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PHY208FALL2008

Week3HW

Due at 11:59pm on Sunday, September 21, 2008

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The next exercise is about reflection and refraction of light

Is Light Reflected or Refracted?

Description: Mostly conceptual questions on index of refraction and Snell's law. Last few parts deal with total internalreflection.

When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occurat the interface separating the two materials. For example, consider a ray of light that travels from air into the water of a lake. Asthe ray strikes the air-water interface (the surface of the lake), it is partly reflected back into the air and partly refracted ortransmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surroundinglandscape and other times the underwater vegetation.

These effects on light propagation occur because light travels at different speeds depending on the medium. The index ofrefraction of a material, denoted by , gives an indication of the speed of light in the material. It is defined as the ratio of thespeed of light in vacuum to the speed in the material, or

.

Part A

When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, thespeed of the light

Hint A.1 Index of refraction

The index of refraction of a material is defined as the ratio of the speed of light in vacuum to the speed in thatparticular material, or

.

Since it is the ratio of two positive quantities that have the same units, the index of refraction is a pure (positive) number.Note that the speed of light in a certain material is inversely proportional to the index of refraction of that material.

ANSWER: increases.

Part B

What is the minimum value that the index of refraction can have?

Hint B.1 How to approach the problem

Remember that the speed of light in a certain material is inversely proportional to the index of refraction of that material.Thus, the minimum value of the index of refraction is calculated for the medium where the speed of light is maximum. Thatoccurs in vacuum where .

ANSWER:

between 0 and 1

The index of refraction of a material is always a positive number greater than 1 that tells us how fast the light travels inthe material. The greater the index of refraction of a material, the more slowly light travels in the material.

An example of reflection and refraction of light is shown in the figure. An incident ray of light traveling in the upper materialstrikes the interface with the lower material. The reflected raytravels back in the upper material, while the refracted ray passesinto the lower material. Experimental studies have shown that theincident, reflected, and refracted rays and the normal to theinterface all lie in the same plane. Moreover, the angle that the

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reflected ray makes with the normal to the interface, called theangle of reflection, is always equal to the angle of incidence.(Both of these angles are measured between the light ray and thenormal to the interface separating the two materials.) This isknown as the law of reflection.

The direction of propagation of the refracted ray, instead, is givenby the angle that the refracted ray makes with the normal to theinterface, which is called the angle of refraction. The angle ofrefraction depends on the angle of incidence and the indices ofrefraction of the two materials. In particular, if we let be theindex of refraction of the upper material and the index ofrefraction of the lower material, then the angle of incidence, , and the angle of refraction, , satisfy the relation

.

This is the law of refraction, also known as Snell's law.

Part C

Now consider a ray of light that propagates from water ( ) to air ( ). If the incident ray strikes the water-air

interface at an angle , which of the following relations regarding the angle of refraction, , is correct?

Part C.1 Find an expression for the ratio of the sines of and

Let the index of refraction of water be and that of air be . Use Snell's law to find an expression for the ratio of the sineof to the sine of .

Express your answer in terms of some or all of the variables , , and .

ANSWER: =

Now, note that for the water-air interface . Therefore, .

ANSWER:

When light propagates from a certain material to another one that has a smaller index of refraction, that is, , thespeed of propagation of the light rays increases and the angle of refraction is always greater than the angle of incidence.This means that the rays are always bent away from the normal to the interface separating the two media.

Part D

Consider a ray of light that propagates from water ( ) to glass ( ). If the incident ray strikes the water-glass

interface at an angle , which of the following relations regarding the angle of refraction is correct?

Part D.1 Find an expression for the ratio of the sines of and

Let the index of refraction of water be and that of glass be . Use Snell's law to find an expression for the ratio of thesine of to the sine of .

Express your answer in terms of some or all of the variables , , and .

ANSWER: =

Now, note that for the water-glass interface . Therefore, .

ANSWER:

When light propagates from a certain material to another one that has a greater index of refraction, that is, , thespeed of propagation of the light rays decreases and the angle of refraction is always smaller than the angle of incidence.This means that the rays are always bent toward the normal to the interface separating the two media.



Part E

Consider a ray of light that propagates from air ( ) to any one of the materials listed below. Assuming that the ray strikes

the interface with any of the listed materials always at the same angle , in which material will the direction of propagation of

the ray change the most due to refraction?

