Question

An object is 20cmto the left of a thin diverging lens that has a 30cmfocal length. (a) What is the image distance i? (b) Draw a ray diagram showing the image position.

Short answer

1. The image distance i is -12cm.
2. The ray diagram is drawn showing the image position.

Step-by-step solution

Listing the given quantities

Object distance p=20cm

Focal length of the lens f = -30cm

Understanding the concepts of lens equation

Here, we need to use the equation of a lens to calculate the image distance for a diverging lens.

We can draw a ray diagram showing the image position using two rays, the ray passing through the center of lens and other ray passing parallel to the principle axis.

Formula:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$

Calculations of the image distance

(a)

The lens equation relates an object distance p, lens focal length f and the image distance i as

$$\begin{gathered} \frac{1}{p}+\frac{1}{i}=\frac{1}{f} \\ \frac{1}{i}=\frac{1}{f}-\frac{1}{p} \\ \frac{1}{i}=\frac{p-f}{fp} \\ i=\frac{pf}{p-f} \\ =\frac{\left(20cm\right)(-30cm)}{(20cm--30cm)} \\ =-12cm \end{gathered}$$

The negative sign shows that the image isvirtualand to the left side of the lens.

The image distance i is -12cm.

Step 4: The ray diagram showing the image position