Chapter 15: Refraction and Lenses

Refraction of Light

If water waves in a ripple tank travel from water with a depth of 1 cm into a shallow area with a depth of .3 cm, the water will slow down. In a ripple tank the depth can be changed by placing a sheet of plexiglass or glass in the water. In the picture to the right the area to the left is deeper water and the the the right is shollow. The picture shows that when the water travels into a shallow area and slows down, the wavelength will shorten. We learned in the last chapter that when waves travel into a new medium the frequency remains the same so if the wavespeed decreases, the wavelength must decrease.

If the wavefronts cross a boundary at an angle and slow down we get a bending of the front. This bending of the path of the waves as they pass from one medium to another is called refraction. It occurs because the leading edge of the front slows down while the front in the faster medium keeps its original speed.

If the path of the wave front changes, the path of a ray will also change. Here bending occurs when the ray crossing the boundary is not perpendicular to the boundary.

Physicists have invented a concept called the index of refraction (symbol is n) to indicate how much the speed is changed when working with light waves. The equation is:

The index of refraction is a ratio, so it has no units. Some times the term index of refraction is called optical density. A material with a higher index of refraction is said to be denser than one with low. Often, materials with high mass density also have high optical density, but this is not always the case.

In air the index of refraction is 1.0003, which indicates the light does not slow down very much. For air we will only use 3 figures or 1.00. Below are some commonly used indices of refraction.

Air: 1.00 Crown glass 1.52 Water 1.33 Flint glass 1.61 Alcohol 1.36 Diamond 2.42

Hint: Since in the universe as we know it the speed of light in a vacuum can’t be exceeded, the index of refraction can never be smaller than 1.00.

Sample problems:
1) If the speed of light in glass is 2.0x108 m/s, what is the index of refraction? Knowns: Equation: [pic]

Solution: [pic]

2) What is the speed of light in water if n = 1.33? Knowns: Equation: [pic]

Solution: [pic]

Although light has properties of both waves and particles, the refraction of light is easiest explained by using its wave properties. When light passes from one medium to another at an angle to the normal, it behaves similar to the water wave and the path of the light is bent. Knowing these three rules about light rays can help you answer conceptual questions involving refraction.

1. If light travels at an angle from a less dense to more dense medium, a ray bends toward the normal.

2. If light travels at an angle from a more dense to less dense medium, a ray bends away from the normal.

3. If the light strikes a boundary at 90o to the surface (at an angle of 0o to the normal) it does not bend.

The light will bend according to the equation: [pic]. This equation is known as Snell’s Law, named after Willebrord Snell, who discovered it in 1621. The variables are shown below.

Medium 1 is the medium in which the light is going toward the boundary.

Medium 2 is the medium in which the light is going away from the boundary.

n1 stands for the index of refraction of medium 1.

n2 stands for the index of refraction of medium 2.

(1 stands for incident angle, the angle to the normal of the light in medium 1.

(2 stands for refracted angle, the angle to the normal of the light in medium 2.

Problem solving strategy for refraction problems:
1) Make a sketch of the situation. Label the known values and use symbols for the unknown values.
2) Make a list of your known and unknown values.
3) Use Snell’s Law to solve for your unknown. If you get undefined or error for an answer, you are probably in a critical angle problem. (See section 5 of this chapter.)
4) Check to see if your answer makes sense. (For example, if the light is going from high to low index, is the refracted angle more than the incident?
Sample problems:
3) Light enters your glass of spring water from the air at an angle of 60.0o to the normal. At what angle will it travel through your water? Knowns: Equation: [pic] Solution: [pic]

4) Al Ligator dropped his flashlight into an aquarium. The light is incident on the crown glass at 40.0o to the normal. At what angle will it travel after it leaves the glass? Knowns: Equation: [pic]

Solution: [pic]

Knowns: Equation: [pic]

Solution: [pic]

5) Light traveling through a piece of glass at 25.0o to the normal has a refracted angle of 41.0o in the air. What is the index of refraction for the glass? Knowns: Equation: [pic]

Solution: [pic]

Total internal reflection and the critical angle

When light travels from a more dense to a less dense medium, there is an incident angle above which the light can’t escape from the more dense medium. This angle is called the critical angle. The critical angle ((c) is reached when the refracted angle would become 90o. The light does not escape, and we end up with what is called total internal reflection. The light going to the surface bounces off, following the law of reflection.

To derive the equation for the critical angle simply use Snell’s law, and apply this special condition. [pic] Since the sine of 90o is equal to 1, the formula for the critical angle reduces to:

Problem solving hint: If you get a problem where you are trying to solve for the refracted angle and your calculator says error, it may be that the light does not get out! You may be past the critical angle.

