Lens Equations

Some textbooks use the variable |p| to represent the object-mirror distance |(d_o)| and the variable |q| to represent the image-mirror distance |(d_i)|.

The formulas to be used in the lenses are as follows: ||\begin{gather} m=\dfrac {h_{i}}{h_{o}} = \dfrac {-d_{i}}{d_{o}} = \dfrac {-l_{f}}{l_{o}} = \dfrac {-l_{i}}{l_{f}}\\[3pt] \dfrac {1}{d_{o}} + \dfrac {1}{d_{i}} = \dfrac {1}{l_{f}}\\[3pt] {d_{i}} = {l_{i}} + {l_{f}}\\[3pt] {d_{o}} = {l_{o}} + {l_{f}}\\[3pt]{l_{i}} \times {l_{o}} = {l_{f}}^2 \end{gather}||

The formulas in the lenses do not use precise units. All the measurements represent length measurements that can be measured in metres or centimetres. Although there is a wide choice of units, it is important to convert all measurements so that the same units of measurement are used.

A stamp measuring |2\ \text {cm}| through a magnifying glass (formed by a converging lens) which has a focal length of |15\ \text {cm}.| The stamp is placed at |9\ \text {cm}| from the magnifying glass. How far away from the magnifying glass is the image? How big is the image?

First we need to identify our variables.||\begin{align}h_{o} &= 2\ \text{cm}\\ d_{o} &= 9\ \text{cm}\\ l_{f} &= +15\ \text {cm}\end{align}||

The focal length is positive since the lens is convergent.

To find the position of the image, an equation allows us to identify the desired variable.
||\begin{align} \frac {1}{d_{o}} + \frac {1}{d_{i}} = \frac {1}{l_{f}} \quad \Rightarrow \quad \frac {1}{d_{i}} &= \frac {1}{l_{f}} - \frac {1}{d_{o}}  \\ \\
\frac {1}{d_{i}}&= \frac {1}{15 \space \text {cm}} \frac {1}{9 \space \text {cm}} \\ \\
\frac {1}{d_{i}}&= \frac {-2}{45} \\\\
d_i&= -22.5 \: \text {cm}\end{align}||
The negative sign of the value of |d_{i}| tells us that the image is virtual. This is to be expected, as the object is located between the focus and the optical centre of the lens.

To find the height of the image, the proportions of the magnification will be used.||\begin{align} \dfrac{h_{i}}{h_{o}} =\dfrac {-d_{i}}{d_{o}} \quad \Rightarrow \quad h_{i} &= \dfrac{h_{o} \times -d_{i}}{d_{o}} \\[3pt] h_i&= \dfrac{2\ \text{cm}\times -(-22.5)\ \text{cm}}{9\ \text{cm}\times h_i}\\[3pt] &= 5\ \text{cm} \end{align}||

Since the sign of |h_i| is positive, the image is straight.

A divergent lens with a focal length of |15 \: \text {cm}| produces an image 3 times smaller than the object. How far away from the lens did we have to place the object?

First we need to identify our variables.||\begin{align}l_{f} &= -15 \: \text {cm} &m &= \dfrac{1}{3}\end{align}||

The negative sign is used for the focal length, as the lens is divergent.

The values of |d_{i}| and |d_{o}| are unknown. However, by using the magnification, it is possible to know the relationship between these two variables. ||\begin{align} m= \dfrac {-d_{i}}{d_{o}} \quad \Rightarrow \quad \dfrac {1}{3} &= \dfrac {-d_{i}}{d_{o}} \\[3pt] \dfrac {d_{o}}{3} &={-d_{i}} \\[3pt] \dfrac {-d_{o}}{3} &={d_{i}} \end{align}||

These variables can then be substituted to find the value of |d_{o}.| ||\begin{align} \frac {1}{d_{o}} + \frac {1}{d_{i}} = \frac {1}{l_{f}} \quad \Rightarrow \quad \frac {1}{d_{o}} + \frac {1}{\frac {-d_{o}}{3}} &= \frac {1}{-15 \: \text {cm}} \\[4pt] \frac {1}{d_{o}} + \frac {-3}{d_{o}} &= \frac {1}{-15 \: \text {cm}} \\[4pt] \frac {-2}{d_{o}} &= \frac {1}{-15 \: \text {cm}} \\[4pt] d_{o} &= 30 \: \text {cm} \end{align}||

The object must therefore be placed at |30\ \text {cm}| from the lens to obtain an image three times smaller.

The magnification |(m)| of an object is the ratio of the height of the image |(h_i)| to the height of the object |(h_o)|.

Some textbooks use the variable |(G)| to represent the magnification |(m)|.

To calculate the magnification, the formula to use is: ||m=\dfrac {h_{i}}{h_{o}} = \dfrac {-d_{i}}{d_{o}} = \dfrac {-l_{f}}{l_{o}} = \dfrac {-l_{i}}{l_{f}}||