Mastering Physics HW 10 Ch 24 - Optical Instruments

HW 10 Ch 24 Optical InstrumentsDue: 11:59pm on Tuesday, November 24, 2015

To understand how points are awarded, read the Grading Policy for this assignment.

A TwoLens System

A compound lens system consists of two converging lenses, one at with focal length ,and the other at with focal length . An object centimeter tall is placed at

.

Part A

What is the location of the final image produced by the compound lens system? Give the x coordinate of theimage.

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

Hint 1. How to handle multiple optics

The image formed by the first lens acts as the object for the second lens.

Hint 2. Find the object distance for the first lens

How far is the object from the first lens?


ANSWER:

Hint 3. Find the image distance from the first lens

Ignoring the second lens, determine where the image is formed just by the first lens. Give its distance fromthe lens.


x = −20.0 cm = +10.0 cmf1x = +20.0 cm = +8.00 cmf2 1.00

x = −50.0 cm

= sob cm

ANSWER:

Hint 4. Find the object distance for the second lens

How far is the image produced by the first lens from the second lens?


ANSWER:

Hint 5. Find the image distance from the second lens

Using the result of the previous hint, determine how far the final image is from the second lens.


ANSWER:

ANSWER:

Correct

Part B

How tall is the image?


Hint 1. How to approach the problem

The total magnification is the product of the magnifications caused by the two lenses seperately: . If you have difficulty finding the individual magnifications, use the other hints.

Hint 2. Find the magnification of the first lens

What is the magnification of the first lens? Recall that magnification is defined in two ways: and

.

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

= sim cm

= sob cm

= sim cm

= 31.8 x cm

mm = m1 m2

m = yimyob

m = − simsob

ANSWER:

Hint 3. Find the magnification of the second lens

What is the magnification of the second lens? Recall that magnification is defined in two ways:

and .


ANSWER:

ANSWER:

Correct

Part C

Is the final image upright or inverted, relative to the original object at ?

ANSWER:

Correct

Now remove the two lenses at and and replace them with a single lens of focal length at . We want to choose this new lens so that it produces an image at the same location as before.

Part D

What is the focal length of the new lens at the origin?


Hint 1. Find the object distance for the third lens

= m1

m = yimyob

m = − simsob

= m2

= 0.236 yim cm

x = −50 cm

upright

inverted

x = +20.0 cm x = −20.0 cmf3 x = 0

How far is the object from the new lens?


ANSWER:

Hint 2. Find the image distance for the third lens

How far is the image from the new lens?


ANSWER:

ANSWER:

Correct

Part E

Is the image formed by the same size as the image formed by the compound lens system? Does it have thesame orientation?

Hint 1. Find the magnification of the third lens

What is the magnification of the third lens? Compare the result with your answer for Parts B and C.


ANSWER:

ANSWER:

= sob cm

= sim cm

= 19.4 f3 cm

f3

= m3

The image is the same size and oriented the same.

The image is the same size and oriented differently.

The image is a different size and oriented the same.

The image is a different size and oriented differently.

Correct

± Understanding Multiple Optics

Learning Goal:

To become familiar with using the image of one instrument as the object of the next and tracing rays through a systemof multiple instruments.

Multiple optics refers to any system of more than one optical instrument through which light passes. Most devicesrelated to optics, such as cameras, microscopes, and telescopes, contain multiple optics systems.

In multiple optics, the image of one optical instrument becomes the object of the next one. Thus, in multiple opticsproblems, you need to find the image created by the first optical instrument that the rays encounter. Then, you will usethat image as the object of the next optical instrument, repeating this pattern until you have followed the rays all theway through the system. It is very important to be alert to the geometry and to signs when you find the object distancefor one instrument from the location of the previous instrument's image. Sometimes, the image is formed on the virtualside of the instrument, leading to a virtual object. This may sound strange, but in practice, its effect on yourcalculations is simply to make the object distance negative instead of positive.

Several optical instruments are placed along the x axis, with their axes aligned along the x axis. A plane mirror islocated at . A converging lens with focal length 5.00 is located at . An object is placed at

.

