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 source of light. A ray of light is a very narrow streak of light that travels along a straight line.  we define Reflection as the return of light back to its original medium upon bouncing off a reflective surfaceEvery 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.

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's distance and di is the image distance 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.

 Given  do = di ,  prove that   θi  = θr . To prove, lets take advantage of the fact that triangles AHI and A'HI  are congruent.  HA equals HA', each triangle has a 90° angle, and both share HI.  From their congruity, we conclude that angles 1 and 2 are equal.   Segment NI is parallel to AA' ; thus,    angle 1 = θi    and   angle 2 = θr ; therefore, θi  = θr Law of Reflection Fig. 1:   The reflected ray appears to be coming from A', the image of A.  Actually, there is nothing behind the mirror.

An example:

 Example 1:  A laser pointer is emitting a narrow beam that makes a 30° angle with a flat mirror.  What is the reflection angle? 30° or 60°?    Draw a diagram for the problem. Solution:  A 30° angle with mirror means a 60° angle with the normal line.  Therefore,  θi = 60°.  Since θi = θr ; therefore, θr = 60° as well. Fig. 2

Another example:

 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: tan α = 2.00 / 0.90, or  α = 65.8°. Fig. 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:

Fig. 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 of the mirror.  The distance between the focal point (F) of a mirror and 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 the following figure:

 Parallel rays of light that are also parallel to the main axis gather (converge) at the focal point of a concave mirror.  It is for this reason that a concave mirror is also called " converging mirror".  Distance VF = 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 Fig. 5

In case of a convex mirror, when parallel rays of light that are also parallel to the main axis reach it, they reflect and they spread apart after reflection as if they are coming from a common point that is behind the mirror.  This common point is called the "virtual focal point."  It is an imaginary point behind the mirror.  See the following figure:

 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 " diverging mirror". Fig. 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 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 on the right). 78

Fig. 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. 910

Fig. 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) along a line going through F, the virtual focal point, travels parallel to the main axis after reflection. 1112

Fig. 9

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

Fig. 10

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

Fig. 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 tiny flat mirror 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

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 back into its original medium upon striking on a reflecting surface.  (b) the bouncing back 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) a 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 that, and at the point of incidence, draw a normal line.  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 common in 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'  IN, 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 (1/2R).  (c) twice 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 hits it, reflect back (a) on itself.  (b) and passes through the center, C.  (c) 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) 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 that distance of 150,000,000km.  A regular light bulb placed at a distance of say 200ft or more has the same effect.  Ray coming from it may be treated as parallel.  Rays from a laser pointer are quite parallel and are ideal for optics experiments.   click here.

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.

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 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 (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 whose diameter is within 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 almost 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' <

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

From all points of the object infinite rays of light emerge radially outward.  If A', the image 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 mirror simply because the mirror is not in their way.  All rays that are incident on the converging mirror will come to the same point after reflection.  The point is not necessarily F.  To find the image of A that we call it A', we choose two of those infinite rays that are convenient for us.  Ray 1 is the ray that travels parallel to the main axis and after reflection passes through F, and Ray 2 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 perpendicular to the main axis is drawn, point B', the image of B, 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, or we may say between F and C.

 do >2f Image Conditions: 1) Real    2) Inverted    3) A'B' < AB   4) f < di < 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.  That verifies that the reflected rays meet at same points that are images of points of the object. Two rays are enough to form the image of point A. 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 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 falls on B.   B , B' and  C are the same, the center of the mirror. 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 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. 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 reflection.  Ray 3 travels toward the mirror as if 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 image forms far from the mirror and much greater than the object.  This is used in car's head lights. Fig. 16

