 # An object is placed 30 cm to the left of a converging lens that has a focal length of 15 cm. Describe what the resulting image will look like (i.e, image distance, magnification, upright or inverted images, real or virtual images)? This article aims to find how resulting images will look, given object distance and focal length. The article uses the concept of the lens equation. In optics, the relationship between the image distance $( v )$, the object distance $( u )$ and focal length $( f )$ of a lens is given by a formula known as the Lens formula. The Lens formula is applicable to both convex and concave lenses. These lenses have negligible thickness. The formula is as follows:

$\dfrac {1}{v} – \dfrac {1}{u} = \dfrac {1}{f}$

If the lens equation gives a negative image distance, then the image is a virtual image on the same side of the lens as the subject. If it gives a negative focal length, then the lens is a diverging rather than a converging lens.

By using the lens equation:

$\dfrac { 1 } { d _ { i } } + \dfrac { 1 } { d _ { o } } = \dfrac { 1 } { f }$

$\Rightarrow \dfrac { 1 } { d _ { i } } + \dfrac { 1 } { 30 } = \dfrac { 1 } { 15 }$

$\Rightarrow d _ { i } = 30 \: cm$

$M = – 1$

When the object is located at $2F$ point, the image will also be located at the $2F$ point on other side of the lens and the image will be inverted. The dimensions of the image are the same as the dimensions of the object.

## Numerical Result

When the object is located at $2F$ point, the image will also be located at the $2F$ point on other side of the lens and the image will be inverted. The dimensions of the image are the same as the dimensions of the object.

## Example

The object is located $50 \: cm$ to the left of the coupler, which has a focal length of $20 \: cm$. Describe what the resulting image will look like (i.e. image distance, magnification, upright or inverted images, real or virtual images).

Solution

By using the lens equation:

$\dfrac { 1 } { d _ { i } } + \dfrac { 1 } { d _ { o } } = \dfrac { 1 } { f }$

$\Rightarrow \dfrac { 1 } { d _ { i } } + \dfrac { 1 } { 50 } = \dfrac { 1 } { 20 }$

$\Rightarrow d _ { i } = 33.33 \: cm$

$M = – 1$

When the object is located at $2F$ point, the image will also be located at the $2F$ point on the other side of the lens, and the image will be inverted. The dimensions of the image are the same as the dimensions of the object.