Sample problem:
6) What is the critical angle at which light from the flashlight will not leave the surface of the lake?

Knowns: Equation: [pic]

Solution: [pic]

Problems:

1) What is the speed of light in diamond?

2) Dinah Might wants to hide a piece of glass, and finds a chemical in which the speed of light is 1.97x108 m/s. Will this chemical work?

3) What is the difference between the speed of light in alcohol, and in water?

4) Below is light traveling through three materials. Rank the speed of light in each of the materials from fastest to slowest. Also rank the index of refraction of each material from highest to lowest. a) b)

5a) The incident wave fronts are shown below. Draw in the refracted wavefronts. At any location you wish, show the normal, the incident and the refracted ray.

6a) The incident ray and 3 wavefronts are shown below. Draw in the refracted and the reflected wavefronts. Also draw the normal, the reflected and the refracted ray.

7) To the right, the reflected wavefronts are shown. Draw in the incident and refracted wavefronts. At the location of your choice on the boundary, draw an incident ray, a reflected ray, a refracted ray and a normal.

8) Qualitatively sketch the path of the light rays below as they enter and exit the glass blocks. You should draw the normal at each interface.

9) Locate the position of the image of the underwater light source as seen by the eye.

10) Light is incident on the glass as shown below. Using a protractor and Snell’s law, determine the path of the light through the layers and back into the air. Draw in the path of the ray on the diagram.

11) Light is incident on flint glass from air at an angle of 45o. At what angle will it travel in the glass? What is the angle between the reflected light and the refracted light?

12) Light passes from air into an unknown liquid. If the incident light is at 66.2o, and the refracted light is at 32.4o, what is the index of refraction of the liquid?

13) A block of unknown material is submerged in water. If the incident light is at 31.0o, and the refracted light is at 27.0o, what is the index of refraction of the block?

14) Use your protractor and Snell’s Law to determine the index of refraction of the block below, which is surrounded by air.

15) The speed of light in an unknown type of glass is 1.90x108 m/s. If light leaving this glass is incident at 33.0o, at what angle will it travel in the air?

16) What is the critical angle for light from: a) crown glass into air? b) diamond into air?

c) flint glass into water? d) water into crown glass?

17) The glass prism to the right has an index of refraction of 1.52. Incident on the left side is a ray of light. Calculate and draw in the path of the light inside the glass and back out into the air. The light is hitting perpendicular to the glass.

18) When Shanda Lear received a ring from her boyfriend, he tells her that the diamond is as true as his love. Not exactly trusting him, she takes it to her physics class and determined that the critical angle for the ring in air is 31.4o. What is the index of refraction for the stone on the ring? (Should she trust what he says?)

19) The equilateral glass prism to the right has an index of refraction of 1.52. Incident on the left side is a ray of light. Calculate and draw in the path of the light inside the glass and back out into the air. The light is hitting at 45o to the normal.

20) How would you set up a demonstration in a ripple tank to show total internal reflection? Sketch a diagram to support your answer. (Below is a real life picture of total internal reflection in a ripple tank.)

21) Conceptual Review

a) How does the incident frequency compare to the refracted frequency? How do you know?

b) How does the incident wavelength compare to the refracted wavelength? How do you know?

c) How does the incident wavespeed compare to the refracted wavespeed if the refracted ray has a larger angle to the normal than the incident ray? How do you know?

Lenses

Lenses use refraction to alter the path of light passing through them. There are four terms commonly used to describe lenses. A converging lens focuses light, and a diverging lens spreads out light. A convex lens is thicker in the center than at the edge, a concave lens is thicker at the edge than in the center. If a lens has a higher index than the surrounding medium, then convex lenses converge light and concave lenses diverge light.
Parallel rays of light were sent to both lenses above. On the convex lens, the light is bent through a point. This point is called the focal point. The distance from the middle of the lens to the focal point is called the focal length. For a convex lens it is positive. On a concave lens, the light is bent such that it appears to be coming from a point. This point is also called the focal point. The focal length for a concave lens is negative. Although the light is actually bent at both surfaces of the lens, for our work, we will consider it to bend only once, at the middle of the lens.