In order to find the location of the final image of the object formed by this system, you will need to trace the raysthrough the system, instrument by instrument. You are strongly advised to draw a picture with the x axis and thelocation of the lens, mirror, and object marked. Then, as you proceed through the problem, you can mark where eachimage is located.

Part A

First, find the location of the image created by the lens by itself (as if no other instruments were present).

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

Hint 1. The thin lens equation

For a thin lens, one has the relation,

where is the image distance, is the object distance, and is the focal length.

ANSWER:

Correct

Part B

x = 0 m x = 12.5 mx = 22.5 m

+ =1s

1s′

1f

s′ s f

= 2.50 x m

Next, find the location of the image created by the plane mirror (after the light has passed through the lens).


Hint 1. The focal length of a plane mirror

Keep in mind that a plane mirror has a focal length of infinity (since a plane mirror has an infinite radius ofcurvature).

Hint 2. Find the object distance

You know from Part A that the image resulting from the passage of light though the lens is located at . If the plane mirror is at , what is the object distance for the plane mirror?

Express your answer in meters to three significant figures.

ANSWER:

ANSWER:

Correct

Proceed with caution! The light reflects off of the mirror and back through the lens a second time!

Part C

What is the location of the final image, as seen by an observer looking toward the mirror, through the lens? Keep inmind that the light must pass back through the lens, and thus you must do one more calculation with the thin lensequation.


Hint 1. Object distance for the lens (second time)

What is the object distance for the lens the second time? Recall that the object is the image made by themirror. You found its coordinate in the last part, so just subtract that from the coordinate of the lens to findthe object distance for the lens the second time.

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

ANSWER:

ANSWER:

x = 2.50 m x = 0 s

= s m

= 2.50 x m

s

= s m

Correct

Part D

Is the final image formed by this system real or virtual?

Hint 1. Real versus virtual

A real image is one that forms at a location that the light actually reaches (imagine a slide projector creatingan image on a screen). A virtual image is one that forms at a location that the light does not actually reach.(Imagine the image created when you look in a mirror hanging on a wall. No light actually makes it past thewall; there only appears to be a copy of you on the other side of the wall!)

Hint 2. Real versus virtual with multiple optics

Since the type of image is determined only by whether the light actually reaches the location of the image,whether the image is real or virtual depends only on the last optical instrument through which the lightpasses. Thus, the light may fail to reach an intermediate image (in this example, the one created by themirror, which was a virtual image) but may still create a real final image, which it will do if the light reachesthe location of the final image created by the system.

ANSWER:

Correct

If you have more than one optical instrument, the total magnification is equal to the product of the individualmagnifications. Keep in mind that the image of one device becomes the object of the next. If the first device creates animage that is ten times the size of the object, and the second creates an image that is twenty times the size of the firstimage (which was its object), then the final image will be two hundred times the size of the original object. This makessense because the second device magnifies further what was already magnified by the first device. To find themagnitude of the magnification of the final image, you will need to consider each instrument and find the individualmagnifications.

Part E

First, find the magnitude of the magnification of the image created when light from the object passesthrough the lens the first time (as if the mirror were not present).


= 20.0 x m

real

virtual

mlens1

Hint 1. Magnification

The magnification of an image is defined as the negative of the ratio of the image distance to the objectdistance. The magnitude determines the relative size of the image as compared to the object, while the signyields the orientation (negative is inverted, positive is upright).

ANSWER:

Correct

Part F

Next, find the magnitude of the magnification of the plane mirror.


ANSWER:

Correct

Of course, this is expected. Plane mirrors don't magnify your image!

Part G

Now find the magnitude of the magnification of the image created when light from the object passes throughthe lens the second time (after reflecting off the mirror).


ANSWER:

Correct

Part H

What is the magnitude of the magnification of the final image?


ANSWER:

= 1mlens1

mmirror

= 1mmirror

mlens2

= 0.5000mlens2

Correct

Part I

Is the final image upright or inverted? Give the orientation relative to the original object.