Hint E.1 How to approach the problem

The direction of propagation of the ray of light will change the most when the difference between the angle of refraction andthe angle of incidence is maximum. Since we are studying a situation where light propagates from air to a material that has agreater index of refraction, we can make use of the results obtained in Part D. We know that, in this case, the angle ofrefraction is always smaller than the angle of incidence. Thus, the difference between the angle of refraction and the angle ofincidence is maximum when the angle of refraction is smallest.

Part E.2 Find an expression for the sine of the angle of refraction

Let the index of refraction of the unknown material be . Use Snell's law to find an expression for the sine of the angle ofrefraction, .

Express your answer in terms of some or all of the variables and .

ANSWER: =

Your result shows that the sine of the angle of refraction is inversely proportional to the index of refraction of theunknown material. Therefore, the angle of refraction is minimum in the material that has the greatest index ofrefraction.

ANSWER: ice ( )

water ( )

turpentine ( )

glass ( )

diamond ( )

The greater the change in index of refraction, the greater the change in the direction of propagation of light. To avoid orminimize undesired bending of the light rays, light should travel through materials with matching indices of refraction.

Is light always both reflected and refracted at the interface separating two different materials? To answer this question, let'sconsider the case of light propagating from a certain material to another material with a smaller index of refraction (i.e.,

).

Part F

In the case of , if the incidence angle is increased, the angle of refraction

Hint F.1 How to approach the question

Recall that, according to Snell's law, the sine of the angle of refraction is directly proportional to the sine of the angle ofincidence. Thus, as the angle of incidence is increased, the angle of refraction changes accordingly. Moreover, since theangle of refraction is greater than the angle of incidence, as you found in Part C, the angle of refraction can reach itsmaximum value sooner than the angle of incidence.

ANSWER: increases up to a maximum value of 90 degrees.

Since the light is propagating into a material with a smaller index of refraction, the angle of refraction, , is always

greater than the angle of incidence, . Therefore, as is increased, at some point will reach its maximum value of

90 and the refracted ray will travel along the interface. The angle of incidence for which is called the critical

angle . For any angle of incidence greater than , no refraction occurs. The ray no longer passes into the second

material. Instead, it is completely reflected back into the original material. This phenomenon is called total internalreflection and occurs only when light encounters an interface with a second material with a smaller index of refractionthan the original material.

Part G

What is the critical angle for light propagating from a material with index of refraction of 1.50 to a material with index of

refraction of 1.00?

Part G.1 Find an expression for the sine of the angle of incidence

Use Snell's law to find a general expression for the sine of the angle of incidence, , for a ray of light that travels from a

material with index of refraction to a material with index of refraction .

Express your answer in terms of , , and .



ANSWER: =

Now find when , = 1.50, and = 1.00.

Express your answer in radians.

ANSWER: =

In conclusion, light is always both reflected and refracted, except in the special situation when the conditions for totalinternal reflection occur. In that case, there is no refracted ray and the incident ray is completely reflected.

The next two problems are applications of snell's law of refraction

Problem 23.14

Description: An underwater diver sees the sun 50 degree(s) above horizontal. (a) How high is the sun above the horizon to afisherman in a boat above the diver?

An underwater diver sees the sun 50 above horizontal.

Part A

How high is the sun above the horizon to a fisherman in a boat above the diver?

ANSWER:

Problem 23.16

Description: The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48.(a) What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48.

Part A

What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

ANSWER:

Underwater Optics

Description: The net refraction of light passing through several materials is calculated and then used to find the apparentdistance of objects underwater.

Your eye is designed to work in air. Surrounding it with water impairs its ability to form images. Consequently, scuba diverswear masks to allow them to form images properly underwater. However, this does affect the perception of distance, as you willcalculate.

Consider a flat piece of plastic (index of refraction ) with water (index of refraction ) on one side and air (index ofrefraction ) on the other. If light is to move from the water into the air, it will be refracted twice: once at the water/plasticinterface and once at the plastic/air interface.

Part A

If the light strikes the plastic (from the water) at an angle , at what angle does it emerge from the plastic (into the air)?

Hint A.1 Angles inside the plastic

There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface)and the angle immediately before the second refraction (the plastic/air interface). To find out how they relate, draw a picturewith the path the light follows in the plastic and the normals to both surfaces. Once you have labeled both angles, keep inmind that the surfaces are parallel, and thus their normals are parallel lines. An important theorem from geometry will giveyou the relationship between the angles.

