Chapter 25

Reflection of Light

Reflection and refraction are phenomena that verify the straight line motion of light.  These two topics will be discussed in this chapter and the next.  Under reflection, we will study the image of objects in mirrors.  In this chapter it is assumed that infinite rays of light emerge from every point of a light source.  A ray of light is a very narrow streak of light that travels along a straight line.   Reflection is defined as the return of light to its original medium upon bouncing off a reflective surface.  Every ray of light that is incident on a shiny flat surface bounces back such that the angles of incident and reflection are equal.  This can be verified by forming the image of an object in a flat mirror.

Real and Virtual Images:

Real images are the result of the intersection of real rays of light.  Real images can be formed on a wall or a screen such as in movie theaters.  Convergent rays of light can intersect and form a real image.

Divergent rays of light do not intersect and cannot form a real image. They appear to be coming from a common point that we call a virtual image.

Image in Flat Mirrors:

It is easy to verify that in a flat mirror, the object and its virtual image are equidistant from the mirror.  You have this experience that when you take one step toward a flat mirror, your image takes one step toward the mirror as well.   In the figure shown, if do is the object distance and di is the image distance, both measured from the mirror, we may write:

do = - di   (The - sign indicates virtual image).

This can be verified by experiment and is used as the basis to prove that angles θi and θr (shown below) are equal.  The equality of these two angles is called the "law of reflection."   θi and θr are called the "angle of incidence" and the "angle of reflection", respectively.  Both angles are measured with respect to the normal line N at the point of incidence I as shown in Fig. 1. 

{To prove the law of reflection, suppose that A'B' is the virtual image of object AB in a flat mirror as shown in Fig. 1.  We are using only point A of the object for our proof.  Suppose AB is the bright filament of a light bulb. 

From Point A, infinite rays of light emerge two of which are selected: rays AH and AI.   (If the diagram is not practiced, your learning of the topic will be less than 20% effective!  Point H is not arbitrary, but Point I is.  Point I can be chosen anywhere on the mirror.)

AH bounces on itself because it hits the mirror perpendicularly.  HA is the reflected ray.  (If you drop a basketball on a horizontal floor, it hits the floor at right angle and bounces back at right angle to reach your hand again).  Ray AI hits the mirror at the Point of incidence I, and reflects as if it is coming from A', the image of A to reach the observer's eyes.  The observer perceives it coming from A'. The reflected rays at I and H are divergent as they travel to the left and do not intersect to form a real image.  Both reflected rays from H and I appear to be coming from A', the virtual image of A. }

If you fully understood the mechanism explained in between the set of braces above, now continue to the following proof:

Given di = do, prove that  θr  = θi .

To prove, lets take advantage of the fact that triangles AHI and A'HI are congruent because: HA' = HA,  both have 90o angles, and both share HI.  From their congruity, we conclude that angles 1 and 2 are equal. 

 From the diagram, angle equalities

 1 = θi   &  2 = θr  also hold ture. Since angles 1 & 2 are equal, θr & θi are also equal.

θr  = θi

Figure 1: The reflected ray from I appears to be coming from A', the image of A.    There is nothing behind the mirror.

Example 1: A laser pointer is emitting a narrow beam that makes a 30o angle with a flat mirror.  Determine the reflection angle. Draw a ray diagram for the problem.

 

Solution:  A 30o angle with mirror means a 60o angle with the normal line.  This means that  θi = 60o. 

Since θr = θi ; therefore,

θr = 60o as well.

 

Figure 2

Example 2: In the figure shown, at what angle α must the laser pointer be held at the top of the wall such that the reflected ray from the mirror hits the floor-wall corner at B?  The height of the wall is h = 1.80m.

 Solution:  Since θi and θr must be equal, triangle AIB must be isosceles and the two triangles ANI and BNI become congruent resulting in the equality of segments AN and BN.   This makes each of the segments to be half of the wall's height or 0.90m. In triangle ANI, we may write:

tanα = 2.00/0.90 ;     α = 65.8o.

Figure 3

 

Image in Spherical Mirrors:

A spherical mirror  is a portion of a sphere that is coated with a reflective material.  If the mirror is shiny inside (the caved-in side), it is called "concave" and if it is shiny outside (the bulged-out side), it is called "convex."  The following figure shows both concave and convex mirrors:

Figure 4

The main axis of a spherical mirror is its axis of symmetry.  The vertex V of a mirror is where the main axis crosses the mirror.  When a parallel set of light rays that are also parallel to the main axis reach a concave mirror, they reflect and the reflected rays pass through a common point that is called the focal point F of the mirror.  The distance from the focal point F of a mirror to its vertex V is called the focal length f of that mirror. Each mirror has a center that of course is the center of the sphere it is made of See Fig. 5.

 

Parallel rays of light that are also parallel to the main axis gather (converge) at the focal point of a concave mirror after reflection.  It is for this reason that a concave mirror is also called a "converging mirror".  Distance FV = f  is called the focal length.  It can be shown that f is 1/2 of R, the radius of the mirror. 

f = R/2.

