## Section47.3Image Formation by Reflection - Algebraic Methods

We can also find the location, orientation, and size of image algebraically rather than by drawing rays. First I will work out formulas that connect object distance, image distance, and radius of curvature of concave mirror. Then we will work out a similar formula for convex mirror. We will see that we can write one relation that will cover both cases if we follow a sign convention.

### Subsection47.3.1Concave Mirror Equation

Consider an object OP located at a distance from a concave mirror so that a real image forms on a screen in front of the mirror as shown in Figure 47.3.1. Although, I will work out a formula for the real image, the same formula will also work for a virtual image with appropriate sign convenstion.

We will refer all horizontal distances from vertex V and all vertical distance from the optical axis OIV. Let $R$ be the radius of the mirror. First notice that upon comparing the angles in triangles $\triangle\text{OPV}$ and $\triangle\text{IQV}\text{,}$ we find that these trinagles form a pair of similar triangles. Therefore, from plane geometry we can state that their corresponding sides must be proportional.

$$\frac{\text{VO}}{ \text{VI} } = \frac{ \text{OP} }{ \text{IQ} }.\label{eq-reflection-1}\tag{47.3.1}$$

Similarly, try to establish that triangles $\triangle\text{OPF}$ and $\triangle\text{SFY}$ are similar. That will give us the following relation.

$$\frac{\text{FO}}{ \text{SY} } = \frac{ \text{OP} }{ \text{FS} }.\label{eq-reflection-2}\tag{47.3.2}$$

By construction, we also have

$$\text{IQ} = \text{FS}. \label{eq-reflection-3}\tag{47.3.3}$$

In our calculations, we will assume that the rays make small angle with the optical axis. This is called paraxial approximation. Paraxial rays make small angle, e.g., $\theta = \angle\text{PVQ}$ at the vertex, therefore we can make the follpwing approximations

\begin{align*} \amp \sin\theta \approx \tan\theta \approx \theta.\\ \amp \cos\theta \approx 1. \end{align*}

Using these approximations we will also have the following important relation.

$$\text{SY} \approx \text{VF}. \label{eq-reflection-4}\tag{47.3.4}$$

From Eqs. (47.3.1) to (47.3.4), we can deduce the following

$$\frac{\text{VO} }{ \text{VI} } = \frac{ \text{FO} }{ \text{VF} }. \label{eq-reflec-concave-5}\tag{47.3.5}$$

Since $\text{FO} = \text{VO} - \text{VF}\text{,}$ we get the following after using this and then dividing both sides by $\text{VO} \text{,}$ and then rearranging terms.

$$\frac{ 1 }{ \text{VO} } + \frac{ 1 }{ \text{VI} } = \frac{ 1 }{ \text{VF} }. \label{eq-concave-mirror-equation}\tag{47.3.6}$$

We now define symbols $p\text{,}$ $q\text{,}$ and $f$ to represent the geometric distances as follows.

\begin{align*} \amp p = \text{VO} = \text{object distance}, \\ \amp q = \text{VI} = \text{image distance}, \\ \amp f = \text{VF} = \text{image distance} = \frac{R}{2}. \end{align*}

Note that the distance $\text{VF}$ is equal to half the radius of the curvature of the mirror. That is, focal length of a concave mirror is

$$f = \frac{R}{2}. \label{concave-mirror-focal-length}\tag{47.3.7}$$

Using these symbols, Eq. (47.3.6) takes the following familiar form, which we will refer to as concave mirror equation.

$$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}, \label{eq-concave-mirror-equation-2}\tag{47.3.8}$$

### Subsection47.3.2Size and Orientation of Image

Refer to Figure 47.3.1 for the drawing of image formation in a concave mirror. To study the size and orientation of the image compared to the object we can study the heights of the object and image points with respect to the symmetry axis. Let height above the axxis be positive and that below the axis be negative. Let

\begin{align*} \amp h_o = \text{OP} = \text{object height}, \\ \amp h_i = -\text{IQ} = \text{object height}, \end{align*}

where note the negative sign to make $h_i$ negative by multiplying the positive length $\text{IQ}$ to negative height corresponding to our drawing in Figure 47.3.1. Now, from similar triangles $\triangle\text{OPV}$ and $\triangle\text{IQV}$ we find

\begin{equation*} \frac{ \text{IQ} }{ \text{OP} } = \frac{ \text{IV} }{ \text{OV} } \end{equation*}

