## Section47.7Thin Lens Equation

We have studied locating image by lenses using ray tracing method. There is also an algebraic method that is very simple to apply and much more convenient than the ray tracing method, especially if you have more than one lens in your system.

Assume we have a thin lens so that all distances are measured from the center of the lens and all heights are measured from the optical axis. Let $p$ denote the object distance, $q$ the image distance, and $f$ the focal length. Then, we will show below that the following equation and sign convention holds for both types of lenses.

\begin{equation} \frac{1}{p} + \frac{1}{q} = \frac{1}{f},\tag{47.7.1} \end{equation}

with the following sign convention assuming that physical rays progress from left to right in the system.

\begin{align*} \amp p \gt 0 \text{ if object to the left of lens,} \\ \amp p \lt 0 \text{ if object to the right of lens,} \\ \amp q \gt 0 \text{ if object to the right of lens (i.e., real,} \\ \amp q \lt 0 \text{ if object to the left of lens (i.e., virtual),} \\ \amp f \gt 0 \text{ for convex lens,} \\ \amp p \lt 0 \text{ for concave lens.} \end{align*}

Let $h_o$ is the height of the object and $h_i$ be the height of the image. They will be postive if above axis and negative if below axis. Then, as before, we define the lateral magnification, often simply called magnification, of the image $m_T$ by

\begin{equation*} m_T = \frac{h_i}{h_o}. \end{equation*}

By analyzing the geometry of image formation one can show that this is also equal to $-q/p\text{.}$ Thus we get a very useful relation.

\begin{equation} m_T = \frac{h_i}{h_o} = -\frac{q}{p}.\tag{47.7.2} \end{equation}

Find the location, orientation and magnification factor of the image in each of the following positions of an object of height 3 cm in front of a convex lens of focal length 10 cm. (a) $p = 50\text{ cm}\text{,}$ (b) $p = 5\text{ cm}\text{,}$ (c) $p = 20\text{ cm}\text{,}$ (d) $p=10\text{ cm}\text{.}$

Hint

Use lens and magnification equations.

(a) $q = 12.5\text{ cm}\text{,}$ $m_T=-\frac{1}{4}\text{,}$ inverted, $0.75\ \text{cm}\text{.}$ (b) $q = -10\text{ cm}\text{,}$ $m_T=+2\text{,}$ same as object, $6\ \text{cm}\text{.}$ (c) $q = 20\text{ cm}\text{,}$ $m_T=-1\text{,}$ inverted, $3\ \text{cm}\text{.}$ (d) $q=\infty\text{,}$ $m_T=-\infty\text{.}$

Solution 1 (a)

(a) In this part we have $p = 50\text{ cm}\text{,}$ and $f = +10\text{ cm}\text{,}$ and we are looking for $q\text{.}$ Therefore,

\begin{align*} \amp \frac{1}{50\ \text{cm}} + \frac{1}{q} = \frac{1}{10\ \text{cm}}\\ \amp \frac{1}{q} = \frac{1}{10\ \text{cm}} - \frac{1}{50\ \text{cm}} = \frac{4}{50\ \text{cm}}. \end{align*}

Inverting this, we get $q = 12.5\text{ cm}\text{.}$ Since $q$ is positive, the image will form on the right of the lens. This will give us a real image. The magnification of the image can be obtained from $p$ and $q\text{.}$

\begin{equation*} m_T = -\frac{q}{p} = -\frac{12.5\ \text{cm}}{50\ \text{cm}} = -\frac{1}{4}. \end{equation*}

The negative magnification means that the image would be inverted. Furthemore, since $|m_T|\lt 1\text{,}$ the image would be shorter than the object. The size of the image is given by

\begin{equation*} |h_i| = |m_T| \times h_o = 0.25 \times 3\ \text{cm} = 0.75\ \text{cm}. \end{equation*}
Solution 2 (b)

(b) Now, $p = 5\text{ cm}\text{,}$ $f = +10\text{ cm}\text{,}$ and $q = ?$ We proceed similarly to (a).

\begin{align*} \amp \frac{1}{5\ \text{cm}} + \frac{1}{q} = \frac{1}{10\ \text{cm}}\\ \amp \frac{1}{q} = \frac{1}{10\ \text{cm}} - \frac{1}{5\ \text{cm}} = - \frac{1}{10\ \text{cm}}. \end{align*}