Figure 5

In case of a convex mirror, when parallel rays of light reach it, they reflect and spread apart (become divergent) after reflection as if they are coming from a common point F that is behind the mirror.  This common point is called the "virtual focal point."  It is an imaginary point behind the mirror.  See Fig. 6.

Parallel rays of light that are also parallel to the main axis spread out (diverge) after reflection from a convex mirror as if they are coming from a point that is behind the mirror.  The point is called the "virtual focal point" of the convex mirror. 

That is why a convex mirror is also called a "diverging mirror".

Figure 6

If parallel rays traveling toward a converging mirror are not parallel to the main axis (the figure on the left), they still come to a common point after reflection, but not at the main focal point F.  We can visualize a plane that passes through F and is perpendicular to the main axis, as shown.  It is called the "focal plane."  Parallel rays that are not parallel to the main axis gather at a point (such as F1) on the focal plane.  In the case of a diverging (convex) mirror, rays reflect in a manner that they appear to have come from a point on the virtual focal plane (as shown in Fig. 7).

 

Figure 7

Important Rays in Mirrors:

Important rays are those that help us draw ray diagrams for the image of objects in spherical mirrors.  Three important rays are introduced below:

1)  A ray parallel to the main axis of a converging mirror (left) passes through F after reflection.  In a diverging mirror (right), a ray parallel to the main axis reflects as if it comes from the virtual focal point. (Fig. 8)

Figure 8

2) A ray that goes through F and reaches a converging mirror (left), travels parallel to the main axis after reflection.  In a diverging mirror, a ray that approaches the mirror (right) as if it will pass through F, the virtual focal point, travels parallel to the main axis after reflection. (Fig. 9)

Figure 9

3)  A ray traveling through the center C of a converging or diverging mirror, reflects back on itself. (Fig. 10)

 

Figure 10

4)  Any ray that is reflected from the vertex of a spherical mirror makes the same angle with the main axis as its incident ray does; in other words, the incident and reflected rays at the vertex are symmetric with respect to the main axis. (Fig. 10.1)

Figure 10.1

An Important Point:

A spherical mirror may be viewed as a collection of an infinite number of extremely small flat mirrors placed in a curved manner beside each other.  For every one of these flat mirrors, there is a different θi that has its own equal  θr .  An exaggeration of the mosaic flat mirrors is shown below.  Note that the radius R drawn to each of the tiny flat mirrors acts as its normal line at the point of incidence.  This is simply because of the fact that the radius of a circle is perpendicular to the circle at any point on its periphery.

  Note that at I1,  θi = θr with the normal line being the radius R going through C.   Also at I2, again θi = θr with the normal line being the radius R going through C.  However the reflected rays pass through F, the focal point.   As you see with keeping θi and θr equal, the reflected rays pass through F that is the midpoint between C and V, or f = R/2.

Note that Ray 3 hits the vertical flat mirror at V with θi  = 0.  This results in θr  = 0 as well causing Ray 3 to reflect back on itself and still pass through F.  Of course, the reason why Rays 1, 2, and 3 pass through F is that not only they are parallel, but also they are parallel to the main axis. (Fig. 11)

Figure 11

Homework:  Draw the above figure for a diverging mirror.

 

Chapter 25 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) Reflection is (a) the return of light to its original medium upon incidence on a reflecting surface   (b) the bouncing back of a light ray such that the angles of incidence and return are equal   (c) a & b.

7) The angle of incidence is the angle that a ray of light makes with (a) a reflecting surface   (b) normal to the reflecting surface at the point of incidence   (c) neither a nor b.   click here.

8) The angle of reflection is the angle that a reflected ray of light makes with (a) the reflecting surface   (b) normal to the reflecting surface at the point of incidence   (c) neither a nor b.

9) The law of reflection states that the angles of incidence θi and reflection θr are (a) perpendicular to each other   (b) parallel to each other   (c) equal.   click here.

Problem: A ray of light makes a 25o angle with water surface at the point of incidence.  Draw a diagram showing the ray, and at the point of incidence, draw the normal line to the water surface.  Answer the following questions:

10) The angle of incidence θi is (a) 25o  (b) 75o  (c) 65o.

11) The angle that the reflected ray makes with water is (a) 25o  (b) 75o  (c) 65o.   click here.

12) The angle of reflection is (a) 25o  (b) 65o  (c) 55o.

13) In Fig. 1, the number of rays that leave the top of object (Point A) (a) is one  (b) are two  (c) are so many, but only two important ones are shown for simplicity.   click here.

14) In Fig. 1, the basis for proof is (a) the fact that in a flat mirror, object and its image are equidistant from the mirror  (b) the fact that in a flat mirror, angles of incidence and reflection are equal   (c) the ray that reaches the observer's eye.

15) In Fig. 1, the mirror is perpendicular to BB'.  We know that di = do.  It is clear that (a) AH' AH    (b) AH' = AH    (c) AH'  AH.   click here.

