Question

A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object.

(a) What is the focal length of the lens? Is the lens converging or diverging?

(b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted?

(c) Draw a principal-ray diagram.

Short answer

1. -48 cm
2. 6.375 mm and the image is erect as m is positive.
3. See step 4 in the solution

Step-by-step solution

Object -image relationship for ((thin lens))

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

s=object distance from the lens, s'=The image distance from the lens, f=The focal length of the lens.

s is (+) in front of the lens, (-) in the back of the lens, s' is (+) in the back of the lens, (-) in front of the lens, f is (+) the lens is convergent, (-) the lens is divergent.

The sign rules for the variables in the equation:

1. Sign rule for the object distance (s): 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 (s dash): When the image is on the same side of the refracting surface as the outgoing light (the refracted light), the image distance is positive; otherwise, it is negative.

Note: the focal length depends on the curvature of the surfaces which forms the lens and depend on the material of the lens. Apply: in most problems, we are asked to get the position of the image forming from the refracted rays through a certain lens by using the focal length of the lens and the position of the object. In other problems, we are given the position of the object and the image to get the focal length of a lens.

Lateral magnification for a thin lens

Solving part (a) of the problem.

s'= -12 cm is negative as it is int the direction of the incoming rays (the image is in front of the lens)

$$\begin{gathered} \frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}=\frac{\mathrm{s}+\mathrm{s}\text{'}}{\mathrm{ss}\text{'}}=\frac{1}{\mathrm{f}} \\ \mathrm{f}=\frac{\mathrm{ss}\text{'}}{\mathrm{s}+\mathrm{s}\text{'}}=\frac{\left(16\mathrm{cm}\right)(-12\mathrm{cm})}{16\mathrm{cm}-12\mathrm{cm}}=-48\mathrm{cm} \end{gathered}$$

Lateral magnification for a thin lens:

$\mathrm{m}=\frac{-\mathrm{s}\text{'}}{\mathrm{s}}=\frac{\mathrm{y}\text{'}}{\mathrm{y}}$

m=The magnification, s=object distance, s'=The image distance, y'=The height of the image, y=The height of the object. m is (+) when the image is erect and(-) when the image is inverted.

Solving part (b) of the problem.

$$\begin{gathered} \mathrm{m}=\frac{-\mathrm{s}\text{'}}{\mathrm{s}}=\frac{--12\mathrm{cm}}{16\mathrm{cmcm}}=0.75=\frac{\mathrm{y}\text{'}}{\mathrm{y}} \\ \mathrm{y}\text{'}=0.75\mathrm{y}=0.75\times 8.5\mathrm{mm}=6.375\mathrm{mm} \end{gathered}$$

The image is erect as m is positive.

Solving part (c) of the problem.