Hey I have most of the parts done I just need help with parts G and I

Consider an object with s = 12; rm cm that produces an image with s' = 15; rm cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter "objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges.

Part A

Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation.

Express your answer in centimeters, as a fraction or to three significant figures.

f =6.67 rm cm

Correct

Part B

Considering the sign of f, is the lens converging or diverging?

converging

diverging

Correct

Part C

What is the magnification m of the lens?

Express your answer as a fraction or to three significant figures.

m =-1.25

Correct

Part D

Think about the sign of s' and the sign of y', which you can find from the magnification equation, knowing that a physical object is always considered upright. Which of the following describes the nature and orientation of the image?

real and upright

real and inverted

virtual and upright

virtual and inverted

Correct

Now consider a diverging lens with focal length f=-15; rm cm, producing an upright image that is 5/9 as tall as the object.

Part E

Is the image real or virtual? Think about the magnification and how it relates to the sign of s'.

real

virtual

Correct

Part F

What is the object distance? You will need to use the magnification equation to find a relationship between s and s'. Then substitute into the thin lens equation to solve for s.

Express your answer in centimeters, as a fraction or to three significant figures.

s =12.0 rm cm

Correct

Part G

What is the image distance?

Express your answer in centimeters, as a fraction or to three significant figures.

s' =

rm cm

Try Again

A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at x=-24; rm cm that is twice as tall as the object.

Part H

What is the image distance?

Express your answer in centimeters, as a fraction or to three significant figures.

s' =24.0 rm cm

Correct

Part I

What is the x coordinate of the object? Keep in mind that a real image and a real object should be on opposite sides of the lens.

Express your answer in centimeters, as a fraction or to three significant figures.

x =

rm cm

Consider an object with s = 12; rm cm that produces an image with s' = 15; rm cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter "objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges.

Part A

Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation.

Express your answer in centimeters, as a fraction or to three significant figures.

f =6.67 rm cm

Correct

Part B

Considering the sign of f, is the lens converging or diverging?

converging

diverging

Correct

Part C

What is the magnification m of the lens?

Express your answer as a fraction or to three significant figures.

m =-1.25

Correct

Part D

Think about the sign of s' and the sign of y', which you can find from the magnification equation, knowing that a physical object is always considered upright. Which of the following describes the nature and orientation of the image?

real and upright

real and inverted

virtual and upright

virtual and inverted

Correct

Now consider a diverging lens with focal length f=-15; rm cm, producing an upright image that is 5/9 as tall as the object.

Part E

Is the image real or virtual? Think about the magnification and how it relates to the sign of s'.

real

virtual

Correct

Part F

What is the object distance? You will need to use the magnification equation to find a relationship between s and s'. Then substitute into the thin lens equation to solve for s.

Express your answer in centimeters, as a fraction or to three significant figures.

s =12.0 rm cm

Correct

Part G

What is the image distance?

Express your answer in centimeters, as a fraction or to three significant figures.

s' =

rm cm

Try Again

A lens placed at the origin with its axis pointing along the x axis produces a real inverted image at x=-24; rm cm that is twice as tall as the object.

Part H

What is the image distance?

Express your answer in centimeters, as a fraction or to three significant figures.

s' =24.0 rm cm

Correct

Part I

What is the x coordinate of the object? Keep in mind that a real image and a real object should be on opposite sides of the lens.

Express your answer in centimeters, as a fraction or to three significant figures.

x =

rm cm

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Consider an object with s = 12\; \rm cm that produces an image with s' =

15\; \rm cm. Note that whenever you are working with a physical object, the

object distance will be positive (in multiple...