16) We may say that two triangles are congruent if (a) two sides and the angle in between of one is equal to two sides and the angle in between of the other  (b) two angles and the side in between of one is equal to two angles and the side in between of the other   (c) the three sides of one have the same lengths as the three sides of the other  (d) a, b, and c.

17) In Fig. 1, which one of the cases in question 16 does apply to Triangles AHI and A'H'I?    (a)?    (b)?    or (c)?

18) The reason case (a) of question 16 applies to the congruity of Triangles AHI and A'H'I, is that (a) AH = AH'   (b) AI = AI because this side is shared by both triangles  (c) the 90 angles are equal in both triangles  (d) a, b, and c collectively.

19) From the congruity of Triangles AHI and A'H'I, one concludes that (a) Angles 1 and 2 are equal   (b) Angles 1 and 2 are unequal   (c) Angles 1 and 2 must each be 90 degrees.   click here.

20) Since AA' is parallel to NI, we may conclude that (a) Angle 1 = θi    (b) Angle 2 = θr    (c) both a &b.

21) From question 20, it is proven that (a) θi = θr    (b) The angle of incidence = angle of reflection    (c) both a & b.

22) In Fig. 1, Ray IA' is (a) real   (b) virtual    (c) can be seen by a person that is standing behind the mirror.   click here.

23) In a flat mirror, di = do,  (a) can be verified by experiment   (b) is a reasonable hypothesis to seek the proof of   (c) both a & b.

24) Neat reflection occurs if (a) the reflecting surface is very very flat and smooth (shiny)  (b) the reflecting surface is rough   (c) both a & b.   click here.

25) If a set of parallel rays are incident on a flat mirror and they all strike the mirror at the same angle, the reflected rays are (a) all parallel  (b) unparallel and each reflect at a different angle  (c) nicely divergent or convergent.

Spherical Mirrors:

26) A spherical mirror is (a) a portion of a circle  (b) a sphere  (c) a paraboloid  (d) b & c  that is coated with a reflecting material.  Note:  If a parabola rotates about its axis, it generates a surface in space that is called a "paraboloid."

27) If the caved-in side of a spherical mirror is reflecting it is a (a) concave  (b) convex.   click here.

28) If the bulged-out side of a spherical mirror is reflecting it is a (a) concave  (b) convex.

29) The main axis of a spherical mirror is (a) its axis of symmetry  (b) the straight line that connects its edges  (c) any line that passes through its center.   click here.

30) The vertex of a spherical mirror is (a) the very top edge of the mirror  (b) the center of the mirror  (c) the point of intersection of its main axis with the mirror.

31) The center of a spherical mirror is (a) of course, the center of the sphere that the mirror is a portion of   (b) the point of  intersection of its main axis with the line that connects its edges  (c) neither a nor b.   click here.

32) The focal point of a spherical mirror is (c) its center   (b) its vertex   (c) midway between its center and vertex.

33) The focal length f  of a spherical mirror is equal to (a) its radius R   (b) half of  radius R/2  (c) twice its radius 2R.

Converging Mirrors:  A converging mirror is concave.   click here.

34) A ray of light that is incident on a converging mirror passes through (a) its center after reflection   (b) its focal point after reflection   (c) halfway between F and C after reflection   (d) a, b, or c may or may not happen.

35) A ray of light that is parallel to the main axis and is incident on a converging mirror passes through (a) its center after reflection  (b) its focal point after reflection  (c) halfway between F and C after reflection  (d) a, b, or c may or may not happen.

36) A ray of light that passes through the center C of a converging mirror reflects (a) back on itself and goes through C again   (b) back and passes through the focal point   (c) reflect back parallel to the main axis.   click here.

37) A ray of light that passes through the focal point F of a converging mirror and reaches it, reflects back (a) on itself   (b) and passes through the center C   (c) and travels parallel to the main axis.

Diverging Mirrors:  A diverging mirror is convex.  For a diverging mirror, F, C, f, and R are virtual.

38) A ray of light that is incident on a diverging mirror reflects (a) such that it appears to be coming from its center   (b) such that it appears to be coming from its focal point   (c) such that it appears to be coming from halfway between F and C   (d) a, b, or c may or may not happen.   click here.

39) A ray of light that is parallel to the main axis and is incident on a diverging mirror reflects such that it appears to be coming from (a) its virtual C   (b) its virtual F   (c) halfway between virtual F and virtual C.

40) A ray of light that appears to be going through the virtual C of a diverging mirror and is incident on it, reflects (a) back on itself and appears to be coming from the virtual C   (b) back and and appears to be coming from virtual F   (c) reflect back parallel to the main axis.   click here.

41) A ray of light that appears to be going through the virtual F of a diverging mirror and is incident on it, reflects back (a) on itself   (b) and appears to be coming from the virtual C   (c) travels parallel to the main axis.

