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.

**Expert Answer**

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**.