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Solve Problems on Mirror and Magnification Formula

Mirror and magnification formulas are fundamental equations used in optics to describe the formation of images by mirrors. These formulas are commonly applied in geometrical optics, which deals with the behavior of light rays using the principles of geometry. This article contains mirror and magnification formula and problems based on it in detail.

Mirror Formula

The mirror formula shows the relationship between the object distance, image distance, and the focal length of the spherical mirror. Hence, the formula is given as,

1/u + 1/v  = 1/f

where, 

• u is the object distance, 
• v is the image distance, and 
• f is the focal length of the mirror

While solving numerical problems using mirror formula one must remember two important things according to the sign convention for spherical mirrors, that are: If object is at left side of the mirror the object distance is taken as negative, if it is in right it is positive. For focal length the sign is depend on type of mirror we are using if mirror is concave then focal length is taken negative and if mirror is convex then focal length is positive .

Mirror formula in terms of Radius of Curvature:

Since, the radius of curvature (R) is two times its focal length (f), that is

R = 2f or  f = R/2

Hence, the Mirror Formula can be written as:

1/u + 1/v = 1/f = 2/R

Magnification

One must have noticed that there is simply an increase or decrease in the size of image by concave or convex mirror, this is a magnification of the object. 

• Magnification is defined as the ratio of the height of the image (h i ) to the height of the object (h o ).

Magnification = Height of image / Height of object

m = H i  / H o

Magnification can also be defined as the ratio of the image distance (v) and the object distance from the mirror (v).

Magnification = -Image distance / Object distance 

Determination of the Nature of the Image formed by a spherical mirror:

• The positive magnitude of magnification shows us that a virtual and erect image is formed.
• The negative  magnitude of magnification shows us that a real and inverted image is formed.

Also, Check

• Mirror Equation
• Image Formation by Spherical Mirror
• Lens Maker’s Formula

Problems on Mirror Formula and Magnification Formula

Problem 1: An object is placed at a distance of 2 times of focal length from the pole of the convex mirror, Calculate the linear magnification.

Let the Focal length of mirror = f So, the object distance, u = -2f  The formula to calculate image distance we use mirror formula as, 1 / v + 1 / u = 1 / f Therefore, 1 / v + 1 / -2f = 1 / f  1 / v  = 1 / f + 1 / 2f   1 / v = 3 / 2f or v = 2f / 3 Magnification is given as,  m = – v / u = -(2f/3) / (-2f)  =   1/3

Problem 2. If the image is a distance of 6 cm and the object is at 12 cm in the front of the concave mirror, Calculate the magnification formed.

Given that, The distance of object, u = – 12 cm The distance of image, v = – 6 cm Since, Magnification is given by,  m = – v / u Therefore, m = – (-6 / -12)  = -0.5 Hence, the image will be diminished by nearly half as size of object. 

Problem 3: In the experiment height of the image is 12 cm whereas the height of the object is 3 cm, would you determine the magnification formed.

Given that,  Height of image = 12 cm  Height of object = 3 cm Magnification in terms of height is given by, m = height of image / height of object = 12 / 3  = 4 Therefore magnification is 4 .

Problem 4: In the case of a concave mirror if the object is placed at the distance of 12 cm. Determine the image distance from the mirror if the height of the object to image ratio is 1:2.

Given that, The object distance, u = -12 cm Ratio of object to image height = 1/2 Magnification = height of image / height of object   = 1/ (1/2)  = 2 Now, magnification in terms of distance of object and image from the mirror, m = – v / u = – v / -12  2 = v / 12   or  v = 12 × 2  = 24 Therefore the distance of image from the mirror is equal to 24 cm .

Problem 5: Calculate the change in the size of the image formed, if the object distance is 18 cm and the distance of the image is 6 cm from the concave mirror. 

Given that, The object distance, u = -12 cm Image distance, v = – 6cm Magnification,  m = – v/u  = – (-6 / -18)  = -1/3 which means that size of image is 1/3 rd of the size of object.

Problem 6: The radius of curvature of the rear view convex mirror of the truck is 6 m. If the car is 8 m from the mirror of the truck. Calculate the distance at which the image is formed.

Given that, Radius of curvature, R = 6 m  Object distance, u = -8 m Focal length is half of Radius of curvature, f = R/2  =  6/2  = 3 m Using mirror formula  1 / v + 1 / u = 1 / f 1 / v + 1 / -8 = 1 / 3 1 / v  = 1 / 3 + 1 / 8 = 11 / 24 v = 24 / 11 m The image is formed at distance of 24 / 11 behind the mirror.