Rays coming from a very distant object are almost parallel.  Rays coming from the Sun, for example, are taken to be almost parallel.  The Sun appears to be small from the distance of 150,000,000km.  A regular light bulb placed at a distance of say 200ft or more has the same effect.  Rays coming from it may be treated as parallel.  Rays from a laser pointer are quite parallel and are ideal for optics experiments.   

42) When parallel rays of light that are also parallel to the main axis of a converging mirror are incident on it, after reflection, they all gather at (a) the center C   (b) the focal point F   (c) the vertex V.   click here.

43) If a spot-like source of light is placed at the focal point F of a converging mirror, the reflected rays are expected to travel (a) parallel to the main axis   (b) normal (perpendicular) to the main axis   (c) in any direction.   click here.

44) The case in the previous question is used in (a) cars headlights   (b) movie theaters   (c) both a & b.

45) When parallel rays of light are incident on a diverging mirror parallel to its main axis, they get reflected and diverged as if they are coming from (a) its virtual center C   (b) its virtual focal point F  (c) neither a nor b.   click here.

 

Image in Converging Mirrors:

There are six cases for image in a converging mirror.  In each case we may use 2 of the 4 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 (reflected) rays of light intersect.  e) A virtual image is formed when the reflected (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 for from the mirror.  A distance of 60m may be considered infinity for a mirror with a diameter of less than half a meter.  Rays coming from distant objects are essentially parallel or close to parallel.  If we hold a converging mirror 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 mirror in front of the Sun and you see the accumulation of the reflected sunrays as a burning spot.  You can change the mirror's orientation and form that spot on the main axis as shown.  The spot on the main axis is the focal point F.

do = .

Image Conditions:    

1) Real   

2) Inverted   

3) A'B' <<AB 

4) di = 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 C1 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.

 

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

From all points of the object infinite rays of light emerge outward.  If only A', the image A, is found, the rest is easier.  Over half of the rays that emerge from A do not reach the mirror simply because the mirror is not in their way.  All of the rays coming from A that reach the converging mirror will come to the same point after reflection.  Such point is not necessarily F.  To find A', the image of A, we choose two rays of those infinite rays that are convenient for us.  Ray 1 is the only ray that travels parallel to the main axis and after reflection passes through F, and Ray 2 is the only one that first goes through F and then travels parallel to the main axis after reflection.  You need to carefully track these to rays.  The reflected rays 1 and 2 intersect at a point that we call A', the image of A.   If from A', a line is drawn perpendicular to the main axis, point B', the image of B, will be obtained.  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, or we may say between F and C.

do >2f.    

Note that points A, C, and A' fall on a straight line.  If you tried to use important Ray 3 (the dotted line), it would go through C, reach the mirror, and reflect back on itself to pass through C again.  It would pass through A' again as well.  This is one verification that all reflected rays meet at same point.

                           Image:

 

 

1) Real             2) Inverted   

3) A'B' < AB     4) f < di< 2f

Figure 13

 

Case III: Object at 2f  (do = 2f )

do = 2f.       Ray 1 travels parallel to the main axis and passes through F after reflection.  Ray 2 passes through F and travels parallel to the main axis after reflection.  The two reflected rays intersect at A' that we call it the image of A.  The line drawn from A' perpendicular to the main axis intersects it at B' that falls on B.  

In this case object B is placed at CImage B' forms at C as well.

                               Image:

1) Real            2) Inverted

3) A'B'=AB     4)  di = 2f

Figure 14

 

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

f < do <2f.    Ray 1 travels parallel to the main axis and passes through F after reflection.  Ray 2  passes through F and travels parallel to the main axis after reflection.  The two reflected 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.

                        Image:

1) Real                2) Inverted   

3) A'B' >AB        4)  di > 2f

Figure 15

 

Case V: Object at f  (do = f )

do = f.   

Ray 1 travels parallel to the main axis and passes through F after reflection.  Ray 3 travels toward the mirror as if it is coming from C and reflects back on itself.  The two reflected rays are parallel and do not intersect.  In practice, the object is placed very close to F so that do is slightly greater than f.  Then the real image forms far from the mirror and much greater than the object.  This case is used in car head lights.                                        Image:

 

 

1) Real                2) Inverted   

3) A'B' >>AB      4)  di = 

Figure 16

When do is slightly greater than f, the image forms far from the mirror and much greater than object as shown in Fig. 17.

Figure 17

Case VI: Object within f  ( do < f )

do < f.   In this case, image is virtual and cannot be formed on a screen.  Ray 1 passes through F after reflection.  Ray 3 goes toward the mirror as if it comes from C and reflects back on itself.  The two reflected rays are divergent leftward.  Extending them rightward, they intersect at A', the virtual image of A.

                               Image:

1) Virtual      2) Upright   

3) A'B' >AB   4) Image behind the mirror

Figure 18

Real and Virtual Images:

In the first 5 cases, A' is formed by the intersection of real rays of light that is called the " real image. A real image can be formed on a screen.

