on the main axis.  Of course, the main axis of a lens is its axis of symmetry.  See Figure 10, below: Fig. 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 passes through the focal point and travels parallel to the main axis after refraction (going through the lens).

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 is where its main axis crosses it.  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: Fig. 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 eye-glasses 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 at infinity ( do = )

Object at infinity means very far from the lens.  A distance of 100m may be considered infinity for a lens whose focal length is 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 almost 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.

 do = ∞ Image Conditions: 1) Real                2) Inverted    3) A'B' <

Case II : Object beyond 2f ( do > 2f )

From all points of the object infinite rays of light emerge radially outward.  If A' the image of point A, the tip of the object, is found, the rest is easy.  Over half of the rays that emerge from A do not reach the lens simply because the lens is not in their way.  All rays that are incident on the converging lens will come to the same point after refraction through the lens.  This gathering point is not necessarily F.  To find the point that we call it A', we choose only two of those infinite rays that are convenient for us.  Ray 1 is the ray that travels parallel to the main axis and after refraction  passes through F, and Ray 2 that passes through F and travels parallel to the main axis after refraction.  You need to carefully track these to rays.  The refracted rays 1 and 2 intersect at a point that we call A', the image of A.  If from A', a line perpendicular to the main axis is drawn, point B', the image of B, the bottom of the object will be determined.  The ray diagram is shown below.  You need to completely redraw it.  As you will note for an object placed beyond 2f, the image forms between f and 2f.

 do >2f Image Conditions: 1) Real        2) Inverted    3) A'B' < AB   4)  f < di < 2f Note that points A, O, and A' fall on a straight line.  If you tried to use important ray 3 ( the dotted line), it would go through O, without refraction, and reach A' where the other two rays intersect. This verifies the fact that refracted rays coming from a point of the object meet at a common point to form the image of  the that point of the object. Fig. 13

Case III: Object at 2f ( do = 2f )

 do = 2f Image Conditions: 1) Real     2) Inverted    3) A'B' = AB   4)  di = 2f Ray 1 travels parallel to the main axis and passes 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' normal to the main axis intersects it at B' the image of B.  B falls on  2F. Fig. 14

Case IV: Object between f and 2f ( f < do < 2f )

 f < do < 2f Image Conditions: 1) Real       2) Inverted    3) A'B' > AB   4)  di > 2f Ray 1 travels parallel to the main axis and passes 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. Fig. 15

Case V: Object at f  ( do = f )

 do = f Image Conditions: 1) Real    2) Inverted    3) A'B' >>AB    4)  di = ∞ " >> " means much greater. 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 do is slightly greater than f.  Then the image forms far from the lens and much greater than the object.  This is the case with  projectors. Fig. 16

The following figure shows such ray diagramWhen AB is slightly passed F, the image forms far from the lens and much greater than AB. Fig. 17

Case VI: Object within f  ( do < f )

 do < f Image Conditions: 1) Virtual    2) Upright    3) A'B' >AB   4)  di  forms behind the lens This is the only case that a virtual image is formed.  A virtual image cannot be formed on a screen.  Ray 1 passes through F after refraction.  Ray 3 goes toward the lens' center without refraction.  The two rays are divergent going to the right.  They should be extended in opposite directions to intersect each other at A'.  A' is the virtual image of A.  It cannot be formed on a screen. Fig. 18

Real and Virtual Images:

In the first 5 cases, A', the image of A is formed because real rays of light intersect to form it and it is called the " real image."  A real image can be formed on a screen.

In case 6 above, A', the image of A is formed by the extension of divergent real rays in the opposite direction, and it is called the " virtual image."  A virtual image cannot be formed on a screen.

