Question

A pepper seed is placed in front of a lens. The lateral magnification of the seed is +0.300. The absolute value of the lens’s focal length is40.0cm. How far from the lens is the image?

Short answer

Absolute value of image distance from the lens is 28cm.

Step-by-step solution

Listing the given quantities

Focal length f = 40cm

Lateral magnification m=+0.300

=3/100

Understanding the concepts of mirror equation

We use the mirror equation to find the object distance from the mirror. Using lateral magnification, we find the image distance. As we are asked ‘how far’ the image is, we just need to know the absolute value; we need not consider the side of the lens.

Formula:

$$\begin{gathered} \frac{1}{p}+\frac{1}{i}=\frac{1}{f} \\ m=-\frac{i}{p} \end{gathered}$$

Step 3: Calculations of the absolute value of image distance from the lens

We are asked for the distance of image from the lens. We need to use the proper sign conventions. For that, we don’t need to know the type of lens.

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

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

The lateral magnification is

$$\begin{gathered} \frac{3}{10}=-\frac{i}{p} \\ i=-\frac{3p}{10} \end{gathered}$$

Substituting i in 1), we get

$\frac{1}{p}-\frac{10}{3p}=\frac{1}{f}$

$$\begin{gathered} \frac{1}{p}-\frac{10}{3p}=\frac{1}{40cm} \\ p=-93.3cm \end{gathered}$$

The lateral magnification is

$m=-\frac{i}{p}$

$$\begin{gathered} i=-0.300\times -93.3cm \\ =28cm \end{gathered}$$

Absolute value of image distance from the lens is 28cm