Therefore,

$$\frac{h_i}{h_o} = -\frac{q}{p}.\tag{47.3.9}$$

The ration $h_i/h_o$ gives both the relative size and relative orientation of the image with respect to the object. It is called magnifcation, or more accurately transverse magnification since the heights are perpendicular to the axis and not along the axis, of the image and denoted by $m_T\text{,}$ or sometimes simply by $m\text{.}$

$$m_T = \frac{h_i}{h_o} = -\frac{q}{p}.\tag{47.3.10}$$

Thus, if $m_T$ is positive, then image has the same orientation as the object and if it is negative the image is inverted. If $|m_T| \gt 1$ ,then image is larger than the object, i.e., magnified, and if $|m_T| \lt 1\text{,}$ then image is smaller than the object, i.e., shrunk.

### Subsection47.3.3Virtual Image by a Concave Mirror

When we use the concave mirror equation (47.3.6) for an object that is within a focal length, i.e., $p \lt R/2 \text{,}$ then we find that $q$ is negative.

\begin{equation*} \frac{1}{q} = \frac{1}{R/2} - \frac{1}{p} = \frac{p - R/2}{ pR/2} \lt 0\ \text{ if }\ p\lt R/2. \end{equation*}

The negative value of $q$ would correspond to distance of the virtual image behind the mirror at a distance $|q|$ from the vertex.

An object of height $4.5\text{ cm}$ is placed $20\text{ cm}$ from a concave mirror of radius of curvature $10\text{ cm}\text{.}$

(a) Find the location of the image.

(b) Find transverse magnification, the size, and orientation of the image.

Hint

(a) Use the concave mirror equation. (b) Use $h_o/h_i = -q/p\text{.}$

(a) $\frac{20}{3}\text{ cm}\text{,}$ (b) $-\frac{1}{3}\text{,}$ inverted, $1.5\text{ cm}\text{.}$

Solution 1 (a)

We use the Concave mirror equation to solve for $q\text{.}$

\begin{align*} \frac{1}{q} \amp = \frac{1}{f} - \frac{1}{p}\\ \frac{1}{q} \amp = \frac{1}{10/2} - \frac{1}{20} = \frac{3}{20} \end{align*}

Inverting and putting the units back in, we get

\begin{equation*} q = \frac{20}{3}\text{ cm}. \end{equation*}
Solution 2 (b)

The transverse magnification will be

\begin{equation*} m_T = \frac{h_i}{h_o} = - \frac{q}{p} = -\frac{20}{3} \times \frac{1}{20} = -\frac{1}{3}. \end{equation*}

The sign here tells us that the image is inverted compared to the object. The size of the image will be

\begin{equation*} |h_i| = h_o |m_T| = 4.5\text{ cm }\times \frac{1}{3} = 1.5\text{ cm}. \end{equation*}

An object of height $4.5\text{ cm}$ is placed $2\text{ cm}$ from a concave mirror of radius of curvature $10\text{ cm}\text{.}$

(a) Find the location of the image.

(b) Find transverse magnification, the size, and orientation of the image.

Hint

(a) Use the concave mirror equation. (b) Use $h_o/h_i = -q/p\text{.}$

(a) $-\frac{10}{3}\text{ cm}\text{,}$ (b) $\frac{5}{3}\text{,}$ same orientation, $7.5\text{ cm}\text{.}$

Solution 1 (a)

We use the Concave mirror equation to solve for $q\text{.}$

\begin{align*} \frac{1}{q} \amp = \frac{1}{f} - \frac{1}{p}\\ \frac{1}{q} \amp = \frac{1}{10/2} - \frac{1}{2} = -\frac{3}{10} \end{align*}