The Lens Formula:

Thin lenses use the same formula as lenss.  It is easy to show that the relation between the object distance do, the image distance, di, and the focal length f of a lens is: This is true for all lenses, converging or diverging. Note in the figure that A'B' = HO and AB = KO.  Also, FO = f . FB = do - f      and    FB' = d i - f . With these in mind, writing the similarity of two pairs of triangles, the above formula can be proven.Proof:  Refer to the figure on the right as a typical case.  ABF is similar to HOF. It results in AB / A'B' = FB / FO     (1) FKO is similar to A'FB' ; thus,      AB / A'B' = FO / FB'     (2)Comparing (1) and (2) results in     FB / FO = FO / FB'   or, (do - f)(di - f ) = f 2     or,      dodi - dof - dif + f 2 = f 2    or, Fig. 19 dodi =  di f + dof  ;dividing through by dodi f , 1 / f   =   1 / do    +   1 / di

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 the second ratio to account for the virtual image.  It is always easier to use the absolute value of magnification and not be bothered by the negative sign in its formula.

Important:

In the lens formula  ( 1/f  = 1/do + 1/di ) ,  do is always positive, because the object itself is real.  If the image is real, di is also positive.  If the image is virtual, di 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 8:  A 10.0-cm 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 state if real or virtual. Also, provide an appropriate ray diagram.Solution:     1/do+1/di = 1/f  1/18 +1/di  =  1/12  ;  di = 36.0cm. Fig. 20 di = 36.0cm.   Since di is on the right, the image is therefore real.   The magnification is: M = - di / do  = -36/18 = - 2.00 ;   Also, M = A'B' / AB ;   -2.00 = A'B'/10.0     A'B' = -20.0 cm.  The image is inverted.

Another example:

 Example 9:  A 5.0-cm 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 state if real or virtual. Also, provide an appropriate ray diagram.Solution:  1/do+1/di = 1/f 1/10.5 +1/di = 1/21   ; di = -21.0cm. Fig. 21 Since di is negative, the image is on the same side as the object (left) and the image is therefore, virtual.    The magnification is:M = - di / do  = -(-21.0)/10.5 = + 2.0 ;   Also, M = A'B' / AB ;   +2.0 = A'B'/5.0    ;      A'B' = + 10.0 cm.    ;     The image is upright.

Another example:

 Example 10: A converging lens forms a real image that is 4.00 times greater than the object at 60.0cm from the lens.  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 necessarily inverted.  M = - 4.00.  But,  M = - di / do ;  - 4 = -60/do  ; do = 15.0cm Fig. 22 1/do+1/di = 1/f  1/15 + 1/60  = 1/f      ; f = 12.0cm AB = 20.0 / 4.00 = 5.00cm

Image in Diverging Lenses:

There is only one case for image in a diverging lens.  No matter where the object is placed around the main axis of a diverging lens, 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:

 Object placed anywhere on one side of the diverging lens along the main axis Image is always: 1) Virtual         2) Upright    3) A'B'

Note that f, the focal distance of a diverging lens is virtual and in solving problems, it should be treated as a negative quantity.

 Example 11:  A 14.0-cm 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 state if it is real or virtual. Also, provide an appropriate ray diagram.Solution:    f = 12.0cm 1/do+1/di =1/f 1/8 +1/di = 1 / (-12)  ;   di = - 4.8cm.  Since di is negative, the image is virtual as expected.   The magnification is: M  = - di /do = -(-4.8)/8  =  0.60 ;   Also, M = A'B' / AB  0.60 = A'B'/14.0 ;    A'B' = 8.4 cm. Fig. 24

Another example:

Example 12:  Show that the general lens formula ( 1/f = 1/do+1/di ) 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 do and  di  are related.  What does the (-) sign mean?  What is the magnification?  Solution:  To be done by students.

Example 13:  Using the general lens formula, show that when object is moved very far away from the lens to say infinity ( do) ,  the image forms at the focal point F.  Solution:  To be done by students.

Example 14:  Using the general lens formula, show that when object is at F, the image forms at infinity.

Solution:  To be done by students.

Example 15:  The distance between the object and its real image in a converging lens is 125.0cm.  The lens has a focal length of  30.0cm.  Knowing that A'B' > AB,  find  do and  di .

 Solution:  From the figure: d i + do = 125.0 and hence d i = 125.0 - do.  Substituting in the lens formula and solving for do yields: 1/do + 1/(125 - do) = 1/30  3750  - 30do + 30do =  125do  - do2 do2 - 125do + 3750 = 0. Fig. 25 Using the quadratic formula yields:  do = 50cm  and  do = 75cm. Only do = 50cm is admissible because  d i = do + 25.0 = 75.0cm and yields a magnification of greater than 1 (A'B'> AB).