Problem 7: A concave mirror produces an image of size n times that of the object and of focal length f. If the image is real then find the distance of the object from the mirror.

Given that Size of image = n × size of object n = Size of image /  size of object  = magnification Since the image is real, it must be inverted hence magnification will be negative,  m = -n Let d is the distance of object then, m = -v/u   -n = -v / d  or v = nd Therefore, the mirror formula: 1 / f = 1/v + 1/u becomes, 1/f = 1/nd + 1/d or 1/f = 1/d(1/n + 1) or 1/d = n/ f(n + 1) Therefore, d = f (n + 1)/ n

Problem 8: Where should the object be placed to obtain a magnification of 1/3? If an object is placed at a distance of 60 cm from a convex mirror, then the magnification produced is 1/2. 

Given that, u = -60 cm m  = 1/2  So, -v/u = 1/2  and  v/60 = 1/2 or v = 30 cm Since, the mirror formula is: 1 / v + 1 / u = 1 / f Therefore, 1 / 30 + 1 / (-60) = 1/f 1/f = ( 2-1 ) / 60 = 1 / 60 f = 60 cm Now for magnification = 1 / 3, – v / u = 1 / 3 or  v = – u / 3  using mirror formula  1 / v + 1 / u = 1 / f 1 / (-u/3) + 1/ u = 1/ 60 -3/ u + 1/u = 1/60 -2/ u = 1/60 or  u = -120 cm object should be placed at 120 cm in front of mirror to get magnification of 1/3.

Problem 9: In the case of a concave mirror, if the object distance is 11 cm, its focal length is 11 cm then, Calculate the image distance.

Given that, Distance of object, u = -11 cm Focal length, f = -11cm Using mirror formula, 1 / v + 1 / u = 1 / f Therefore, 1 / v + 1 / -11 = 1/ -11 So, 1/v = 0 or  v = infinity This means that image will be at infinity if object is present at the focal length.

Problem 10: If the object distance is 32 cm in front of the concave mirror, the focal length of the mirror is 16 cm. State the nature and the size of the image formed.

Given that, Object distance, u = -32 cm Focal length , f = -16 cm For image distance use mirror formula, 1 / v + 1 / u = 1 / f Therefore, 1/ v + 1/ -32 = 1/ -16 or  1/ v = 1/ -16 + 1/ 32 or 1/ v = (-2 + 1) / 32 So, v = -32 cm Hence the image is located 16 cm in front of the mirror. and the image formed is real and inverted.  Size of image will be same as that of object, as it is located at center of curvature.

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Concave mirror – problems and solutions

1. An object is placed 10 cm from a concave mirror . The focal length is 5 cm. Determine (a) The image distance (b) the magnification of image

The focal length (f) = 5 cm

The object distance ( d o ) = 10 cm

Formation of image by concave mirror :

The image distance :

1/ d i = 1/f – 1/ d o = 1/5 – 1/10 = 2/10 – 1/10 = 1/10

d i = 10/1 = 10 cm

The image distance is 10 cm.

The magnification :

m = – d i / d o = -10/10 = -1

1 means that the image is the same as the object.

The minus sign indicates that the image is inverted. If the sign is positive than the image is upright.

2. A 5-cm-high object is placed in front of a concave mirror with a radius of curvature of 20 cm. Determine the image height if the object distance is 5 cm, 15 cm, 20 cm, 30 cm.

The radius of curvature ( r ) = 20 cm

The focal length (f) = R/2 = 20/2 = 10 cm

The object height (h o ) = 5 cm

a) the focal length (f) = 10 cm and the object distance ( d o ) = 5 cm

The image distance ( d i ) :

1/ d i = 1/f – 1/ d o = 1/10 – 1/5 = 1/10 – 2/10 = -1/10

d i = -10/1 = -10 cm

The minus sign indicates that the image is virtual or the image is behind the mirror.

The magnification of image ( m ) :

m = – d i / d o = -(-10)/5 = 10/5 = 2

The plus sign indicates that the image is upright.