The following figure shows such ray diagramWhen AB is near F from the center side, but not exactly on F, the image forms far from the mirror 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 mirror 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 reflection.  Ray 3 goes toward the mirror as if it comes from the center C and reflects back on itself.  The two reflected rays are divergent in traveling leftward.  We extend them in the opposite direction till they intersect 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 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. Note in the figure that A'B' = HV and AB = KV.  Also, FV = 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 HVF. It results in AB / A'B' = FB / FV   (1) FKV is similar to FA'B' ; thus, AB / A'B' = FV / FB'   (2) Comparing (1) and (2) results in FB / FV = FV / FB'  or, (do - f)(di - f ) = f 2   or, Fig. 19 dodi - dof - dif + f 2 = f 2    or, 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.  A sign convention may be made as follows:  For a real image both do and di are in front of the mirror and positive and the negative sign in front of the fraction bar makes magnification negative showing that the image is inverted and real.  For a virtual image do is positive and di, behind the mirror, is negative.  This makes the magnification positive showing that the image is upright and virtual.  Other sign conventions are possible. It is easier to use the absolute value of magnification and not be bothered by its sign.

Important:

In the mirror formula  ( 1/f  = 1/do + 1/di ) ,   we let do to be 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 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 object is placed at 15.0cm from a converging mirror perpendicular to its main axis. The radius of the mirror is 20.0cm.   Find the image distance, magnification, size, and state if 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, the image is therefore real.   The magnification is: M = - di / do  = -30/15 = - 2.00 ;   Also, M = A'B' / AB ;   -2.00 = A'B'/4.0     A'B' = -8.0 cm.  The image is inverted. Fig. 20

Another example:

 Example 4:  A 3.0-cm 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, and state if 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.   Since di is negative, the image is therefore, virtual.   The magnification is: M = - di / do  = -(-10)/5 = + 2.0 ;   Also, M = A'B' / AB ;  +2.0 = A'B'/3.0 A'B' = + 6.0 cm. The image is upright. Fig. 21

Another example:

 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 of the mirror. and the object size if the image size is 24.0cm. Solution:  Since the image is real, it is necessarily inverted.  M = - 4.00. We may write M = - di / do ; - 4 = -60/do  ; do = 15.0cm 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 Fig. 22

Image in Diverging Mirrors:

There is only one case for image in a diverging mirror.  Since the center and focal point of a diverging mirror is behind the mirror and virtual, no matter where the object is placed in front of the 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.  It is normally stated that " objects in this mirror appear farther than their actual distances."  This is because a diverging mirror forms images smaller than the actual sizes of the objects.  A ray diagram is shown below:

 Object placed anywhere in front of the mirror Image is always: 1) Virtual 2) Upright    3) A'B'

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 14.0-cm 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 distance, magnification, size, and state 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 therefore 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. The image is upright. Fig. 24

Another example:

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 RNote that as  R , then 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, the image forms at infinity.  Solution:  To be done by students.

Example 10:  The distance between the 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,  do = - di .  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's height is AB = 4.0cm.

 Solution: R = 60.0cm ;  f = 30.0cm   From the figure: d i - do = 25.0 and hence d i = do + 25.0.  Substituting in the mirror formula and solving for do yields: 1/do + 1/( do + 25  ) = 1/3030do + 750 + 30do = do2 + 25do do2 - 35do - 750 = 0,  Using the quadratic formula yields:  do = 50cm  and  do = -15cm are the possible answers.  Only do = 50cm is admissible because object is real.  This results in  d i = do + 25.0 = 75.0cm. Fig. 25 M = -75.0 /50.0 = -1.50 from which and M = A'B'/AB   we get:: A'B' = - 6.0cm. The image is inverted.

Another example:

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  Rfdo di , and the image size, if the objects height is AB = 4.5cm.

 Solution: R = 40.0cm  ;   f = 20.0cm    From the figure: d i + do = 30.0 and hence d i = 30 - do.  Substituting in the mirror formula and solving for do yields:  1/do - 1/( 30 - do ) = 1/20  600 - 20do - 20do = 30do  - do2 do2 - 70do + 600 = 0,  Using the quadratic formula yields:  do =  60cm  and  do = 10cm are the possible answers. Fig. 26 Only do = 10.0cm causes a virtual image.  This results in d i = 30.0 - do = 20.0cm.  Note that in reality we should write di = -20.0cm because di is technically behind the mirror.  Also M = 2.00  and  A'B' = 9.00cm

Another example:

Example 13:  The distance between the object and its virtual image in a diverging mirror is 40.0cm.  The mirror has a diameter of 192.0cm.  Find  Rfdo di , and the image size, if the objects height is AB = 12.0cm.  Draw a ray diagram before solving.