Inverting and putting the units back in, we get

\begin{equation*} q = -\frac{10}{3}\text{ cm}. \end{equation*}

In the case of real image forming in front of the mirror, $q$ was positive. Now, we get a negative $q\text{.}$ The image here will form behind the mirror at a distance $\frac{10}{3}\text{ cm}\text{.}$

Solution 2 (b)

The transverse magnification will be

\begin{equation*} m_T = \frac{h_i}{h_o} = - \frac{q}{p} = - \left(-\frac{10}{3} \right) \times \frac{1}{2} = \frac{5}{3}. \end{equation*}

The positive sign here tells us that the image has same vertical orientation as the object. Since $m_T \gt 1\text{,}$ the image is larger than the object. The size of the image will be

\begin{equation*} |h_i| = h_o |m_T| = 4.5\text{ cm }\times \frac{5}{3} = 7.5\text{ cm}. \end{equation*}

Where should a $3\text{-cm}$ tall object be placed in front of a concave mirror of radius $20\text{ cm}$ so that its image is real and $2\text{ cm}$ tall?

Hint

Set up two equations in $p$ and $q\text{,}$ and then solve them simultaneously.

$25\text{ cm}\text{.}$

Solution

Real image from a concave mirror is inverted. Therefore, we take $h_i = -2\text{ cm}$ and $h_o = 3\text{ cm}\text{.}$ This gives

\begin{equation*} -\frac{q}{p} = \frac{h_i}{h_o} = -\frac{2}{3}. \end{equation*}

Therefore

$$3 q = 2 p.\label{eq-Placing-an-Object-to-Obtain-Real-Image-of-Given-Size-from-a-Concave-Mirror-1}\tag{47.3.11}$$

Using the radius of curvature of the mirror we set up the second equation.

$$\frac{1}{p} + \frac{1}{q} = \frac{2}{R} = \frac{2}{20} = \frac{1}{10}.\label{eq-Placing-an-Object-to-Obtain-Real-Image-of-Given-Size-from-a-Concave-Mirror-2}\tag{47.3.12}$$

Using Eq. (47.3.11), we eliminate $q$ from this equation to get (I divided throughout by 3 first.)

\begin{equation*} \frac{1}{3p} + \frac{1}{2p} = \frac{1}{30} \end{equation*}

This can be solved easily to get $p = 25\text{ cm}\text{.}$ First factor out $1/p\text{,}$ then the factor multiplying $1/p$ will be $5/6\text{,}$ which can be sent to the other side and simplified.

### Subsection47.3.4Convex Mirror Equation

Consider an object OP located at a distance from a convex mirror as shown in Figure 47.3.5.

As for the concave mirror, we will refer all horizontal distances from vertex V and all vertical distance from the optical axis OIV. First notice that upon comparing the angles in triangles $\triangle\text{OPF}$ and $\triangle\text{YVF}\text{,}$ we find that these trinagles form a pair of similar triangles. Therefore, from plane geometry we can state that their corresponding sides must be proportional.

$$\frac{ \text{OF} }{ \text{VF} } = \frac{\text{OP}}{ \text{VY} }.\label{eq-convex-mirror-equation-1}\tag{47.3.13}$$

Similarly, try to establish that triangles $\triangle\text{OPV}$ and $\triangle\text{IQV}$ are similar. That will give us the following relation.

$$\frac{ \text{VO} }{ \text{VI} } = \frac{\text{OP}}{ \text{IQ} }.\label{eq-convex-mirror-equation-2}\tag{47.3.14}$$

In paraxial approximation, $\text{VY} \approx \text{IQ}\text{.}$ Therefore, we will have

\begin{equation*} \frac{ \text{OF} }{ \text{VF} } = \frac{ \text{VO} }{ \text{VI} }. \end{equation*}

Now, from the figure, we notice that

\begin{equation*} \text{OF} = \text{OV} + \text{VF} = \text{VO} + \text{VF}. \end{equation*}

Therefore,

\begin{equation*} \frac{ \text{VO} + \text{VF} }{ \text{VF} } = \frac{ \text{VO} }{ \text{VI} }. \end{equation*}