Another example:

Example 16:  The distance between the object and its virtual image in a converging lens is 32.0cm.  The lens has a focal length of 24.0cm.  Find  do di , and the image size, if the objects height is AB = 4.5cm.

 Solution:  From the figure: d i - do = 32.0 and hence d i = 32 + do.  Substituting in the lens formula and solving for do yields: 1/do - 1/( 32 + do ) = 1/24  768 + 24do - 24do = 32do  + do2 do2 + 32do - 768 = 0,  Using the quadratic formula yields:   do =  -48cm  and  do = 16cm are the possible answers.  do = 16.0cm is acceptable (being positive), and d i = 48.0cm.  In the formulas, we should treat d i  negative.    d i = - 48.0cm. Fig. 26 M  = - di /do = -(-48)/16  =  3.00 ;   Also, M = A'B' / AB  3.00 = A'B'/4.5 ;    A'B' = 13.5 cm.

Another example:

Example 17:  The distance between the object and its virtual image in a diverging lens is 24.0cm.  The lens has a focal length of 48.0cm.  Find  do, di , 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.

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 R1 and R2 are the radii of curvature of the sides of the lens, n is its refraction index, with f being the focal length.  In this formula, if R1 or R2 is positive, the corresponding side is convex (bulged out) and adds to the converging property of the lens.  If R1 or R2 is negative, the corresponding side is concave (caved in) and adds to the diverging property of the lens.  n, the refraction index,  is usually 1.5 for clear glass.

Example 18:  The radii of curvature of a convex-convex 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 radius it has.  Using the lens makers formula yields:  1/f = (n-1)( 1/R1 + 1/R2) 1/f  = (1.5-1)(1/24 + 1/36) =  5/144   ;  f = 29 cm Fig. 27

Example 19:  The radius of curvature of the convex side of a convex-flat lens is 30.0cm.  The refraction index of  glass is 1.50,  Find its focal length.

 Solution:  The diagram shows that the flat side has a radius of curvature of ∞.  Not that 1/∞ = 0. Using the lens makers formula yields:  1/f = (n-1)( 1/R1 + 1/R2) 1/f  = (1.5-1)(1/30 + 1 / ∞) =  1/60   ;  f = 60.0cm Fig. 28

Example 20: The radius of curvature of the concave side of a concave-flat lens is 25.0cm.  The refraction index of  glass is1.50,  Find its focal length.

 Solution:  Note that the radius of the concave side is always treated as negative. The diagram shows that the concave side has a radius of curvature of R1 = -25.0cm  and that of the flat side is R2 =  ∞.   Not that 1/∞ = 0. Using the lens makers formula yields:  1/f = (n-1)( 1/R1 + 1/R2) 1/f  = (1.5-1)(-1/25 + 1 / ∞) =  -1/50  ;  f = -50.0cm Fig. 29

Example 21: The radius of curvature of the concave side of a concave-convex 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 radius of the concave side is always treated as negative. The diagram shows that the concave side has a radius of curvature of R1 = -25.0cm  and that of the convex side is R2 = 40.0cm.  Using the lens makers formula yields:  1/f = (n-1)( 1/R1 + 1/R2) 1/f  = (1.5-1)(-1/25 + 1 / 40) = -15/2000  ;  f = -133cm Fig. 30

The Convergence Power ( C ) of a Lens:

The power of a lens is defined as the the reciprocal of its focal length.  The shorter the focal length of a lens, the greater its converging or diverging power.  The formula is where if f is in meters in SI, then C is expressed in " diopters."

Example 22:  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 23:  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 it.  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 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 24:  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 distance, f,  and the negative power, C, of this lens.