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

The Mirror Formula:

It is easy to show that the relation between the object distance do, the image distance, di, and the focal length f of a mirror is:

This is true for all mirrors, flat, converging or diverging.

Figure 19

 

Referring to Fig. 19, check and see if you agree that

A'B' = HV & AB = KV.  Also, FV = f  ;  FB = do- f&  FB' = d i- f.

With these in mind, writing the similarity of two pairs of triangles, the mirrors formula can be proven.

Proof:  Refer to Fig. 19 as a typical case. 

ABF is similar to HVF.  It results in    AB/A'B' = FB/FV    (1)

FKV is similar to FA'B'.  It results in  AB/A'B' = FV/FB'   (2)

Comparing 1 & 2 results in  FB/FV = FV/FB'  or,  (do - f)/f = f/(di - f )  or,

(do - f)(di - f ) = f 2   or,   dodi - dof - dif + f 2 = f 2  or,   dodi =  di f + dof. 

 Dividing through by dodi f ,  we get:   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. 

Some texts give a negative sign to these ratios to account for the image being real or virtual and also inverted or upright.  Generally, in mirrors and lenses, an inverted image is real, and an upright image is virtual.

In Chapters 25 and 26, will use the |M| rather than placing a negative sign in front of each ratio. 

In solving the chapter's examples and problems, and only when substituting numbers in the formulas, any real quantity must be given a (+) sign and any virtual quantity must be given a (-) sign.  It will work perfectly everywhere. 

In the mirror formula  (1/f  = 1/do + 1/di), we let do to be always positive, because the object itself is always 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 mirror, f is real and positive.   For a diverging mirror, f is virtual and negative.  Overall, anything real is positive and anything virtual is negative.

Example 3:  A 4.0-cm (in height) object is placed at 15.0cm from a converging mirror perpendicular to the main axis (Fig. 20).  The radius of the mirror is 20.0cm.   Find the image distance, magnification, size (height), and state of the image (real or virtual).  Also, provide an appropriate ray diagram.  

 Solution:   f = R/2 ;   f = 10.0cm.

1/f = 1/do+1/di  ;  1/10 = 1/15 + 1/di 

di = 30.0cm.  

Since di is positive, image is real

 |M| = di /do = 30/15 = 2.00.

Also, M = A'B'/AB ; 2.00 = A'B'/4.0     A'B' = 8.0cm.

Figure 20

 

Example 4:  A 3.0-cm (in height) object is placed at 5.0cm from a converging mirror perpendicular to its main axis. The radius of the mirror is 20.0cm.   Find the image distance, magnification, size (height), and state (real or virtual).  Also, provide an appropriate ray diagram.

Solution: f = R/2 ; f = 10.0cm ;  1/f = 1/do+1/di ; 

1/10 = 1/5 +1/di ;  di = -10.0cm.  (-) sign means virtual image and upright.

 Magnification:

 M = di /do = 10/5 = 2.0  

 Since M = A'B'/AB as well 

 2.0 = A'B'/3.0

 A'B' = + 6.0 cm.

Figure 21

 

Example 5:  A converging mirror forms a real image that is 4.00 times greater than the object at 60.0cm from the mirror.  Find the radius R of the mirror and the object size if the image size (height) is 24.0cm.

Solution: Since the image is real, it is necessarily inverted.  |M| = 4.00. 

 |M| = di /do ;   4 = 60/do  ;   do = 15.0cm.   Now,  1/f = 1/do+1/di

1/f = 1/15 + 1/60 ;  f = 12.0cm ;  R = 2f ;  R = 24.0cm do = 24.0/4.00 = 6.00cm.

Figure 22

Image in Diverging Mirrors:

There is only one case for image in a diverging mirror.  The C and f of a diverging mirror are behind the mirror and therefore virtual.  No matter where an object is placed in front of a diverging mirror, the image is always upright, virtual, smaller than the object, and forms behind the mirror within the virtual focal distance.  This type of mirror is used in stores, at the cross-section of hallways, and in cars as side mirrors for rearview.  A ray diagram is shown below:

Object Position: Anywhere in front of the mirror

 

For a larger image, only the top half of the mirror is drawn.  Ray 1 that travels parallel to the main axis reflects as if it is coming from the virtual F.   Ray 2  going toward F reflects parallel to the main axis.  The reflected rays 1and 2 are divergent and do not intersect in front of the mirror.  Their extension in the opposite direction meet within the virtual focal length at A'.  Since A' is above the main axis, the image is upright.  It is also virtual, and smaller than the object.

              Image:

 

Figure 23

1) Virtual            2) Upright   

3) A'B' <AB       4) forms behind the mirror within the virtual focal distance.

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

Example 6: A 14.0cm high object is placed at 8.0cm from a diverging mirror perpendicular to its main axis. The radius of the mirror is 24.0cm.  Find the image position, magnification, size (height), and determine if real or virtual.  Also, provide an appropriate ray diagram.