The image height ( h i ) :

m = h i / h o

h i = h o m = (5 cm)(2) = 10 cm

The image height is 10 cm.

b) The focal length (f) = 10 cm and the object distance ( d o ) = 15 cm

1/ d i = 1/f – 1/ d o = 1/10 – 1/15 = 3/30 – 2/30 = 1/30

d i = 30/1 = 30 cm

The plus sign indicates that the image is real or the image is 30 cm in front of the mirror, on the same side as the object.

m = – d i / d o = -30/15 = -2

The minus sign indicates that the image is inverted.

The image is 2 times larger than the object.

The image height (h i ) :

c) The focal length (f) = 10 cm and the object distance ( d o ) = 20 cm

The image distance (d i ) :

1/ d i = 1/f – 1/ d o = 1/10 – 1/20 = 2/20 – 1/20 = 1/20

d i = 20/1 = 20 cm

The positive sign indicates that the image is real or the image is 20 cm in front of the mirror, on the same side as the object.

m = – d i / d o = -20/20 = -1

The negative sign means the image is inverted.

The image height ( hi ) :

h i = h m = (5 cm)(1) = 5 cm

d) The focal length (f) = 10 cm and the object distance ( d o ) = 30 cm

1/ d i = 1/f – 1/ d o = 1/10 – 1/30 = 3/30 – 1/30 = 2/30

d i = 30/2 = 15 cm

The plus sign indicates that the image is real or the image is 15 cm in front of the mirror, on the same side as the object.

The magnification of image ( m) :

m = – d i / d o = -15/30 = -0.5

The image is 0.5 smaller than the object.

h i = h o m = (5 cm)(0.5) = 2.5 cm

3. An image an by a concave mirror is 4 times greater than the object. If the radius of curvature 20 cm, determine the object distance in front of the mirror!

The magnification of image ( m ) = 4

The focal length (f) = r /2 = 20/2 = 10 cm

Wanted : The object distance ( d o )

m = – d i / d o

4 = – d i / d o

– d i = 4 d o

d i = – 4 d o

1/f = 1/ d o + 1/ d i

1/10 = 1/ d o + 1/4 d o

4/40 = 4/4 d o + 1/4 d o

4/40 = 5/4 d o

(4)(4s) = (5)(40)

16 d o = 200

d o = 12.5 cm

The object distance = 12.5 cm.

4 . A 1-cm high object is placed 10 cm from a concave mirror with the focal length, f = 15 cm. Determine :

A. The image distance ?

B. The image height?

C. The properties of image formed by the concave mirror?

The object height (h) = 1 cm

The object distance (d o ) = 10 cm

The focal length of the concave mirror (f) = 15 cm

A. The image distance (d i )

1/f = 1/d o + 1/d i

1/d i = 1/f – 1/d o = 1/15 – 1/10 = 2/30 – 3/30 = -1/30

d i = -30/1 = -30 cm

The negative sign indicates that the image is virtual or the image is behind the mirror.

B. The image height (h i )

The magnification of the image (M) :

M = -d i /d o = h i /h o

M = -(-30)/10 = 30/10 = 3 times

M = h i / h o

3 = h i / 1 cm

h i = 3 (1 cm)

the image height is 3 cm. The plus sign indicates that the image upward.

C. The properties of the image :

Virtual, upward, larger than object

5 . The magnification of the image, according to the image below.

The object distance (d o ) = 60 cm

The focal length (f) = 20 cm

Wanted : The image magnification (M)

1/d i = 1/f – 1/d o = 1/20 cm – 1/60 cm = 3/60 cm – 1/60 cm = 2/60 cm

d i = 60/2 cm = 30 cm

M = d i /d o = 30 cm / 60 cm = 1/2 times

6 . If the object is placed 6 cm from a concave mirror, the image distance is 12 cm as shown in figure below. Whhat is the image distance if the object is moved from the original position 1 cm away from the mirror.

The object distance (d o ) = 6 cm

The image distance (d i ) = 12 cm

Wanted : if the object distance (d o ) = 7 cm then the image distance is …

1/f = 1/do + 1/di = 1/6 + 1/12 = 2/12 + 1/12 = 3/12

f = 12/3 = 4 cm

The focal length is positive, means that the focal point is real or the rays pass through the point.

1/d i = 1/f – 1/d o = 1/4 – 1/7 = 7/28 – 4/28 = 3/28

d i = 28/3 = 9.3 cm

7. A dentist observes and checks the patient’s teeth using a mirror with an 8 cm radius. In order for the hole to be seen clearly by the doctor, what is the distance between the patient’s teeth and the mirror?

