This**Â article aims**Â to find how resulting images will look, given **object distance**Â and **focal length**. The article uses the concept of the **lens equation**. In optics, the relationship between the image distance $ ( v ) $, the **object distance**Â $ ( u ) $ and **focal length**Â $ ( f ) $ of a lens is given by a formula known as the **Lens formula.**Â 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**Â lens equation gives**Â a **negative image distance**, then the image is a**Â virtual image**Â on the same side of the lens as the subject. If it gives a **negative focal length**, then the lens is a**Â diverging**Â rather than a converging lens.

**Expert Answer**

By **using the lens equation:**

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

\[ \Rightarrow \dfrac { 1 } { d Â _ { i } } + \dfrac { 1 } { 30 } = \dfrac { 1 } { 15 } Â \]

\[ \Rightarrow d _ { i } = 30 \: Â cm \]

\[ M = – 1 \]

When the **object is located**Â at $ 2F $ point, the **image**Â will also be **located**Â 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**Â at $ 2F $ point, theÂ **image**Â will also be **located**Â 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:**

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

\[ \Rightarrow \dfrac { 1 } { d _ { i } } + \dfrac { 1 } { 50 } = \dfrac { 1 } { 20 } Â \]

\[ \Rightarrow d _ { i } = 33.33 \: cm \]

\[ M = – 1 Â \]

When the **object is located**Â at $ 2F $ point, the **image**Â will also be **located**Â 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**.