Solution: f = R/2 ;  f = 12.0cm ;  1/f = 1/do+1/di ;  -1/12 = 1/8 +1/di  di = - 4.8cm.

 Since di is negative, the image is    virtual as expected. 

 Magnification:

 |M| = di /do = 4.8/8 = 0.60

 Also,  |M| = A'B' AB

 0.60 = A'B'/14.;    A'B' = 8.4 cm.

 The image is upright.

Figure 24

 

Example 7:  Show that the general mirror formula (1/f = 1/do+1/di ) is valid for flat mirrors as well.   A flat mirror may be viewed as a spherical mirror of radius R ∞.    A R , 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 8:  Using the general mirror formula, show that when object is moved very far away from the mirror to say infinity ( do  ) ,  the image forms at the focal point F.  Solution:  To be done by students.

 

Example 9:  Using the general mirror formula, show that when object is at F, image forms at infinity.  Solution:  To be done by students.

 

Example 10:  The distance between an object and its image in a flat mirror is 78.0cm.  How far from the mirror is the object placed?  Neglect the thickness of the glass of the mirror.

Solution: In flat mirrors,  di = - do.  This means that the object and its image are equidistant from the mirror.  The negative sign shows that the image is virtual.  do = 78.0cm/2 = 39.0cm.

 

Example 11:  The distance between an object and its real image in a converging mirror is 25.0cm.  The mirror has a diameter of 120.0cm.  Find   Rfdo, di, and the image size if the object size (height) is 4.0cm.

Solution: R = 60.0cm  or,  f = 30.0cm ;  From the figure: d i - do = 25.0 ;  d i = do + 25.0.  Substituting in the mirror formula and solving for do yields:

1/do + 1/(do+ 25) = 1/30 ;  multiplying through by 30do(do+ 25) gives us

30do + 750 + 30do = do2 + 25do   or,   do2 - 35do - 750 = 0.

Using Quadratic formula yields  

do= 50cm  &  do= -15cm as possible answers.  Only do = +50.0cm is acceptable because object is real. 

This results in d i = do + 25.0 = 75.0cm.

|M| =75.0 /50.0 = 1.50 ;

Also, from |M| = A'B'/AB, we get:

 A'B' = 6.0cm   ;   We already know that real image is inverted.

Figure 25

 

Example 12: The distance between the object and its virtual image in a converging mirror is 30.0cm.  The mirror has a diameter of 80.0cm.  Find   Rf, dodi , and the image size if the objects height (size) is 4.5cm.

Solution: R = 40.0cm ;  f = 20.0cm

From Fig. 26: d i+ do = 30.0cm   or,    d i = 30 - do.

Substituting in the mirror formula and solving for do yields:

 1/do - 1/(30 - do) = 1/20 ;  multiplying through by 20do(30 - do) results in

 600 - 20do -20do = 30do - do2     or,   do2 - 70do +600 = 0.

 Quadratic formula yields solutions: 

do= 60cm  &  do= 10cm.

do = 10.0cm is acceptable because it puts AB within f to form a virtual image. 

This results in d i = 30.0 - do = 20.0cm

Also M = 2.00  and  A'B' = 9.00cm

 

Figure 26

 

Example 13:  The distance between an object and its virtual image in a diverging mirror is 40.0cm.  The mirror has a diameter of 192.0cm.  Find R, f, dodi, and the image size, if the objects height is 12.0cm.  Draw a ray diagram before solving.  Hint: The focal length of a diverging mirror is negative.

Solution:  Left for students.  Answers:  96.0cm, 48.0cm, 24.0cm, 16.0cm , and 8.0cm.

 

Chapter 25 Test Yourself 2:    

1) In the ray diagrams for image in spherical mirrors, 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) This question is eliminated.

3)  Only (a) 1   (b) 2   (c) 3  of the 4 important rays is (are) sufficient to form the image of an object in a spherical mirror.

4) An incident ray parallel to the main axis of a converging mirror passes through (a) C   (b) F  (c) V after reflection.   click here.

5) An incident ray parallel to the main axis of a diverging mirror reflects as if it is coming from (a) the virtual C   (b) the virtual F   (c) V of the diverging mirror.

6) An incident ray going through C of a converging mirror reflects back (a) on itself an passes through C, again  (b) reflects and passes through F  (c) reflects and may take any direction.   click here.

7) An incident ray aiming toward C of a diverging mirror reflects back (a) on itself as if it is coming from the virtual C  (b) reflects as if coming from the virtual F   (c) reflects back and may take any direction.

8) An incident ray going through F of a converging mirror reflects back (a) on itself and passes through F again  (b) reflects and passes through C  (c) reflects and travels parallel to the main axis.   click here.

9) An incident ray appearing to pass through the virtual F of a diverging mirror reflects back (a) on itself an passes through F   (b) reflects and passes through C   (c) reflects and travels parallel to the main axis.

10) When an object is placed beyond the C of a converging mirror, its image forms (a) at F   (b) at C    (c) between F and C.  