 Solution:  Points A, B, A', and B' are made on a vertical flat sheet of cardboard adjusted to show the 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.  From the two similar triangles AHF and BKF, we may write:  (a) 25/10. = (45+f) / f    ;    25f = 450 +10.f    ;   f = 30.cm(b) C = 1 / f       ;   C = 1 / (- 0.30m) = - 3.33 diopters. Fig. 31

Another example:

Example 25:  A student is farsighted and uses a pair of glasses each of which power is +1.50 diopters.  What is the focal length of each lens in cm?

Solution:  C = 1/f    ;    f = 1/C    ;    f = 1 / (1.50m-1) =  0.667m    ;    f = 66.7cm

Convergence Theorem:

If a number of thin lenses are placed beside each other  with their main axes aligned, the total power is equal to the sum of individual powers.  We may write:

C = C1 + C2 + C3 + . . .

Example 26:  A concave (diverging) lens with a focal length of 40.0cm is combined with a convex (converging) lens with a focal length of 25.0cm with their main axes aligned.  What is the focal length of the combined lens?

Solution:  C = C1 + C2   ;  1/f = 1/f1 + 1/f2  ;  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 27:  A student combines a concave lens and a 5.00-diopter 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:  We have: C1 = ?    ;    C2 = 5.00 diopters    ;    C = 1/(0.500m) = 2.00 diopters.

C = C1 + C2  ;  2 = C1+5  ;  C1 = -3.00 dptrs  ;  f1 = 1/(-3.00m-1;   f1 = -0.333m   ;   f1 = -33.3cm

Optical Instruments:

A few optical instruments of interest are discussed below.

Telescope:

There are two types of telescopes: refractors and reflectors.  A refractor uses two converging lenses and works completely on the basis of refraction.

A reflector uses a converging mirror and a converging lens. We will discuss the two-lens telescope or simply the refractor here.

Refractor Telescopes:

A refractor telescope has a lens that is exposed to light coming 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 "eye-piece."  The focal length, fo,  of the objective lens is a few meters and that of the eye-piece, fe,  is a few centimeters.  The magnification power of a refractor may be found from the formula  M = fo / fe.  A ray diagram showing the formation of the first and second images is shown below: Fig. 32

A'B', the first real and inverted image formed at Fo is not labeled for clarity and lack of space in the figure.   The observer adjusts the eye-piece such that the first image is not exactly at Fe , the focal point of the eye-piece, but it is within the focal length of the eye-piece.  The Eye-piece forms a virtual image (A''B'') from that first real image.  The observer simply sees A''B''.    A'B' , formed at Fo, shown but not labeled, is a real image that acts as a virtual object for the eye-piece.  This is the first time, we have mentioned a term like "virtual object."

Human Eye:

For the anatomy of human eye, you may refer to your text or an anatomy text.  The main point here is that human eye has a converging lens whose focal length can vary by the eye muscles.  This causes a clear image to be formed on the retina where photosensitive nerves transmit the image to the brain for processing and recognition.  As an object gets closer to human eye (s), the eye muscles push from all sides to make the eye lens more converging so that the image forms at the retina.  If the image is not formed on the retina for any reason, the image 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 for observing  nearby objects more convex, a convex lens must be added to the eye lens.  Therefore the remedy for farsightedness is a converging lens. Fig. 33( a ) Fig. 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 it more convergent by using a convergent lens as shown in Fig. (c). Fig. 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 to form 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 for observing far away objects less convex, a concave (diverging)  lens must be added to the eye lens.  Therefore the remedy for nearsightedness is a diverging lens. Fig. 34( a ) Fig. 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 it less convergent by using a divergent lens as shown in Fig. (c). Fig. 34( c )

3) Astigmatism:

When eye is not capable of forming the image of a point as a 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 custom-made 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.  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 36 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) If the object is placed perpendicular to the main axis of a lens, (a) it justifies the perpendicularity of the image to the main axis as well.  (b) we may draw the image of it 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) O 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 30º-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 45º-angle.

8) A ray going through F of a converging 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 diverging 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.  Verify your answer by drawing an appropriate ray-diagram.

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.  Verify your answer by drawing an appropriate ray-diagram.

