Ray diagrams, the thin lens equation and magnification

In a nutshell

​Lenses change the direction of light rays to converge or diverge. They are used to form real or virtual images. The result of light passing through a lens can be represented using a ray diagram. ​

Equations

Variable definitions

Rules for drawing ray diagrams

There are three rules for rays refracted by a lens. They describe how light rays behave after they pass through the lens.

Rules for rays

Tip: Only two of these rays need to drawn to find the point at which they cross. The easiest rays to draw are 1 and 3 but feel free to use 2 if to double check!

Describing an image

​​PROCEDURE

Ray diagrams

Ray diagrams can be constructed for light from an object passing through a converging or diverging lens.

CONVErging RAY DIAGRAM

diverging RAY DIAGRAM

There are a set of steps to follow for drawing a ray diagram for each lens. 

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Converging lens

PROCEDURE

1.	Draw a ray of light going from the top of the object to the lens, parallel to the axis. The first rule for rays says that this will pass through the principal focus (F). Draw the ray going through the principal focus.	​
2.	Draw a ray going from the top of the object passing through the middle of the lens. The third rule for rays says that this should continue on the same path. Draw the ray continuing on this path.
3.	Mark the point where the two lines meet. This is the top of the image.
4.	Repeat the past 3 steps for the bottom of the image. If the bottom of the object is on the axis, the bottom of the image will also be on the axis.

Note: F is used to label the principal focus on both sides. 2F means twice the focal length away from the lens. The axis can be thought of as like the x-axis of a graph.

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The distance the object is placed from the lens will affect the type of image created. If the object is placed closer to the lens than the principal focus, a virtual image will be formed.

An image can be determined as virtual as the rays on the refracted side of the lens won't meet at a point - they diverge. To construct a virtual image, continue the path of the rays (as dotted lines) on the object side of the lens. They should meet at a point.

Diverging lens

PROCEDURE

1.	Draw a ray of light going from the top of the object to the lens, parallel to the axis. The first rule for rays says that this will appear to have come from the principal focus (F). Draw a dotted line from the left-hand side's principal focus to the ray. Then draw a normal line continuing on this path.	​
2.	Draw a ray going from the top of the object passing through the middle of the lens. The third rule for rays says that this should continue on the same path. Draw the ray continuing on this path.
3.	The two rays won't meet on the other side of the lens. Instead mark where the two lines intersect (meet) on the object's side.
4.	Repeat the past 3 steps for the bottom of the image. If the bottom of the object is on the axis, the bottom of the image will also be on the axis.

Note: Diverging lenses always create virtual images, no matter the distance from the lens.

Magnification

The magnification is measured as the relative distance of the image compared to the distance of the object.

​$$m = \frac{v}{u}$$​​

Note: You can also find the magnification by dividing the image height over the object height!

Thin lens equation

The following equation can be applied to all thin converging and diverging lenses:

​$$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$$

Note: If the values are positive, the image is real.

Example:

A student investigates the focal length of a thin lens by using it to project an image onto a screen. The object is set $$35\,cm$$ from the centre of the lens and the screen moved until the inverted image is in focus. This position is found to be $$55\, cm$$ from the lens. Determine the focal length of the lens. 

Firstly, state the variables in the question:

​$$u = 35 \ cm\newline v = 55 \ cm$$

Next, state the equation needed:

​$$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$$

Finally, substitute in and solve:

​$$\dfrac{1}{f} = \dfrac{1}{35 \times 10^{-2}} + \dfrac{1}{55 \times 10^{-2}}\newline \\[0.1in] \dfrac{1}{f} = 4.675\newline \\[0.1in] {f} = \dfrac{1}{4.675}\newline \\[0.1in] f = 0.21 \ m$$​​

The focal length is $$\underline{0.21 \ m}$$​​