11) When an object is placed at C of a converging mirror, its image forms (a) at F   (b) at C    (c) between F and C. 

12) When an object is placed at  (very far) from a converging mirror, its image forms (a) at F  (b) at C   (c) between F and C.   click here.

13) When an object is placed between F and C of a converging mirror, its image forms (a) at F  (b) at C   (c) beyond C. 

14) When an object is placed within f of a converging mirror, its image forms (a) infinitely far away  (b) at C   (c) behind the mirror and is virtual.   click here.

15) A converging mirror forms a virtual image when the object is placed (a) at F    (b) at C   (c) within f.

16) A diverging mirror 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 spherical mirror 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 mirror, Fig. 7, the reflected 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 mirror, Fig. 7, the reflected rays are divergent.  The extension of the reflected rays in the opposite direction could come to a point (a) at the virtual F   (b) on the virtual focal plane     (c) both a & b.   click here.

20) The mirror used at the intersection of roads and hidden driveways is (a) diverging   (b) converging   (c) flat.

21) The farther an object is from a diverging mirror, the closer its image to the (a) virtual C   (b) virtual F   (c) Vertex.

22) The formula that relates object distance do to image distance di  and focal length f of a spherical mirror is (a) 1/do+1/di = 1/   (b) 1/do-1/di = 1/f           (c) 1/do-1/di = -1/f     (d) a, b, & c.

23) In the mirrors 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| of mirrors 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 larger than the object   (b) the image is smaller than the object   (c) the image is equal to the object.   click here.

26) We may write: (a) |M| = do/di = A'B'/AB    (b) |M| = di/do = A'B'/AB      (c) both a & b.

Problem I: A 6.00cm high object is placed at 25.0cm from a converging mirror with a diameter of 40.0cm.  Draw an appropriate ray-diagram and answer the following questions:

27) The focal length of the mirror is (a) 20.0cm    (b) 10.0cm    (c) 15.0cm

28) The image forms at (a)16.7cm   (b)7.14cm   (c)-15.0cm from the mirror.

29) |M| is (a) 1.50    (b) 2.50    (c) 0.667.  click here.

30) The object is placed (a) beyond C    (b) at C    (c) between F and C.

31) The image is formed (a) beyond C    (b) at C    (c) between F and C.

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 mirror with a diameter of 60.0cm.  Draw an appropriate ray-diagram and answer the following questions:

33) The focal length of the mirror is (a) 15.0cm    (b) 30.0cm    (c) 25.0cm.

34) The image forms at (a) 45.0cm   (b) 9.50cm   (c) 30.0cm from the mirror.

35) |M| is (a) 4.50    (b) 1.00    (c) 2.33.   click here.

36) The object is placed (a) beyond C    (b) at C    (c) between F and C.

37) The image is formed (a) beyond C    (b) at C    (c) between F and C.

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 mirror with a radius of 20.0cm.  Draw an appropriate ray-diagram and answer the following questions:

39) The f of the mirror is (a) 10.0cm    (b) 5.00cm    (c) 4.00cm.   click here.

40) The image forms at (a) 60.0cm   (b) 17.5cm   (c) 30.0cm from the mirror.

41) |M| is (a) 10.0    (b) 0.20    (c) 5.00.   click here.

42) The object is placed (a) beyond C    (b) at C    (c) between F and C.

43) The image is formed (a) beyond C    (b) at C    (c) between F and C.

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 mirror whose radius is 20.0cm.  Draw an appropriate ray-diagram and answer the following questions:

45) The focal length f is (a) 10.1cm    (b) 10.0cm    (c) 9.90cm.   click here.

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

47) |M| is (a) 10.0   (b) 100   (c) 1000.   click here.

48) The object is placed (a) beyond C    (b) at C    (c) almost at F.

49) The image is formed (a) way beyond C   (b) at C   (c) between F and C.

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 from a converging mirror with a radius of 24.0cm.  Draw an appropriate ray-diagram and answer the following questions:

51) The f of the mirror is (a) 10.1cm    (b) 12.0cm    (c) 9.90cm.   click here.

52) The image forms at (a) -24.0cm from the mirror (behind)   (b) 24.0cm from the mirror   (c) 12.0cm from the mirror.

53) |M| is (a) 1.00   (b) 3.00   (c) 2.00.    click here.

54) The object is placed (a) within F    (b) at C    (c) almost at F.

55) The image is formed (a) behind the mirror, upright and virtual    (b) at C    (c) between F and C.

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 at 18.0cm from a diverging mirror with a radius of 24.0cm.  Draw an appropriate ray-diagram and answer the following questions:

57) The mirror's virtual focal length is (a)-24.0cm   (b)-36.0cm   (c)-12.0cm.

58) The image forms (a) 36.0cm   (b) 7.20cm   (c) 10.8cm behind the mirror.