Simplifying this equation we get

\begin{equation*} \frac{ 1 }{ \text{VO} } + \frac{ 1 }{ \text{VF} } = \frac{ 1 }{ \text{VI} }. \end{equation*}

Let's rearrange this equation so that object and image distance are on one side of the equation.

$$\frac{ 1 }{ \text{VO} } - \frac{ 1 }{ \text{VI} } = -\frac{ 1 }{ \text{VF} }.\label{eq-convex-mirror-equation-3}\tag{47.3.15}$$

Let us introduce the following symbols for our discussion.

\begin{align*} \amp p = \text{VO} = \text{object distance}, \\ \amp q = \text{VI} = \text{image distance}, \end{align*}

From ray drawing, we know that the focal point F is at a distance of half the radius $R$ of the mirror from the vertex V.

$$VF = \frac{R}{2}.\tag{47.3.16}$$

Then, Eq. (47.3.15) becomes the following equation. We will refer to this equation the covex mirror equation.

$$\frac{ 1 }{ p } - \frac{ 1 }{ q } = -\frac{ 1 }{ R/2 }.\label{eq-convex-mirror-equation-using-radius}\tag{47.3.17}$$

#### Subsubsection47.3.4.1Sign Convention

Equation (47.3.17) for the convex mirror differs from the concave equation, Eq. (47.3.8) in signs of image distance and focal length. We can write both the concave mirror equation and the convex mirror equation in the same form but with a sign convention.

Let $q'=-q\text{,}$ and $f'=-R/2\text{.}$ Then,

$$\frac{ 1 }{ p } + \frac{ 1 }{ q' } = \frac{ 1 }{ f' },\label{eq-convex-mirror-equation-using-focal-length}\tag{47.3.18}$$

where $q'=-|q'| \lt 0$ and $f'=-R/2 \lt 0$ for the convex mirror. We can use this same equation or the cncave mirror as well if we make sure that $f' = R/2 \gt 0$ but $q' \gt 0$ if the image is real and $q'=-|q'| \lt 0$ if the image is virtual.

#### Subsubsection47.3.4.2Size and Orientation of Image by Convex Mirror

I will not go through the geometric arguments, but suffice to say that, the algebraic relation here is same as that for the concave mirror. Let $h_o$ be the object height from the axis, and $h_i$ the image height. Let $p$ the object distance and $q$ the image distance, which will be negative here. Then, we have following for the transverse magnification.

$$m_T = \frac{h_i}{h_o} = -\frac{q}{p}.\tag{47.3.19}$$

An object of height $4.5\text{ cm}$ is placed $20\text{ cm}$ from a convex mirror of radius of curvature $10\text{ cm}\text{.}$

(a) Find the location of the image.

(b) Find transverse magnification, the size, and orientation of the image.

Hint

(a) Use the convex mirror equation. (b) Use $h_o/h_i = -q/p\text{.}$

(a) $5\text{ cm}$ behind the mirror, (b) $\frac{1}{4}\text{,}$ same orientation as object, $1.1\text{ cm}\text{.}$

Solution 1 (a)

We use the convex mirror equation to solve for $q\text{.}$ We note that $f = -R/2 = -10/2 = -5\text{ cm}$ here.

\begin{align*} \frac{1}{q} \amp = \frac{1}{f} - \frac{1}{p}\\ \frac{1}{q} \amp = -\frac{1}{5} - \frac{1}{20} = -\frac{1}{5} \end{align*}

Inverting and putting the units back in, we get

\begin{equation*} q = -5\text{ cm}. \end{equation*}

The negative sign is expected for convex mirror since the image is behind the mirror. The image will be $5\text{ cm}$ from the vertex behind the mirror.

Solution 2 (b)

The transverse magnification will be

\begin{equation*} m_T = \frac{h_i}{h_o} = - \frac{q}{p} = -\frac{-5}{20} = +\frac{1}{4}. \end{equation*}

The positive $m_T$ means that the image has the same orientation as the object. The size of the image will be

\begin{equation*} |h_i| = h_o |m_T| = 4.5\text{ cm }\times \frac{1}{4} = 1.1\text{ cm}. \end{equation*}