Q56E - University Physics with Modern Physics

Question

A thin lens with a focal length of 6.00 cm is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.

Step-by-Step Solution

Answer

1. The angular magnification obtained is 4.1667.
2. The object distance from the lens should be 4839 cm

Step 1: Object-Image Relationship and sign rules for the variables

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

Where s = 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.

Step 2: Calculate the angular magnification

$$\begin{gathered} M=\frac{s\text{'}}{f} \\ =\frac{25}{6} \\ =4.1667 \end{gathered}$$

Step 3: Calculate the required object distance from the lens

Use the formula,

                                              $$\begin{gathered} \frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f} \\ \Rightarrow \frac{1}{s}=\frac{1}{f}-\frac{1}{s^{\text{'}}} \\ \Rightarrow \frac{1}{s}=\frac{1}{6}-\frac{1}{-25} \\ =\frac{31}{150cm} \\ \Rightarrow s=\frac{150}{31} \\ =4.839 cm \end{gathered}$$

 

 

 

Thus, the angular magnification obtained is 4.1667 and the object distance from the lens should be 4.839 cm.