Solution:  Left for students.  Answer:  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) If the object is placed perpendicular to the main axis, (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)  (a) 1   (b) 2   (c) 3  of the 3 important rays is (are) sufficient to form the image of an object in a spherical mirror.

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

5) A 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.

6) A 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) A ray going through C of a diverging mirror reflects back (a) on itself as if coming from virtual F.  (b) reflects as if coming from virtual C.  (c) reflects back and may take any direction.

8) A 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) A 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 C of a converging mirror, its image forms (a) at F.  (b) at C.   (c) between F and C.  Verify your answer by drawing an appropriate ray-diagram.

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

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

14) When an object is placed within f of a converging mirror, its image forms (a) at .  (b) at C.   (c) behind the mirror and is virtualVerify your answer by drawing an appropriate ray-diagram.  click here.

15) The only case a converging mirror forms a virtual image is 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 the object distance, do, to the image distance, di, and the focal length, f ,of a spherical or flat mirror 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 mirror's 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 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.     (c) both a & b.

Problem(I): A 6.00cm high object is placed at 25.0cm from a converging mirror whose diameter is 40.0cm.  Draw an appropriate ray-diagram. 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 from the mirror.  (b) 7.14cm from the mirror.  (c) -15.0cm from the mirror.

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 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 whose diameter is 60.0cm.  Draw an appropriate ray-diagram. 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 from the mirror.  (b) 9.50cm from the mirror.  (c) 30.0cm from the mirror.

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

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

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

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 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. Answer the following questions:

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

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

47) The absolute value of  magnification 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 whose radius is 24.0cm.  Draw an appropriate ray-diagram. Answer the following questions:

51) The focal length 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) 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 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: A 3.50cm high object is placed at 18.0cm from a diverging mirror whose radius is 24.0cm.  Draw an appropriate ray-diagram. Answer the following questions:

57) The virtual focal length of the mirror is (a)  -24.0cm.   (b) -36.0cm.  (c)   -12.0cm.    click here.

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

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 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 the 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.  You may verify this by drawing a few ray diagrams and placing the object at different positions.

61) The image is formed (a) behind the mirror, upright and virtual.   (b) within the virtual focal point.   (c) both a & b.

62) The image size is (a) 1.40cm.   (b) 10.5cm.   (c) 7.00cm.     click here.

Problem(VI): 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. 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 mirror.  (b) 40.0cm from the mirror.  (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(VII): 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. Answer the following questions:

66) The image is real; therefore, the object and its image are (a) at 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) The equation that may be written is (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 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(VIII): 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. Answer the following questions:

74) The image is virtual; therefore, the object and its image are (a) at 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 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 = 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 = 5 do.    (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 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 and upright / 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 image size. (b) State if image is real / virtual and upright / 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 and upright / inverted.. Also, provide an appropriate ray diagram.

6) The distance between the 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 an appropriate ray diagram and find  Rfdo di , and the image size, if the object's height is AB = 3.0cm.

8) The distance between the object and its virtual image in a spherical mirror is 72.0cm.  The mirror has a diameter of 108.0cm.  Find  R,  fdo di , and the image size, if the object's size is AB = 4.5cm, and 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.  Find  Rfdo di , and the image size, if the object's size is AB = 24.0cm, and it is known that the image is smaller than the object.

10) Using the mirrors equation for a concave mirror, show that (a) when object is at infinity, its image forms at F, (b) when the object is at F, the image forms  very faraway from the mirror, and (c) when the object is at 2f, the 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.

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)  54.0cm, 27.0cm, 18.0cm, 54.0cm, 13.5cm

9) 80.0cm, 40.0cm, 30.0cm, 120.0cm, 6.00cm