# Lens equation – Concave lens and examples

The lens equation is:

`1/F=1/(do)+1/(di)`

Where

`F` | `=` | Principal focal length | |

`do` | `=` | Object distance | |

`di` | `=` | Image distance |

But to continue, we need to be careful and clear in our understanding of what is negative and what is positive.

## Concave single lens

In all cases and situations of the concave lens the image will always be.

SMALLER

UPRIGHT

VIRTUAL (cannot be projected on a screen)

Between principal focal point and the concave lens

and to this list we should now add:

**Examples**

**Question 1**

If an object is `12cm` from a concave lens with a focal length of `-6cm` what is the image distance from the lens and is the image upright or upside down?

**Answer**

Immediately we see a minus principal focal length (we presume this is what the person who set the exam paper means by just saying focal length) then we know we must be dealing with a concave lens.

Knowing we are dealing with a concave lens we know the image will be upright.

Draw a quick sketch to remind ourselves:

From this quick reminder we have

`1/F=1/(do)+1/(di)`

`1/-6=1/12+1/(di)`

`-0.16dot6=0.083dot3+1/(di)`

`1/(di)=-0.25`

Multiply both sides by `di` and divide both sides by `-0.25` to get `di` on its own.

`1/-0.25=di`

`di=-4`

and looking at the sketch we know all concave images are upright.

Answer = `-4` and UPRIGHT

**Question 2**

A concave lens of focal length `15cm` forms an image `10cm` from the lens. How far is the object from the lens?

**Answer**

As we are given the fact that this is a concave lens we should draw a quick sketch to remind ourselves.

We have to assume two things

i. When focal length of `15cm` is mentioned we must assume that this means principal focal length and being a concave lens the focal length must be `-15cm`

ii. Although an image is formed at `10cm` because this is a concave lens this must be `-10cm`

From this quick sketch we can follow with the lens equation which gives us.

`1/F=1/(do)+1/(di)`

`1/-15=1/(do)+1/-10`

`-0.066dot6=1/(do)-0.1`

`1/(do)=-0.066dot6+0.1`

`1/(do)=0.0333`

Multiply both sides by `do` and divide both sides by `0.0333` to get `do` on its own.

`1/0.0333=do`

`do=30cm`

**Example 3**

A nail is placed a distance of `40cm` from a diverging lens and an image is produced at a distance of `10.91cm` from the lens. What is the distance of the principal focal point from the lens?

**Answer**

As soon as we can, when we read that there is a diverging lens we should draw a quick sketch to remind ourselves of the details of this lens:

We should realise that the image distance is not just `10.91cm` but minus `10.91cm`. We should also realise that the answer we are seeking should provide us with a minus number for the focal point.

Now we should recall the lens formula:

`1/F=1/(do)+1/(di)`

`1/F=1/40+1/-10.91`

`1/F=0.025-0.091659`

`1/F=-0.066659`

Multiply both sides by `F` and divide both sides by `-0.66659` we get.

`F=1/-0.066659`

`F=-15`

As we hoped, we get a minus principal focal length of `-15`