Hint 1. Magnification in multiple optics (orientation)

If you have more than one optical instrument, the total magnification is just the product of the individualmagnifications. The image of one device becomes the object of the next. If, for example, the first devicecreates an image that is inverted (negative magnification) and the second creates one that is also inverted(negative magnification), then the final image will be upright (positive magnification). Mathematically, thismakes sense because the product of two negative numbers is positive. Conceptually, this makes sensebecause if the first device inverts the image, and the second also inverts the image, then the first deviceflips the image upside down, and the second flips it back again, resulting in an image that is rightside up.

Hint 2. Find the orientation from the lens (first time)

Is the image created by the lens (as if the mirror were not present) inverted or upright?

ANSWER:

Hint 3. Find the orientation from the mirror

Is the image created by the mirror (of the image created by the lens) inverted or upright, compared to theimage created by the lens?

ANSWER:

Hint 4. Find the orientation from the lens (second time)

Is the image created by the lens (of the image created by the mirror) inverted or upright, compared to theimage created by the mirror?

ANSWER:

= 0.5000m

upright

inverted

upright

inverted

upright

inverted

ANSWER:

Correct

Corrective Lenses Conceptual Questions

The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance atwhich an object can be seen clearly) are measured for six different people.

near point ( )

far point ( )

Avishka 40

Berenice 30 300

Chadwick 25 500

Danya 25

Edouard 80 200

Francesca 50

Part A

Which, if any, of these people are nearsighted (myopic)?

List the first letter of all correct answers in alphabetical order. For example, if Avishka and Edouard are theonly nearsighted ones, enter AE.

Hint 1. Being nearsighted (myopic)

A person with normal vision can focus clearly on objects an infinite distance away. A person is considerednearsighted, or myopic, if their fully relaxed eye focuses only out to a finite distance. This is the result ofthe eye converging the light from infinity in too short a distance.

ANSWER:

Correct

Nearsightedness can be corrected by diverging lenses, which form virtual images at the individual's far pointwhen presented with objects at infinity.

upright

inverted

cm cm

∞

∞

∞

BCE

Part B

Which, if any, of these people require bifocals to correct their vision?

List the first letter all correct answers in alphabetical order. For example, if Avishka and Edouard are theones needing bifocals, enter AE.

Hint 1. Being farsighted (hyperopic)

Those with normal vision can focus on objects at a normal reading distance, defined to be 25 . A personis considered farsighted, or hyperopic, who is unable to focus on objects within this distance. This problemis the result of the eye not converging the light from a nearby object strongly enough.

Hint 2. Bifocals

A person who is both myopic and hyperopic will need bifocals to see both far and near objects clearly.Bifocals typically have a relatively small converging lens set in the bottom portion of an otherwise diverginglens. To see nearby objects, for example to read, the person looks through the bottom converging portion ofthe lens.

ANSWER:

Correct

Farsightedness can be corrected by converging lenses, which form virtual images at the individual's near pointwhen presented with objects at 25 . Bifocals include both converging and diverging lenses to correct forboth farsightedness and nearsightedness.

Part C

Which, if any, of these people's vision can be corrected using only converging lenses?

List all correct answers in alphabetical order. For example, if Avishka and Edouard are the ones whosevision can be thus corrected, enter AE.

ANSWER:

Correct

Part D

Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision.

Rank these from largest to smallest power required. To rank items as equivalent, overlap them.

Hint 1. Power

cm

BE

cm

AF

The power of a lens is the inverse of its focal length. Lenses with small focal lengths have large power andcan correct more serious vision defects.

Hint 2. Deviation from "normal" near point

The near point of a normal eye is 25 . The larger the difference of a person's near point from this normalvalue, the more serious the vision defect.

ANSWER:

Correct

Nearsightedness and Farsightedness

A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely faraway. There exist, however, certain conditions under which the range of vision is not so extended. For example, anearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannotfocus on objects closer than a certain point (the near point). Note that even though the presence of a near point iscommon to everyone, a farsighted person has a near point that is much farther from the eye than the near point of aperson with normal vision.Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, theeye converges the light coming from the image formed by the corrective lens rather than from the object itself.

cm

Part A

When glasses (or contact lenses) are used to correct nearsightedness, where should the corrective lens form animage of an object located at infinity in order for the eye to form a clear image of that object?