12) When an object is placed on the left at (very far) from a converging lens, its image forms on the right (a) at F.  (b) at 2F.   (c) between F and 2F. Verify your answer by drawing an appropriate ray-diagram.   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 (a) at F.  (b) at 2F.   (c) beyond 2F.  Verify your answer by drawing an appropriate ray-diagram.

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 virtualVerify your answer by drawing an appropriate ray-diagram.   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.  The focal plane is a plane that passes through F and is to the main axis.  (c) both a & b.    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, do, to the image distance, di, and the focal length, f ,of a thin lens is (a) (1/do) + (1/di) = (1/f).    (b) (1/do) * (1/di) = (1/f).    (b) (1/do) - (1/di) = (1/f).

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 less 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(II): A 6.00cm high object is placed at 25.0cm on the left of a converging lens whose focal length is 10.0cm.  Draw an appropriate ray-diagram. 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 from the lens.  (b) 7.14cm from the lens.  (c) -15.0cm from the lens.

29) The absolute value of  magnification 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(III): A 9.00cm high object is placed at 30.0cm from a converging lens with a focal length of 15.0cm.  Draw an appropriate ray-diagram. 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 from the lens.  (b) 9.50cm from the lens.  (c) 30.0cm from the lens.

35) The absolute value of  magnification 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(IV): A 2.00cm high object is placed at 12.0cm from a converging lens with a focal length of 10.0cm.  Draw an appropriate ray-diagram. 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 from the lens.  (b) 17.5cm from the lens.  (c) 30.0cm from the lens.

41) The absolute value of  magnification 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(V): A 1.50cm high object is placed at 10.1cm from a converging lens whose focal length is 10.0cm.  Draw an appropriate ray-diagram. Answer the following questions:

45) The image is (a) real.   (b) virtual.   (c)  much greater than the object.    (d) both a & c.    click here.

46) The image forms at (a) 343cm from the lens.  (b) 575cm from the lens.  (c) 1010cm from the lens.

47) The absolute value of  magnification 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(VI): A 3.50cm high object is placed at 8.00cm on the left of a converging lens whose focal length is 12.0cm.  Draw an appropriate ray-diagram. 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 from the lens (left).  (b) 24.0cm from the lens (right).  (c) 12.0cm from the lens.

53) The absolute value of magnification 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: A 3.50cm high object is placed on the left at 18.0cm from a diverging lens whose focal length is 12.0cm.  Draw an appropriate ray-diagram. 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 on the left of the lens at (a) 36.0cm from the lens.  (b) 7.20cm from the lens.  (c) 10.8cm from the lens.

59) The absolute value of magnification 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 and 2F are 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.  The closer the object to the div. lens, the closer its image to the div. lens as well.  The farther an object from a div. lens, the closer its image to the virtual focal point.  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(*): 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 ray-diagram. Answer the following questions:

63) The absolute value of magnification is (a) 5.00.   (b) 0.200.   (c) neither a nor b.   click here.

64) The image is formed (a) 20.0cm from the lens.  (b) 40.0cm from the lens.  (c) 60.0cm from the lens.

65) The focal length of the lens is (a) 10.0cm.  (b) 6.67cm.  (c)  21.5cm.    click here.

Problem(**): 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 ray-diagram. Answer the following questions:

66) The image is real; thus, the object and image are (a) both on the left of the lens.  (b) one on the left and the other on the right  the lens.    click here.

67) The 56.0cm that is the distance from the object to its image is  (a)  di + do.   (b) di - f.   (c) di - do.  To understand this clearly, a ray diagram must be drawn.

68) The equation that may be written is (a) di + do = 56.0cm.    (b) di - f = 56.0cm.   (c) di - do = 56.0cm.

69) To solve for the unknowns di and do , (a) another relation between di and do must exist.  (b) these two unknowns can be solved with only one equation.  (c) neither a nor b.     click here.

70) The second equation, from the magnification information, is (a) di = do.    (b) di = 5 do.    (c) di = 6do.

71) Combining the first and second equations, results in: (a) do = 4.00cm.    (b) do = 9.00cm.    (c) do = 8.00cm.

72) The value of di is (a) 48.0cm.    (b) 28.0cm.    (c) 18.0cm.    click here.