Hint A.2 Important theorem from geometry

If two parallel lines are cut by a transversal (a third line not parallel to the first two), then alternate interior angles arecongruent.

Part A.3



Part A.3 Find the angle in the plastic

Using Snell's law, find , the angle inside the plastic (i.e., the angle to the normal immediately after the first refraction at

the water/plastic interface).

Hint A.3.a Snell's law

Snell's law states that , where is the angle of incidence, is the index of refraction for the

medium from which light is incident, is the angle of refraction, and is the index of refraction for the medium into

which light emerges.

Express your answer in terms of , , and . Remember that the inverse sine of a number should be entered as

asin(x) in the answer box.

ANSWER: =

Now, when you calculate , don't forget that .

Express your answer in terms of , , , and . Remember that the inverse sine of a number should be entered

as asin(x) in the answer box.

ANSWER:

=

Notice that does not appear in this equation.

Humans estimate distance based on several different factors, such as shadows and relative positions. The most importantmethod for estimating distance, triangulation, is performed unconsciously. Triangulation is based on the fact that light fromdistant objects strikes each eye at a slightly different angle. Your brain can then use that information to determine the angle

as shown in the figure. In the figure, points L and R representyour left and right eyes, respectively. The distance between youreyes is , and the distance to the object, point O, is .

Part B

What is the distance to the object in terms of and ?

Express your answer in terms of and .

ANSWER:

=

Part C

If the distance to the object is more than about 0.4 , then you can use the small-angle approximation . What is

the formula for the distance to the object, if you make use of this approximation?


ANSWER:

=



=

Your eyes determine by assuming that and (in the figure) are equal. This is true, unless the light rays are bent before

they reach your eyes, as they are if you're wearing a scuba mask underwater.

Underwater, the situation changes, as shown in the figure..Youreyes will calculate an apparent distance using the angle that

reaches your eyes, instead of the correct geometric angle . This

is the same that you calculated in Part A. Note that there are

no important geometric considerations arising from the refractionexcept the substitution of for , because the refraction takes

place so close to your eyes. If the problem discussed someonelooking out of the porthole of a submarine, the geometry wouldbecome more complicated.

Part D

Now use the expression found in Part C for the distance between your eyes and the object at point O, and find the ratio of theapparent distance to the real distance, . Remember that the apparent distance is the distance calculated by your eyes using

the angle instead of the angle . Since we are dealing with small angles, you may use the approximations

and .

Part D.1 Use the small-angle approximations

From Part A, you have the expression

.

Apply both small-angle approximations to this equation to get a simpler expression for .

Express your answer in terms of , , and .

ANSWER: =

Part D.2 Find

Because of the refraction your eyes use instead of . Once you've used the small-angle approximations, plug your

equation for into the equation from Part C, . Since you are putting into the equation instead of , this

gives the apparent distance .

Express your answer in terms of , , , and .

ANSWER: =


ANSWER: =

Part E

Given that and , by what percent do objects underwater appear closer than they actually are?

Hint E.1 How to approach the problem

To find the fraction by which objects appear closer, simply find the difference between the apparent distance and the truedistance, then divide by the true distance. You can then convert this fraction to a percent to get the final answer. Forexample, if an object is 1 away and it appears to be 0.6 away then it is 40% closer ( %).



Express your answer to two significant figures.

ANSWER: %

These problems deal with ray optics and lenses.

Problem 23.5

Description: A student has built a l-long pinhole camera for a science fair project. She wants to photograph the WashingtonMonument, which is 167 m (550 ft) tall, and to have the image on the film be h high. (a) How far should she stand from theWashington...

A student has built a 19.0 -long pinhole camera for a science fair project. She wants to photograph the WashingtonMonument, which is 167 m (550 ft) tall, and to have the image on the film be 5.10 high.

Part A

How far should she stand from the Washington Monument?

ANSWER: m

A Laser and a Lens

Description: A laser is projected through a lens onto a screen. Find the location of the point on the screen.

A laser is mounted as shown in the figure (distances and are given) above the axis of a converging lens with positive focal

length . The laser beam travels parallel to the axis of the lens. A large screen is placed at a distance to the right of the lens.

The laser beam passes through the lens and makes a dot on the screen at a distance , measured upward from the axis of the

lens. Assume that a positive value means that the dot is above the axis, while a negative value means that the dot is below theaxis.

Part A

Find the distance .

Hint A.1 How to approach the problem

Make your own diagram and draw the refracted ray. Then use the geometry of similar triangles to find the position of the doton the screen.

