Physics, 18.02.2020 05:33 kamarionnatillman13

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 =

Answers: 1

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Answer from: zoeatlowapple

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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Answer from: lefthandeddolan

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

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.

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 from: Quest

insects!

explanation:

Answer from: Quest

u can delete ur account.

1. go on google and write

2 open this and click on your profile pic and if your profile is not appearing, then u have to log in.

3. when u click on your profile pic, there appears four options on the top.

4. click on the 1st option and scroll down and go at the bottom.

5. click on edit profile.

6. there appears a option 'prefernces' and at that option at the right side , 'edit' is written.

7. click on the edit option.

8. there is a option " i want to delete my account" and click on that.

hope this u

and ur welcm

: )

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

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