A. less than 4 cm in front of a concave mirror

B. less than 4 cm in front of a convex mirror

C. more than 4 cm in front of the concave mirror

D. more than 4 cm in front of the convex mirror

Radius of mirror (r) = 8 cm

The focal length of mirror (f) = r / 2 = 8 / 2 = 4 cm

Wanted : The distance between the patient’s teeth and the mirror

The mirror used is a concave mirror or a convex mirror? In order for the tooth hole to be clearly visible by the doctor, the mirror used should be able to enlarge the image of the tooth and the image must be upright. Convex mirror always produces inverted images and the size of the image is smaller than the size of the object. Conversely a concave mirror can produce an upright image if the object distance (d) is smaller than the focal length (f). If the object distance is greater than the focal length (f) then the concave mirror produces an inverted image.

The focal length (f) of the concave mirror is 4 cm, therefore the patient’s teeth should be less than 4 cm in front of a concave mirror.

The correct answer is A.

8. A concave mirror has a radius of curvature of 24 cm. If the object is placed 20 cm in front of the mirror then determine the properties of the image.

A. Real, upright and enlarged

B. Real, inverted and enlarged

C. Virtual, upright and enlarged

D. Virtual, inverted and smaller

Radius of curvature (r) = 24 cm

Focal length (f) = R/2 = 24/2 = +12 cm

The focal length of the concave mirror is positive or real because the light passes through the focal point of the mirror.

Object distance (d) = 20 cm

Wanted : Properties of image

Image is virtual or real? Calculate the image distance (s’):

1/d’ = 1/f – 1/d

1/d’ = 1/12 – 1/20

1/d’ = 5/60 – 3/60

1/d’ = 2/60

The image distance signed positive means that the image is real because it is passed by light.

Image enlarged ? Upright or inverted? First calculate the image magnification (M):

M = -d’ / d = -30/20 = -1.5

M > 1 means the image is enlarged, M has a negative sign means an inverted image. So the image properties are real, inverted, enlarged.

The correct answer is B.

9. A spherical mirror produces an image has size 5 times greater than the object on a screen, 5 meters away from the object. The mirror is…..

A. concave with the focal length of 25/24 m

B. convex with the focal length of 25/24 m

C. concave with the focal length of 24/25 m

D. convex with the focal length of 24/25 m

Magnification of image (M) = 5 times

The distance between object and image = 5 meters

The size of the image produced by a convex mirror is always smaller than the size of the object, therefore, the mirror is a concave mirror.

Object distance (d) = x

Image distance (d’) = x + 5

Image magnification (M) = 5 times

The formula of image magnification :

The formula of the focal length (f) :

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Concave Mirror: Problems with Answers for AP Physics 2

Here, using the mirror equation and magnification formula for curved mirrors a number of problems on the concave mirror are solved which is helpful for the AP Physics 2 exam.

We tried to illustrate all properties of the image formed by a concave mirror using equations and ray diagrams for a deeper understanding. 

Summary of concave spherical mirrors

Mirror equation is \[\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\] Where $f, d_i, d_o$ are the focal length, image distance and the object distance, respectively. 

The magnification formula for spherical mirrors is also written as \[M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}\] where $h_i, h_o$ are the image and object heights, respectively. 

Sign Rules for Concave mirror:

• The focal length is positive $f>0$. 
• The image distance must be inserted positive $d_i>0$ if the image is formed in front of the mirror.
• The image distance must be inserted negative $d_i<0$ if the image is formed back of the mirror.

Note : In front of the mirror is where the object is placed. 

Concave Mirror Solved Problems:

Problem (1): A pencil is placed 6 cm in front of a concave mirror having a radius of curvature of 40 cm.  (a) What is its focal length? (b) Determine the position of the image formed by this mirror.  (c) Draw a ray diagram and verify your previous results. (d) What do we conclude from this problem?

Solution: (a) Recall that the focal length $f$ of a spherical mirror is related to the radius of curvature $R$ by the following formula \[f=\frac R2\] Substituting the known value into the above, we get \[f=\frac R2=\frac {40}{2}=20\quad {\rm cm}\]  (b) Using the mirror equation below and solving for unknowns such as image distance $d_i$ or object distance $d_o$, we complete our calculation. \[\frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{f}\] To use mirror equation, we must note the sign rules involved in this equation: 

• $f<0$ for convex mirror and $f>0$ for concave mirror.
• The image distance is always formed behind the convex mirror so $d_i<0$. 
• In a concave mirror, if the image is formed on the same side of the object so $d_i>0$ otherwise $d_i<0$ must be inserted. 