Lens Ray Diagrams

First a little vocabulary about images. There are two types of images, real and virtual. A real image can be seen on a screen. It is formed by light that is actually focused at the location of the image. A virtual image can not be seen on a screen. Light is not actually focused at that spot. A virtual image appears behind the lens. Erect means that both image and object have the same orientation. If the object is right side up, so is the image. Inverted means the image is flipped over. If the object was right side up, the image will be upside down. Magnification refers to the size of the image as compared to the object. If the image is larger than the object, it has a magnification greater than one. If the image is smaller, the magnification is less than one.
Ray diagrams are used to show four things about an image; the location, type, orientation, and magnification. There are an infinite number of rays that can be used to locate an image. However, three of them are very convenient because they can be drawn without using a protractor. All three leave from the top of the object and follow the paths shown below. The spot where the rays meet on the right is the location of the top of the image. All rays from the base of the object will meet at a point on the optic axis even with the top of the image. In this case the image is inverted and real. Ray 1: Parallel to the optic axis and bends through (for convex) or away from (for concave) the primary focal point. Ray 2: Through (for convex) or toward (for concave) the secondary focal point and bends parallel to the optic axis. Ray 3: Passes through the center of the lens and is not bent.

Steps for solving ray diagram problems: 1. Draw an optic axis. (A long horizontal line.) Hint: Use a pencil for ray diagrams! 2. Choose a scale for your problem such that the drawing will fit and take up at least half of the width of your paper. (Letting 1 cm on your paper equal 1, 2, 4, 5 or 10 cm usually works well.) 3. Place the lens in the center of the paper. (With experience you will learn when the center is not best.) Do this by drawing a vertical line that will act as the middle of the lens. Then sketch in an outline of the shape of the lens. It is not important that the lens is perfectly shaped, only the centerline will be used. 4. Draw an upright arrow for the object on the left side of the lens. The object may be to any scale. (1, 2, or 3 cm in height usually works best.) The side you place the object (left) on is behind the lens. 5. Locate focal points f1 and f2, and placed small dots at the spots. F1 is on the right side (front) of the lens for convex lenses, and on the left side (behind) for concave. 6. Draw the 3 rays from the top of the object, bending rays 1 and 2 at the middle of the lens. (Only two of the rays are necessary, but the third gives you a way to check.) 7. At the spot where the rays meet, draw a vertical arrow for the image. If the rays do not meet in front of the lens, trace the refracted rays back behind with dashed lines until the dashed lines meet. If the rays meet below the axis, point the arrow down. If they meet above the axis, point the arrow up. 8. For the image distance (di), measure from the lens to the image and multiply it times your scale. 9. For the image height (hi), measure the length of the image arrow and multiply it by the scale you chose for the object. 10. For the magnification (M), divide the image height by the object height.
Sample problems:
1) Draw a ray diagram for an object 6.0 cm high, placed 20. cm from a convex lens with a focal length of 15 cm. When done, identify the type of image (real or virtual, erect or inverted), di, hi, and M.

[pic]
2) Draw a ray diagram for an object 4.0 cm high, placed 10. cm from a convex lens with a focal length of 20.cm. When done, identify the type of image (real or virtual, erect or inverted), di, hi, and M. [pic]

3) Draw a ray diagram for an object 10.0 cm high, placed 10. cm from a concave lens with a focal length of -20. cm. When done, identify the type of image (real or virtual, erect or inverted), di, hi,, and M. [pic]
Since concave lenses have negative focal lengths, f1 is on the left, and f2 is on the right. The problems have shown that convex lenses produce both real and virtual images. If the object is more than a focal length from the lens, the image is real. If the object is less than a focal length, the image is virtual. Concave lenses only produce virtual, smaller images.

Part 3: The thin lens equation

Two equations are needed to solve the problems mathematically. The first is called the thin lens equation:

The second is the equation for magnification: [pic] A negative answer indicates the image is erect and virtual. If you are asked to solve for magnification, you may report the absolute value. You may use any units of length in the problems, but they must all be the same. Here is a summation of the sign convention for lenses:

Sample problems:
4) a) Where will the image of a 15 cm tall candle appear if placed 50. cm from a lens with 20. cm focal length? b) How tall is the image? (This problem is an example of one that is easy to do with common denominators.)

5) A 35 mm lens (f = 35.0 mm) is used to take a picture of a 185 cm tall car, located 5.00 m from the camera. a) How far from the lens should the film be placed. b) What is the height of the image on the film? (This is an example of a problem that is not easy to do with common denominators.)

6) A candle is placed 30.0 cm to the left of a lens, and the image appears 10.0 cm to the left of the lens. What is the focal length of the lens, and what type of image is it? Since the image is behind the lens, it must be virtual. Since the image distance is 1/3 the object distance, the image must be 1/3 the size of the object. Since the magnification sign is negative, the image must be erect. The focal length is negative so the lens is diverging.

Problems:
1) Which of the following lenses are converging? diverging? convex? concave? (To figure out if converging or diverging, bring in a parallel ray of light above and below the center. Where the ray strikes a surface, sketch in a normal and bend the light appropriately. If the light is spread out it is a diverging lens, bent toward the center, converging.)