Hint 1. Range of vision in nearsightedness

To be effective, the corrective lens should form an image at a point located within the range of clear vision.Recall that a nearsighted person can form a clear image of any object located as far from the eye as the farpoint, but not farther.

ANSWER:

Correct

This effect is achieved by the use of a diverging lens, as shown in the figure.

Part B

If a nearsighted person has a far point that is 3.50 from the eye, what is the focal length of the contactlenses that the person would need to see an object at infinity clearly?

Express your answer in meters.


Once you have determined the object distance and the image distance, you can use the thinlens equationto find the focal length of the contact lenses. Notice that the information found in Part A will help you findthe image distance.

The lens should form the image at the near point.

The lens should form the image at the far point.

The lens should form the image at a point closer to the eye than the near point.

The lens should form the image at a point farther from the eye than the far point.

df m f1


At what distance from the contact lens is the object?

Hint 1. Contact lenses

Contact lenses are placed directly against the eye, so the lensobject distance is the same as theeyeobject distance.

ANSWER:

Hint 3. Find the image distance

At what distance from the contact lens should the image form in order for the eye to focus clearly on it?


Hint 1. Contact lenses as corrective lenses

As you found in Part A, a corrective lens should form the image at the far point. Also note thatcontact lenses are placed directly against the eye, so the eyeobject distance is the same as thelensobject distance.

ANSWER:

ANSWER:

Correct

Part C

When glasses (or contact lenses) are used to correct farsightedness, where should the corrective lens form animage of an object located between the eye and the near point in order for the eye to form a clear image of thatobject?

do

= do

df

−df∞−∞

di

= di m

= 3.50 f1 m

Hint 1. Range of vision in farsightedness

To be effective, the corrective lens should form an image at a point located within the range of clear vision.Recall that a farsighted person can form a clear image of any object located as close to the eye as the nearpoint, but not closer.

ANSWER:

Correct

This effect is achieved by the use of a converging lens, as shown in the figure.

Part D

If a farsighted person has a near point that is 0.600 from the eye, what is the focal length of the contactlenses that the person would need to be able to read a book held at 0.350 from the person's eyes?



Once you have determined the object distance and the image distance, you can use the thinlens equationto find the focal length of the contact lenses. Notice that the information found in Part C will help you findthe image distance.


At what distance from the contact lens is the object?

The lens should form the image at the near point.

The lens should form the image at the far point.

The lens should form the image at a point closer to the eye than the near point.

The lens should form the image at a point farther from the eye than the far point.

m f2m

do


Hint 1. Contact lenses

Contact lenses are placed directly against the eye, so the lensobject distance is the same as theeyeobject distance.

ANSWER:

Hint 3. Find the image distance

At what distance from the contact lens should the image form in order for the eye to focus clearly on it?


Hint 1. Corrective lenses

Use the information you obtained in Part C. That is, keep in mind that the corrective lens should formthe image at the near point and that contact lenses are placed directly against the eye.

ANSWER:

ANSWER:

Correct

Problem 24.44

Your task in physics laboratory is to make a microscope from two lenses. One lens has a focal length of 2.2 , theother 1.1 . You plan to use the more powerful lens as the objective, and you want the eyepiece to be 15 fromthe objective.

Part A

For viewing with a relaxed eye, how far should the sample be from the objective lens?

Express your answer to two significant figures and include the appropriate units.

ANSWER:

= do m

di

= di m

= 0.840 f2 m

cmcm cm

All attempts used; correct answer withheld by instructor

Part B

What is the magnification of your microscope?

Express your answer using two significant figures.

ANSWER:

Correct

Problem 24.15

A magnifier has a magnification of 9× .

Part A

How far from the lens should an object be held so that its image is seen at the nearpoint distance of 25 ?Assume that your eye is immediately behind the lens.

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Incorrect; Try Again; 3 attempts remaining

Score Summary:Your score on this assignment is 47.0%.You received 2.35 out of a possible total of 5 points, plus 0 points of extra credit.

= 1.08 s cm

= 120M

cm

2.8 cm

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Mastering Physics HW 10 Ch 24 - Optical Instruments