Question

The focal length of a simple magnifier is 8.0 cm. Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observers near point, 25.0 cm in front of the eye? (b) If the object is 1.0 mm high, what is the height of its image formed by the magnifier?

Short answer

1. The object’s distance from the lens should be 6.06 cm.

b. The height of the image formed by the magnifier is 4.123 mm.

Step-by-step solution

Object-Image Relationship, sign rules for the variables, and Lateral magnification for a thin lens

$\frac{1}{s}+\frac{1}{s^{\text{'}}}=\frac{1}{f}$

Where object distance from the lens

s' = image distance from the lens

f = focal length of the lens

Sign rules for the variables:

1. Sign rule for the object distance: when the object is on the same side of the refracting surface as the incoming light, object distance s is positive; otherwise, it is negative.
2. Sign rule for the image distance: when the image is on the same side of the refracting surface as the outgoing light, image distance s' is positive; otherwise, it is negative.

Calculate the object’s distance from the lens

1. Use the formula,

$$\begin{gathered} \frac{1}{s}+\frac{1}{s^{\text{'}}}=\frac{1}{f} \\ \Rightarrow \frac{1}{s}=\frac{1}{f}-\frac{1}{f^{\text{'}}} \\ =\frac{1}{8}-\frac{1}{-25} \\ =\frac{33}{200}\mathit{c}\mathit{m} \\ \Rightarrow s=\frac{200}{33} \\ =6.06\mathbf{}\mathit{c}\mathit{m} \end{gathered}$$

Calculate the height of the image formed by the magnifier

(b)

$$\begin{gathered} m=\frac{-s^{\text{'}}}{s} \\ =\frac{-\left(-25\right)}{6.06} \\ =4.125=\frac{y^{\text{'}}}{y} \\ \Rightarrow y^{\text{'}}=4.125*1mm=4.125mm \end{gathered}$$

Thus, the object’s distance from the lens should be 6.06 cm and the height of the image formed by the magnifier is 4.125 mm.