In this problem, the mirror is concave so focal length must be put with a plus sign $f>0$ in the mirror equation. \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{20}&=\frac{1}{d_i}+\frac{1}{6}\\ \\ \Rightarrow \frac{1}{d_i}&=\frac{1}{20}-\frac{1}{6}\\\\&=\frac{6-20}{20\times 6}\\ \\ &=\frac{-14}{20\times 6}\end{align*} Flipping both sides of above relation, we get \[d_i=-\frac{120}{14}=8.57\quad {\rm cm}\] As you can see, we obtained a negative value for image distance for an object in front of a concave mirror. Thus, we conclude that the image must be formed on the opposite side of the mirror, or in other words, behind the mirror. 

This image which is formed on the opposite side of the object is called a virtual image. 

Therefore, the image of the above pencil is located about 9 cm behind the mirror. 

(c) In the following figure, using two rays we obtained the image of the pencil. 

(d) In this problem, the object sits between the focal point and the concave mirror. As the ray diagram above showed, once there was such a case, we obtain a virtual, upright, and magnified image. But how much bigger? see next problem. 

Problem (2): A candle 6 cm tall is placed at a distance of 8 cm in front of a concave mirror whose radius of curvature is 20 cm. Determine the position, size, orientation, and nature (real or virtual) of the image. 

Solution : Known data is object' height $h_o=6\,{\rm cm}$, object distance $d_o=10\,{\rm cm}$, and the radius of curvature of the mirror $R=20\,{\rm cm}$. The focal length is obtained as below \[f=\frac R2=10\quad {\rm cm}\] As you can see, $d_o<f$ that is the object is located between the focal point and the concave mirror. In such cases, we have a virtual, upright, and bigger image (see the previous problem). 

Applying the mirror equation get the position of the image $d_i$ as below \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{10}&=\frac{1}{d_i}+\frac{1}{8}\\ \\ \Rightarrow \frac{1}{d_i}&=\frac{1}{10}-\frac{1}{8}\\\\&=\frac{8-10}{10\times 8}\\ \\ &=\frac{-2}{80}\end{align*} Reversing both sides of the above, we have $d_i=-40\,{\rm cm}$. Because image distance $d_i$ is negative, the image is on the opposite side of the mirror and hence is virtual. 

To quantify how much bigger the image is in the curved mirrors, we use the magnification formula for curved mirrors as below \[M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}\] Where $h_i$, $h_o$ are the image and object heights. \[M=-\frac{d_i}{d_o}=-\frac{-40}{8}=5\] Thus, the object is magnified as big as 5 times. 

Problem (3): A certain concave spherical mirror has a focal length of 10 cm. An object is located 10 cm in front of it. Find the properties of the image. 

Solution : Here, the object is at the focal point. Use the mirror equation to find the image position as below \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{10}&=\frac{1}{d_i}+\frac{1}{10}\\ \\ \frac{1}{d_i}&=0 \\ \\ \Rightarrow d_i&=\infty \end{align*} Thus, the image is formed at infinity!. Using the magnification formula, we can see that it is also infinite. 

From this problem, we conclude that when an object is placed at the focal point, its image is formed at infinity with infinite in size. 

Problem (4): In front of a concave mirror whose radius of curvature is 20 cm we place a candle, 5 cm tall, at a distance of 25 cm away from it. Determine (a) the position,(b) the size, and (c) the nature (real or virtual) and orientation of the image. 

Solution : All we need to find the image properties is to apply the mirror equation and magnification formula. In this problem, the radius of curvature of the concave curved mirror is given which is related to the focal length by the following formula \[f=\frac R2=\frac {20}{2}=10\,{\rm cm}\] The object is placed outside the center of the curvature of the concave mirror because $d_o>2f$. Now, we want to find the properties of the image formed by this object in this position. 

(a) First of all, using the mirror equation we find the image distance from the concave mirror but recall that, according to mirror sign rules, for the concave mirror, the focal length must be inserted as $f>0$. \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{10}&=\frac{1}{d_i}+\frac{1}{25}\\ \\ \frac{1}{d_i}&=\frac{1}{10}-\frac{1}{25}\\\\&=\frac{25-10}{10\times 25}\\\\&=\frac{15}{25} \\ \\ \Rightarrow d_i&=\frac{25}{15}=16.7\,{\rm cm} \end{align*} Because $d_i>0$, the image is in front of the mirror and is real (on the same side of the object). In other words, the image is formed between the focal point and the center of the curvature. 