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 mirrors, F and C are virtual and behind the mirror; therefore, there are no bench marks as to where an object is placed in front of a diverging mirror.  The closer the object to a div. mirror, the closer its behind-the-mirror-image to the div. mirror as well.  The farther an object from a div. mirror, the closer its image to the virtual focal point f.  You may verify this by drawing a few ray diagrams and placing the object at different positions.  Experimentally, you may move your hand to and away from a diverging mirror and observe the image positions.

61) The image is formed (a) behind the mirror, upright and virtual    (b) within the virtual focal length f     (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 mirror.  The image formed on a screen is 5.00 times greater than the object.  Draw an appropriate ray-diagram and answer the following questions:

63) |M| is (a) 5.00    (b) 0.200    (c) neither a nor b.     click here.

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

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

Problem VIII: An object is 40.0cm from its real image in a converging mirror.  The image is 6.00 times greater than the object.  Draw an appropriate ray-diagram and answer the following questions:

66) The image is real; therefore, the object and its image are (a) on the same side in front of the mirror   (b) one in front and the other behind the mirror  (c) both behind the mirror.   click here.

67) The 40.0cm 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) We may write: a) di +do= 40.0cm   b) di - f = 40.0cm   c) di - do= 40.0cm.

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 2nd equation, using |M| is (a) di = do    (b) di = 5do    (c) di = 6do.

71) Solving the two equations, we get: (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 IX: An object is 32.0cm from its virtual image in a converging mirror.  The image is 3.00 times greater than the object. Draw an appropriate ray-diagram and answer the following questions:

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

75) The 32.0cm that is (a) di + do   (b) di - f    (c) di - do.   To understand this clearly, a ray diagram must be drawn.   click here.

76) The correct equation is (a) di + do = 32.0cm    (b) di - f = 32.0cm    (c) di - do = 32.0cm.

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

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.

Problems:

1) A 178-cm tall lady is in front of a flat mirror placed on a wall.  Her eyes are 168cm above the floor.  Find the (a) minimum height of the mirror and (b) the distance its lower edge must have from the floor in order to show her full height image.  Draw a ray diagram for the problem.

2) A 6.00-cm object is placed at 18.0cm from a converging mirror perpendicular to its main axis. The radius of the mirror is 32.0cm.    Find (a) the image distance, the absolute value of magnification, and image size.  (b) State if image is real, virtual, upright, or inverted.  Also, provide an appropriate ray diagram.

3) A 4.0-cm object is placed at 10.0cm from a converging mirror perpendicular to its main axis. The diameter of the sphere of the mirror is 48.0cm.   Find (a) the image distance, the absolute value of magnification, and the image size.  (b) State if image is real, virtual, upright, or inverted.  Also, provide an appropriate ray diagram.

4) A converging mirror forms a real image that is 5.00 times greater than the object at 80.0cm from the mirror.  Find (a) the radius of the mirror and (b) the object size if the image size is 24.0cm.

5) A 19.0-cm object is placed at 11.0cm from a diverging mirror perpendicular to its main axis. The radius of the mirror is 27.0cm.   Find (a) the image distance, the absolute value of magnification, and image size.  (b) State if image is real, virtual, upright, or inverted.  Also, provide an appropriate ray diagram.

6) The distance between an object and its image in a flat mirror is 108.0cm.  How far from the mirror is the object placed?  Neglect the thickness of the glass of the mirror.

7) The distance between an object and its real image in a converging mirror is 60.0cm.  The mirror has a diameter of 64.0cm.  Draw the related ray diagram and find Rfdodi, and the image size, if the object's height is 3.0cm.

8) The distance between an object and its virtual image in a spherical mirror is 72.0cm. The mirror has a diameter of 108.0cm.  Determine (a) the mirror type, and (b) the values of R, fdodi, and the image size if the object's size is 4.5cm.   It is known that the image is greater than the object.

9) The distance between the object and its virtual image in a spherical mirror is 150.0cm.  The mirror has a diameter of 160.0cm.  Determine (a) the mirror type, and (b) the values of R,  f do  di , and the image size, if the object's size is 24.0cm.   It is known that the image is smaller than the object.

10) Using the mirrors formula for a concave mirror, show that (a) when object is at infinity, its image forms at F, (b) when object is at F, its image forms infinitely far away, and (c) when the object is at 2f, its image forms at 2f as well.

11) Using the mirrors equation, show that for a flat mirror, the image (a) is always at the same distance to the mirror as the object is, and (b) is always virtual.

Answers:

1) 89.0cm, 84.0cm

2) [144cm, 8.00, 48.0cm], Real, Inverted

3) [-60.0cm, 6.00, 24.0cm],Virtual, Upright

4) 26.7cm, 4.80cm

5) [-6.06cm, 0.551, 10.5],Virtual, Upright    6) 54.0cm

7)  32.0cm, 16.0cm, 20.0cm, 80.0cm, 12.0cm

8)  (a) Converging  (b) 54.0cm, 27.0cm, 18.0cm, 54.0cm, 13.5cm

9)  (a) Diverging  (b) 80.0cm, 40.0cm, 30.0cm, 120.0cm, 6.00cm