Therefore, $q = -10\text{ cm}\text{,}$ which is negative, meaning the image is on the left side of the lens, the same side as the object. The magnification of this image

\begin{equation*} m_T = -\frac{q}{p} = -\frac{-10\ \text{cm}}{5\ \text{cm}} = +2. \end{equation*}

The positive magnification means that the image would be upright, the same orientation as the object. Since $|m_T|>1\text{,}$ the image would be larger than the object, i.e. the image will be magnified. The size of the image is given by

\begin{equation*} |h_i| = |m_T| \times h_o = 2 \times 3\ \text{cm} = 6\ \text{cm}. \end{equation*}
Solution 3 (c)

(c) In this case, $p = 20\text{ cm}\text{,}$ $f = +10\text{ cm}\text{,}$ $q = ?$

\begin{align*} \amp \frac{1}{20\ \text{cm}} + \frac{1}{q} = \frac{1}{10\ \text{cm}}\\ \amp \frac{1}{q} = \frac{1}{10\ \text{cm}} - \frac{1}{20\ \text{cm}} = \frac{1}{20\ \text{cm}}. \end{align*}

Therefore, $q = 20\text{ cm}\text{,}$ which is positive, meaning the image is on the right side of the lens.

\begin{equation*} \text{Magnification,}\ \ m_T = -\frac{q}{p} = -\frac{20\ \text{cm}}{20\ \text{cm}} = -1. \end{equation*}

The negative magnification means that the image would be inverted. Since $|m_T|=1\text{,}$ the image would be of the same size as the object. The size of the image would be given by

\begin{equation*} |h_i| = |m| \times h_o = 1 \times 3\ \text{cm} = 3\ \text{cm}. \end{equation*}
Solution 4 (d)

Here, we are placing the object at the first focal point of the lens. We expect the outgoing rays from the lens will be parallel in all directions. This will mean the image will be very far away, i.e., at infinity. Let's see how the algebra works out.

\begin{equation*} \frac{1}{10\text{ cm}} + \frac{1}{q} = \frac{1}{10\text{ cm}} \rightarrow \frac{1}{q} = 0. \end{equation*}

Therefore, $q=\infty\text{.}$ From $m_T=-q/p\text{,}$ we get $m_T=-\infty\text{.}$ That means if you are a little bit inside a focal length, image will be virtual, have same vertical orientation, and will be magnified considerably. however, if you were a little but further than the focal point, the image will be real, far away and inverted in orientation. So, the negative sign in $m_T=-\infty$ is a little bit deceptive since if you are right at the focal point, the image will be at infinty, which means there will be no image at all.

Find the location, orientation and magnification factor of the image in each of the following positions of an object of height 3 cm in front of a concave lens of focal length 10 cm. (a) $p = 50\text{ cm}\text{,}$ (b) $p = 5\text{ cm}\text{,}$ (c) $p = 20\text{ cm}\text{,}$ (d) $p = 10\text{ cm}\text{.}$

Hint

Use lens and magnification equations.

(a) $q = -8.33\text{ cm}\text{,}$ $m_T=+\frac{1}{6}\text{,}$ same vertical orientation, $0.5\ \text{cm}\text{.}$ (b) $q = -3.33\text{ cm}\text{,}$ $m_T=+\frac{2}{3}\text{,}$ same vertical orientation, $2\ \text{cm}\text{.}$ (c) $q = -6.67\text{ cm}\text{,}$ $m_T=+\frac{1}{3}\text{,}$ same vertical orientation, $1\ \text{cm}\text{.}$ (d) $q = -5\text{ cm}\text{,}$ $m_T=+\frac{1}{2}\text{,}$ same vertical orientation, $1.5\ \text{cm}\text{.}$

Solution 1 (a)

(a) In this part we have $p = 50\text{ cm}\text{,}$ and $f = -10\text{ cm}\text{,}$ and we are looking for $q\text{.}$ Therefore,

\begin{align*} \amp \frac{1}{50\ \textrm{cm}} + \frac{1}{q} = \frac{1}{-10\ \textrm{cm}}\\ \amp \frac{1}{q} = - \frac{1}{10\ \textrm{cm}} - \frac{1}{50\ \textrm{cm}} = - \frac{6}{50\ \textrm{cm}}. \end{align*}