Hint A.2 Drawing the refracted ray

Since the incident ray is parallel to the axis of the lens, the refracted ray passes through the focal point on the right side of thelens.



Part A.3 Find a pair of similar triangles

Once you have drawn your own diagram with the refracted ray shown, similar to the picture shown here, consider the twosimilar triangles ACF and EFG that the refracted ray forms with the axis of the lens, one on each side of the focal point F.The triangle ACF has sides of length and . What are the

lengths of the corresponding sides of the triangle EFG?

ANSWER: and

and

and

and

Now use triangle similarity to find . Note that your final answer should refer to the location of the dot on the screen.

If the dot on the screen is above the axis of the lens, the corresponding length should be positive. If the dot on the

screen is below the axis of the lens, the corresponding length should be negative.

Express your answer in terms of some or all of the variables , , , and . Note that not all of the variables may be

required for the answer.

ANSWER: =

Notice that the sign of the height depends upon the sign of . If the focal point is in front of the screen, the dot will

be below the axis of the lens. If the focal point is on the screen, the image will be at the level of the axis of the lens, and ifthe focal point is behind the screen, the dot will be above the axis of the lens.

Now consider a specific case. Let the laser be 50 centimeters to the left of the lens and at height 15 centimeters above the axisof the lens. The lens has focal length 30 centimeters, and the screen is 1 meter to the right of the lens.

Part B

What is the position of the dot on the screen?

Express your answer in centimeters, to two significant figures.

ANSWER: =

Part C

If the laser is moved farther from the lens, what happens to the dot on the screen?

Hint C.1 How to approach the problem

Consider the expression found in Part A. Does the position of the dot on the screen depend on the distance of the laser fromthe lens?

ANSWER: It moves up.It moves down.It does not move.

Part D

If the laser is moved up, farther above the axis of the lens, what happens to the dot on the screen? Assume that the ray stillstrikes the lens.

Hint D.1 How to approach the problem



Recall the expression found in Part A. If you double the value of , how is affected?

ANSWER: It moves up.It moves down.It does not move.

Focusing with the Human Eye

Description: A man looks at three objects at different distances. Explain how he must adjust the lenses in his eyes to bring thevarious objects into focus.

Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: asquirrel sitting on a rock next to him, a tree a few meters away, and a distant mountain. As Joe is taking in the view, he thinksback to what he learned in his physics class about how the human eye works.

Light enters the eye at the curved front surface of the cornea,passes through the lens, and then strikes the retina and fovea on theback of the eye. The cornea and lens together form a compond lenssystem. The large difference between the index of refraction of airand that of the aqueous humor behind the cornea is responsible formost of the bending of the light rays that enter the eye, but it is thelens that allows our eyes to focus. The ciliary muscles surroundingthe lens can be expanded and contracted to change the curvature ofthe lens, which in turn changes the effective focal length of thecornea-lens system. This in turn changes the location of the imageof any object in one's field of view. Images formed on the foveaappear in focus. Images formed between the lens and the foveaappear blurry, as do images formed behind the fovea. Therefore, tofocus on some object, you adjust your ciliary muscles until theimage of that object is located on the fovea.

Part A

Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be__________ the distance between the front and back of his eye.


Draw a picture of the object (the tree), the lens, and the image it produces. Be sure to include the focal point of the lens.Where must the fovea be in your sketch if this object is in focus? Is the focal point between the lens and the fovea, on thefovea, or behind the fovea?

ANSWER: greater thanless thanequal to

Part B

Joe's eyes are focused on the tree, so the squirrel and the mountain appear out of focus. This is because the image of thesquirrel is formed ______ and the image of the mountain is formed _____.

Hint B.1 Image of the squirrel

The squirrel is closer to the lens (the eye) than the tree. As long as Joe's eyes stay fixed on the tree, their focal length doesnot change. Using the lens equation, determine whether the image of the squirrel is closer to the lens than the image of thetree or farther away.

Hint B.2 Image of the mountain

The mountain is farther from the lens (the eye) than the tree. As long as Joe's eyes stay fixed on the tree, their focal lengthdoes not change. Using the lens equation, determine whether the image of the mountain is closer to the lens than the image ofthe tree or farther away.