73) The focal length is (a) 6.0cm.    (b) 8.0cm.    (c) 6.86cm.

Problem(***): 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 ray-diagram. Answer the following questions:

74) The image is virtual; therefore, the object and image are (a) both on the same side of the lens.  (b) one in front and the other behind the lens.  click here.

75) The 16.0cm that is the distance from the object to its image is  (a)  di + do.   (b) di - f.   (c) di - do.  To understand this clearly, a ray diagram must be drawn.    click here.

76) The equation that may be written is (a) di + do = 16.0cm.    (b) di - f = 16.0cm.   (c) di - do = 16.0cm.

77) The second equation, from the magnification information, is (a) di = 3do.    (b) di = 5 do.    (c) di = do.

78) Combining the first and second equations, results in: (a) do = 4.00cm.    (b) do = 9.00cm.    (c) do = 8.00cm.

79) The value of di is (a) 48.0cm.    (b) 24.0cm.    (c) 18.0cm.    click here.

80) The focal length is (a) 6.0cm.    (b) 12.0cm.    (c) 15.0cm.

81) The lens makers formula is (a) 1/f = (n-1)[1/R1+1/R2].   (b) 1/f = (n+1)[1/R1+1/R2].   (c) 1/f = (n)[1/R1+1/R2 ].

82) The radii of curvature of a convex-concave lens are 20.0cm and 40.0cm, respectively.  with n = 1.50, its focal length is (a) 40.0cm.    (b) 60.0cm.    (c) 80.0cm.    click here.

83) The radii of curvature of a convex-convex lens are 20.0cm and 40.0cm, respectively.  with n = 1.50, its focal length is (a) 30.0cm.    (b) 26.7cm.    (c) 80.0cm.

84) The radii of curvature of a concave-concave lens are 80.0cm and 80.0cm, respectively.  with n = 1.50, its focal length is (a) -80.0cm.    (b) 60.0cm.    (c) 90.0cm.    click here.

85) The radius of curvature of a convex-flat 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 flat-concave 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 convergence power formula is (a) C = 1/f2.    (b) C = 1/f3.    (c) C = 1/f.

88) The Convergence power is in diopters if the focal length is in (a) meters.  (b) cm.  (c) mm.    click here.

89) The converg. power of a convex lens whose focal length is 40.0cm is (a) 2.0diopters.  (b) 3.0diopters.  (c) 2.5diopters.

90) The converg. power of a concave lens with f = -25cm is (a) -2.0diopters.  (b) -4.0diopters.  (c) -5.0diopters.

91) The radii of curvature of a concave-convex lens are 25.0cm and 50.0cm, respectively.  with n = 1.50, its convergence power is (a) -2.0diopters.  (b) -4.0diopters.  (c) -1.0diopters.   click here.

92) The radii of curvature of a concave-concave lens are 30.0cm and 60.0cm, respectively.  with n = 1.50, its convergence power is (a) -2.5diopters.  (b) -4.0diopters.  (c) -1.0diopters.

93) The radii of curvature of a convex-convex lens are 120.0cm and 60.0cm, respectively.  with n = 1.50, its convergence power is (a) -2.5diopters.  (b) 1.25diopters.  (c) -1.0diopters.   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 in contact and 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:  A flat-concave lens is placed in a horizontal table 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, to find the focal lens of the combined lens system, answer the following questions:

97) The focal length (f1) of the glass part is (a) -25.0cm.   (b) 50.0cm.   (c) -50.0cm.   click here.

98) The focal length (f2) of the water part is (a) -25.0cm.   (b) 75.8cm.   (c) -50.0cm.

99) The total convergence power is (a) -0.68diopters.    (b) 0.68diopters.    (c) 3.87diopters.

100) The focal length of the combined lens 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 = fe/fo.   (b) M = fo*fe.  (c) M = fo/fe.

104) In refractor, fo and fe are respectively of the order of (a) cm & cm.   (b) m & cm.  (c) m & m.    click here.

105) The magnification of a refractor with fo = 2.20m and fe = 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 not-converging-enough 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 point.    click here.

114) Dispersion is the (a) the 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.