(b) Magnification in curved mirrors is the ratio of the image size to the object size or the ratio of the image distance to the object distance with a negative. \[M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}\]

From the first expression, we can find the magnification of the mirror as \[M=-\frac{d_i}{d_o}=\frac{16.7}{25}=-0.6\] The image is 0.6 the size of the object. 

The image height is also obtained by the second expression as below \begin{align*} M&=\frac{h_i}{h_o}\\\\ -0.6&=\frac{h_i}{5}\\\\ \Rightarrow  h_i&=(-0.6)(5)=-3\,{\rm cm}\end{align*} The minus sign indicates that the image is inverted. 

(c) Because $d_i>0$, the image is real. Since the magnification is negative $M<0$, the image is inverted. 

All of the above findings can also be obtained by a concave mirror ray diagram as shown in the figure below. 

The notes we learn from this problem are that when an object is placed outside the center of curvature of a concave mirror, then the image is real, inverted, reduced in size, and is formed between focal point $f$ and center of curvature $C$. 

For more practicing, you can also check out the following pages Converging Lens Problems with Solution Diverging Lens Problems with Solution  

Author : Dr. Ali Nemati Date Published: 5/14/2021  

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Ray Optics: Reflection and Mirrors

Reflection and mirrors: problem set overview.

There are 15 ready-to-use problem sets on the topic of Reflection and Mirrors. The problems target your ability to use the law of reflection, to understand the relationship between image distance and object distance for plane mirrors, and to use the mirror equation and magnification ratio to solve problems that relate object and image characteristics to the focal length of concave and convex mirrors. Problems range in difficulty from the very easy and straight-forward to the very difficult and complex.   

• The Law of Reflection

Light rays follow a rather predictable pattern when it comes to reflection off a plane mirror surface. The angle at which the light ray approaches the mirror surface is equal to the angle at which it departs from the mirror. This is known as the law of reflection . In physics, the angles of approach are measured with respect to the normal line to the surface. The normal line is the imaginary line that is perpendicular to the mirror at the point that the light ray strikes the mirror. The angle between the normal line and the approaching or incident ray is known as the angle of incidence . Similarly, the angle between the reflected ray and the same normal line is known as the angle of reflection . According to the law of reflection, the angle of incidence is equal to the angle of reflection. A more detailed and exhaustive discussion of the law of reflection and associated terms can be found at The Physics Classroom Tutorial . Alternatively, you can try our video titled The Law of Reflection .  

Characteristics of Plane Mirror Images

Objects placed in front of plane mirrors will have a corresponding image located behind the mirror. The distance from the image to the mirror is always identical to the distance from the object to the mirror. So if a person stands 2.0 meters in front of the mirror, then the image will be located an identical 2.0 meters behind the mirror. Such an image is a virtual image . When viewing such a virtual image in the mirror, it would seem as though light is coming from a location 2.0 meters behind the mirror. If you were to walk behind the mirror and look at this so-called virtual image location, there would be nothing physical present there. It only seems to the observer as thought light is coming from this location to the eye when viewing the image of the person in the mirror. A more detailed and exhaustive discussion of plane mirror image characteristics can be found at The Physics Classroom Tutorial . Alternatively, you can try our video titled Image Formation for Plane Mirrors .  

Curved Mirror Mathematics

Most of the problems in this unit pertain to curved mirrors - both the concave and the convex varieties.The two equations of relevance for these problems are the mirror equation and the magnification equation. The mirror equation relates the image distance to the object distance and the focal length. The mirror equation is

1/f = 1/d o + 1/d i

The variable d o represents the object distance or the distance between the mirror surface and the object. The variable d i represents the image distance or the distance between the mirror surface and the image. The variable f stands for the focal length of the mirror. In some problems, the focal length is not stated; rather, the radius of curvature of the spherical mirror is stated. The radius of curvature ( R ) is simply twice the focal length value ( R = 2•f ). Like any equation in physics, the mirror equation can be used to solve for an unknown variable through algebraic substitution and rearrangement. Given that there are three quantities present in the mirror equation, two of them must be known in order to solve for the third unknown quantity.