Inverting this, we get $q = -8.33\text{ cm}\text{.}$ Since $q$ is negative, the image will form on the same side of the lens as the object. This will give us a virtual image. The magnification of the image can be obtained from $p$ and $q\text{.}$

\begin{equation*} m_T = -\frac{q}{p} = -\frac{-[50\ \textrm{cm}/6] }{50\ \textrm{cm}} = +\frac{1}{6}. \end{equation*}

The positive magnification means that the image would have same vertical orientation as the object. Furthemore, since $|m_T|\lt 1\text{,}$ the image would be shorter than the object. The size of the image is given by

\begin{equation*} |h_i| = |m_T| \times h_o = \frac{1}{6} \times 3\ \textrm{cm} = 0.5\ \textrm{cm}. \end{equation*}
Solution 2 (b)

(b) Now, $p = 5\text{ cm}\text{,}$ $f = -10\text{ cm}\text{,}$ and $q = ?$ We proceed similarly to (a).

\begin{align*} \amp \frac{1}{5 \ \textrm{cm}} + \frac{1}{q} = \frac{1}{-10\ \textrm{cm}}\\ \amp \frac{1}{q} = - \frac{1}{10\ \textrm{cm}} - \frac{1}{5 \ \textrm{cm}} = - \frac{3}{10\ \textrm{cm}}. \end{align*}

Therefore, $q = -3.33\text{ cm}\text{,}$ which is negative, meaning the image is on the left side of the lens, the same side as the object. The magnification of this image

\begin{equation*} m_T = -\frac{q}{p}= -\frac{-[10\ \textrm{cm}/3] }{5 \ \textrm{cm}} = +\frac{2}{3}. \end{equation*}

The positive magnification means that the image would be upright, the same orientation as the object. Since $|m_T| \lt 1\text{,}$ the image would be shorter than the object, i.e. the image will be diminished in size. The size of the image is given by

\begin{equation*} |h_i| = |m_T| \times h_o = \frac{2}{3} \times 3\ \textrm{cm} = 2\ \textrm{cm}. \end{equation*}
Solution 3 (c)

(c) In this case, $p = 20\text{ cm}\text{,}$ $f = -10\text{ cm}\text{,}$ $q = ?$

\begin{align*} \amp \frac{1}{20 \ \textrm{cm}} + \frac{1}{q} = \frac{1}{-10\ \textrm{cm}}\\ \amp \frac{1}{q} = - \frac{1}{10\ \textrm{cm}} - \frac{1}{20 \ \textrm{cm}} = - \frac{3}{20\ \textrm{cm}}. \end{align*}

Therefore, $q = -6.67\text{ cm}\text{.}$ The negative image distance means the image is on the left side of the lens, i.e. on the same side as the object.

\begin{equation*} \text{Magnification,}\ \ m_T = -\frac{q}{p} = -\frac{-[20\ \textrm{cm}/3] }{20 \ \textrm{cm}} = +\frac{1}{3}. \end{equation*}

The positive magnification means that the image would have the same vertical orientation as the object. The size of the image would be given by

\begin{equation*} |h_i| = |m| \times h_o = \frac{1}{3} \times 3\ \textrm{cm} = 1\ \textrm{cm}. \end{equation*}
Solution 4 (d)

(c) In this case, $p = 10\text{ cm}\text{,}$ $f = -10\text{ cm}\text{,}$ $q = ?$

\begin{align*} \amp \frac{1}{10 \ \textrm{cm}} + \frac{1}{q} = \frac{1}{-10\ \textrm{cm}}\\ \amp \frac{1}{q} = - \frac{1}{10\ \textrm{cm}} - \frac{1}{10 \ \textrm{cm}} = - \frac{1}{5\ \textrm{cm}}. \end{align*}

Therefore, $q = -5\text{ cm}\text{.}$ The negative image distance means the image is on the left side of the lens, i.e. on the same side as the object.

\begin{equation*} \text{Magnification,}\ \ m_T = -\frac{q}{p} = -\frac{-5\text{ cm} }{10 \ \textrm{cm}} = +\frac{1}{2}. \end{equation*}

The positive magnification means that the image would have the same vertical orientation as the object. The size of the image would be given by

\begin{equation*} |h_i| = |m| \times h_o = \frac{1}{2} \times 3\ \textrm{cm} = 1.5\ \textrm{cm}. \end{equation*}

Note that unlike the case with the convex lens, placing the object at the first focal point did not result in image at infinity!