Question

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

Part A

Calculate the image position

s'=

Part B

Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.

h =

Answer 1

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

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

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

Answer 2

The image of an object closer to the converging lens than the focal length

is larger than the object.

Part A: The image position away from the lens and on the same side of the

lens is $\underline{19.\overline{259} \ cm}$

Part B: The image height is $\underline{h = 1. \overline{481} \, cm}$

Reasons:

The given parameters are;

Height of the object = 1.0 cm-tall

Distance of the object in front of the lens, u = 13 cm

Focal length of the converging lens, f = 40 cm

Part A:

The lens formula is;

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

Therefore;

$\dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{v}$

Which gives;

$\dfrac{1}{40} - \dfrac{1}{13} = -\dfrac{27}{520} = \dfrac{1}{v}$

$The \ image \ position, \ v = -\dfrac{520}{27} = -19\dfrac{7}{27} \approx -19.\overline{259}$

The image position, v ≈ $-19.\overline{259}$ cm. away from and on the same side as the

object relative to the lens.

Part B

The image height is given by the magnification relation;

$\dfrac{Object \ height}{Image \ height} = -\dfrac{Object \ distance}{Image \ distance}$

Therefore;

$\dfrac{1.0 \ cm}{Image \ height} = -\dfrac{13 \ cm}{\left(-\dfrac{520}{27} \ cm\right)}$

$Image \ height, \, h = \dfrac{\left(\dfrac{520}{27} \, cm\right) \times 1.0 \, cm}{13 \, cm} =\dfrac{40}{27} \ cm = 1. \overline{481} \, cm$

$\underline{h = 1. \overline{481} \, cm}$

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