A curved mirror usually causes an image to be either magnified or reduced in size relative to the size of the object. The magnification ratio is a number which expresses the amount of magnification or reduction. The magnification ratio is simply the ratio of the image size to the object size. It is often calculated using the equation

M = h i / h o

In this equation, the variable M represents the magnification, the variable h i represents the image height, and the variable h o represents the object height. It ends up that the ratio of the image to object heights is equivalent to the ratio of the image to the object distance. And so the magnification equation is often written as

M = h i / h o = - d i / d o

The negative sign in the above equation is related to what could be the most problematic aspect of this topic. The variables in these two equations can be either positive or negative. The positive and negative nature is determined by the actual characteristics of the images which are formed and the mirrors which are used in the specific problems. The table below summarizes the so-called sign conventions for the six variables of these two equations.

When reading a problem, it is important to give attention to cues within the problem in order to determine the sign on the given quantity. For example, the distance from the focal point to a mirror is often stated. This is simply a distance value corresponding to the absolute value of the focal length. Whether the focal length is positive or negative is dependent upon whether the mirror is concave or convex. A careful reading of the problem and an understanding of the sign convention on focal length (as stated in the table above) allows one to make the decision about the sign on f. As a second example, some problems describe an image being located a stated distance from a curved mirror. The stated value is simply the absolute value of the image distance. Whether the d i value is positive or negative depends upon whether the image is in front of or behind the mirror. A careful reading of the problem statement along with an understanding of the sign convention for image distance (as stated in the table above) allows one to make the decision about the sign on d i . These types of decisions are critical to your success on thes problems. Making the correct decisions has nothing to do with your mathematical skills; rather, they are tests of your conceptual understandings and your willingness to read a problem carefully and to give attention to details which may be important.

There are a few instances in this problem set in which the mirror equation must be used to solve for an unknown variable but only one of the other two variable values are known. Such problems usually have a statement of the effect: "the image is real and three times the size of the object." Such a statement reveals information about the magnification of the image. Since the ratio of the image to object height is equal to the (negative of the) ratio of the image distance to object distance, we can say that size and height can be treated synonymously. Stating that the image is three times the size of the object is stating that the ratio h i /h o is either +3 or -3. Determining whether h i /h o is +3 or -3 demands an understanding of the sign conventions (as discussed in the above table). The h o value is always positive (for our purposes). The h i value is positive for upright images and negative for inverted images. Since this statement asserts that the image is real (and thus inverted), a -3 value must be assigned to the h i /h o ratio. Since hi/ho is equal to -d i /d o , the -3 value can be equated with -d i /d o . This stream of logic allows one to write an expression for d i in terms of d o . This expression for d i in terms of d o can be substituted into the mirror equation in order to transform it into a single equation with a single unknown. Customary algebraic manipulations can then be performed in order to solve for d i or for d o .

The table below summarizes the process of transforming a verbal statement into a mathematical equation which ultimately is used to substitute into the mirror equation.

Habits of an Effective Problem-Solver

An effective problem solver by habit approaches a physics problem in a manner that reflects a collection of disciplined habits. While not every effective problem solver employs the same approach, they all have habits which they share in common. These habits are described briefly here. An effective problem-solver...

• ...reads the problem carefully and develops a mental picture of the physical situation. If needed, they sketch a simple diagram of the physical situation to help visualize it.
• ...identifies and records the known and unknown quantities in an organized manner. Equates given values to the symbols used to represent the corresponding quantity - e.g., d o = 24.2 cm; d i = 16.8 cm; f = ???.
• ...plots a strategy for solving for the unknown quantity; the strategy will typically center around the use of physics equations and be heavily dependent upon an understanding of physics principles.
• ...identifies the appropriate formula(s) to use, often times writing them down. Where needed, they perform the needed conversion of quantities into the proper unit.
• ...performs substitutions and algebraic manipulations in order to solve for the unknown quantity.

Read more...  

Additional Readings/Study Aids:

The following pages from The Physics Classroom Tutorial may serve to be useful in assisting you in the understanding of the concepts and mathematics associated with these problems.

• Image Characteristics for Plane Mirrors
• What Portion of a Mirror is Required to View an Image?
• Image Characteristics for Concave Mirrors
• The Mirror Equation - Concave Mirrors
• Image Characteristics for Convex Mirrors
• The Mirror Equation - Convex Mirrors

Watch a Video

We have developed and continue to develop Video Tutorials on introductory physics topics. You can find these videos on our YouTube channel . We have an entire Playlist on the topic of Reflection and Mirrors .