Please help the instructions for this project are in this link. (http://www.weba

Question

Please help the instructions for this project are in this link. (http://www.webassign.net/userimages/Module%208%20-%20Optical%20Bench%20-%20Instructions.pdf?db=v4net&id=454761)

The reason the colored rays are dispersed so that you can see individual colored rays, and each color of light has a different focal point, is because the ____________is different for each color of the spectrum.

This phenomenon is referred to as ____________

of light.

Part II: Ray Diagram

Step 2

Calculate is the image distance.

A. 2.0

B. 6.67    

C.10

Given the magnification, calculate the image height.

A. 2

B. -1.33    

C. 1.33

Step 3

Calculate is the image distance.

A. 6

B. -6    

C. 12

Given the magnification, calculate the image height.

A. 3

B. 6    

C. -3

Referring to the images in Steps 2 & 3:

(a) How can you tell the images are real?

.
(b) What does the negative sign on the magnification mean?

.

Step 4

As you moved the focal point from -4 to -8 on the optical axis, the lens

Step 5

Calculate is the image distance.

A. -2.25

B. 2.25    

C. 4.50

Calculate the image height.

A. 0.75

B. -0.75    

C. -3

When you moved the focal point along the optical axis progressively closer to the lens and then through to the other side, you saw the rays change and an image appeared to the left of the lens. Complete the following sentences for the image that appeared to the left of the lens.

(a) What kind of image was it?

(b) How could you tell?

Explanation / Answer

Question 1:

each colour has different wavelength.

hence the correct answer is :

The reason the coloured rays are dispersed so that you can see individual coloured rays, and each colour of light has a different focal point, is because the wavelength is different for each colour of the spectrum.

question 2:

this phenomenon is referred to as dispersion of light.

part ii.

step 2:

given :

focal length=f=4

object distance u=-10

using lens equation:

if image distance is v, then

(1/v)-(1/u)=1/f

==>v=6.67

hence option B is correct.

image height/object height=image distance/object distance

==>image height=(6.67/(-10))*2

=-1.33

hence option B is correct.

step 3:

given:

focal distance=f=3

object height=h=3

object distance =u=-6

let image distance be v.

using lens equation:

(1/v)-(1/u)=1/f

==>v=6

hence option A is correct.

image height=object height*(image distance/object distance)

=3*(6/(-6))=-3

hence option C is correct.

part a:

as images are on the other side of the lens where the object is, the images are real.

part b:

magnification is negative means the image is inverted.

step 5:

focal length=f=-3

object height =3

object distance u=-9

let image distance be v.

using lens equation:

(1/v)-(1/u)=1/f

=>v=-2.25

hence option A is correct.

image height=object height*(image distance/object distance)

=3*(-2.25/(-9))=0.75

hence option A is correct.

part a:

image was virtual.

part b:

it was on the same side of the lens as the object.

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