ANSWER: between the lens and fovea / between the lens and foveabetween the lens and fovea / behind the foveabehind the fovea / between the lens and foveabehind the fovea / behind the fovea

Part C

Joe now shifts his focus from the tree to the squirrel. To do this, the ciliary muscles in his eyes must have _____ the curvatureof the lens, resulting in a(n) _______ focal length for the cornea-lens system. Note that curvature is different from radius ofcurvature.

ANSWER:



ANSWER: increased / increasedincreased / decreaseddecreased / increaseddecreased / decreased

Part D

Finally, Joe turns his attention to the mountain in the distance but finds that he cannot bring the mountain into focus. This isbecause he is nearsighted. But when Joe puts on his glasses, he can see the mountain clearly. To adjust for hisnearsightedness, his glasses must contain _____ lenses.

Hint D.1 Focusing on distant objects

The image of a distant object like the mountain always forms at (or very close to) the focal point of the fovea-lens system.When Joe is looking at the most distant object he can see clearly, where is the focal point?

Hint D.2 The role of corrective lenses

Nearsightedness and farsightedness are both caused by the fact that the ciliary muscles cannot make the focal length of thelens arbitrarily large or small. The corrective lenses must make the image of the distant mountains form someplace that hiseyes are naturally able to focus on.

ANSWER: convergingdiverging

A Two-Lens System

Description: Calculate the image size and position for a two-lens system. Then, find the lens that would put the image in thesame position if it replaced the two lenses.

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 the image.

Hint A.1 How to handle multiple optics

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

Part A.2 Find the object distance for the first lens

How far is the object from the first lens?

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

ANSWER: =

Part A.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 from the lens.


ANSWER: =

Part A.4 Find the object distance for the second lens



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


ANSWER: =

Part A.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: =

Part B

How tall is the image?

Hint B.1 How to approach the problem

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

Part B.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.

ANSWER: =

Part B.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: =

Part C

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

ANSWER: uprightinverted

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?

Part D.1 Find the object distance for the third lens

How far is the object from the new lens?




ANSWER: =

Part D.2 Find the image distance for the third lens

How far is the image from the new lens?


ANSWER: =


ANSWER: =

Part E

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

orientation?

Part E.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: 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.

A Microscope for Biology

Description: Find the length of a microscope tube needed to view an object in focus, given its distance from the objective andthe objective focal length.

In a two-lens system, the image produced by one lens acts as the object for the next lens. This simple principle finds applicationsin many optical instruments, including some of common use such as the microscope and the telescope.

Part A

The microscope available in your biology lab has a converging lens (the eyepiece) with a focal length of 2.50 mounted onone end of a tube of adjustable length. At the other end is another converging lens (the objective) that has a focal length of 1.00

. When you place the sample to be examined at a distance of 1.30 from the objective, at what length will you need to

adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye?

Note that to view the sample with a completely relaxed eye, the eyepiece must form its image at infinity.


Since the microscope is a two-lens system, there are two steps to follow to solve the problem. First calculate the location ofthe image formed by the objective. Then consider this image to be the object for the eyepiece and find its distance from theeyepiece. The length of the tube will be the sum of the image distance for the objective and the object distance for theeyepiece.

Part A.2 Find the image distance for the objective

What is the image distance for the objective, given that its focal length is 1.00 and the object distance is 1.30 ?

Hint A.2.a The thin-lens equation

The image distance can be found from the thin-lens equation

,

where is the object distance and is the focal length of the lens.



Express your answer in centimeters.

ANSWER: =

Part A.3 Find the object distance for the eyepiece

What is the object distance for the eyepiece, given that it forms an image at infinity and its focal length is 2.50 ?

Part A.3.a Find where the object is located

At what point should the object for the eyepiece be located if the eyepiece is to form an image at infinity?

ANSWER: at the center of curvature of the eyepieceat the focal point of the eyepieceat a point between the focal point and the center of curvature of the eyepieceat infinityat the focal point of the objective


ANSWER: =

The length of the tube is then the sum of the image distance for the objective and the object distance for the eyepiece.


ANSWER: =

Problem 23.68

Description: A slide projector needs to create a h-high image of a h1-tall slide. The screen is l from the slide. (a) What focallength does the lens need? Assume that it is a thin lens. (b) How far should you place the lens from the slide?

A slide projector needs to create a 98.0 -high image of a 2.20 -tall slide. The screen is 294 from the slide.

Part A

What focal length does the lens need? Assume that it is a thin lens.

ANSWER: cm

Part B

How far should you place the lens from the slide?

ANSWER: cm

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