2) From a point on the top of the bulb, draw ten rays that pass through the lens and converge to a single point on the screen, forming part of the image of the bulb.

3) Consider how the real image is formed. a) What parts of the lens (top, middle, bottom, all) are struck by light from the top of the bulb?

b) What parts of the lens (top, middle, bottom, all) focus light from the top of the bulb to the point on the screen?

c) Is the entire lens needed to form the image? Explain.

d) If a mask is added which blocks light to half of the lens, in what ways will the image change and/or stay the same? Add rays to the diagram below to aid your explanation.

4) Carefully draw ray diagrams to find the image distances and the magnification of the following object distances for a 10.0 cm focal length converging lens. On your diagram indicate whether the images are erect or inverted, real or virtual. a) an object 14 cm away b) an object 5.0 cm away

5) Carefully draw ray diagrams to find the image distances and obtain the magnification for setups a) and b). On your diagram indicate whether the images are erect or inverted, real or virtual. a) an object 10.0 cm away from a concave lens where f = -15.0 cm b) an object 8.0 cm away from f = -4.0 cm lens

6) Using the thin lens equation, solve for the 4 image distances in problems 4 and 5.

7) a) Make a ray diagram to locate the image formed by the 0.92 m long baseball bat that is 5.0 m from the lens. (Because of the distances involved, this will not be to scale.)

b) The image of the bat is 0.25 m tall. Solve for the image distance for the focal length of the lens.

c) Draw an eye at a position that could see the image of the bat.

8) a) Make a ray diagram to locate the image of the tree. Draw rays from both the top and bottom of the tree.

b) The tree is 22.0 meters away from the lens, forming an image 3.0 cm high. The lens focal length is 15.0 cm. Determine the height of the tree. (This drawing is not to scale so you must solve using your equations.)

9) The diagram below shows a real image on the left and an object on the right. Make a ray diagram using both the top and bottom of the object and image. Sketch in a lens and label the focal points.

10) a) Make a ray diagram to locate the image of the paramecium (or hairy ear). Draw rays from both the top and bottom of the paramecium
b) Draw an eye at a position that could see the image of the paramecium.

c) The mutant paramecium is 1.0 cm tall and placed 3.0 cm from a lens that has a focal length of 5 cm. Determine the image height.

11) A lens and a virtual image of a person are pictured below. Draw a qualitatively correct ray diagram that shows the size and location of the actual person.

12) a) Make a ray diagram to locate the virtual image of the “A”.

b) Does the extension of the “A” above the lens affect the image? Explain.

c) Draw an eye at a position that would be able to see the image of the “A”.
d) The “A” is 40.0 cm tall, 1.0 meters from the lens, and its virtual image is 18 cm tall. Solve for the focal length of the lens.

13) A nearsighted (myopic) person can focus on nearby objects, but cannot see distant objects clearly. The ray diagram below shows the formation of a sharp image in front of the retina.

What kind of eyeglass lens should be used so that the image will form at the retina? Draw in the lens and corrected image location on the diagram.
14) Mr. Walker’s slide projector has a 12.0 cm focal length. If the slide is 12.6 cm from the lens, where should the screen be placed to have the sharpest image? b) If George Washington is 2.0 cm tall on the slide, how tall will he be on the screen? c) How is the slide positioned in the projector?

15) A newspaper is held 6.0 cm from a convex lens of 15.0 cm focal length. What is the image distance of the print? b) How many times is the print magnified?

16) Jim Naseum collects stamps. He wants to own a magnifying glass that makes the stamps look 4.0 times larger when they are held 3.5 cm from the lens. What focal length does he need?

17) Rosie (Ray Diagram’s cat) and the stump below are 1.10 meters from top to bottom. He is photographing them with a 50.0 mm lens from a distance of 350. cm. How far from the lens should the film be for the sharpest focus? b) How tall will the cat and stump appear on the negative?

18) In the human eye, the distance from the lens to the retina is around 2.50 cm. What is the focal length of the lens when viewing something 12.0 meters away? b) When viewing something 25 cm away? c) In which of these would the lens be the thickest?

19) Dick Shunary is using a Xerox machine to reduce a document to 1/3 (.333) of its original size. If the copy paper is 30.0 cm away from the lens, how far must the lens be from the document? b) What is the focal length of the lens?

----------------------- f2 ray 2

ray 3

i

o

Answers: image is real and inverted di = 12 cm x 5 = 60. cm hi = 3.0 cm x 6 = 18 cm [pic]

b) The refracted ray and two wavefronts are shown below. Draw in the incident ray, the normal and all of the missing wavefronts.

[pic]

v = 2.0x108 m/s

c = 3.0x108 m/s

Unknown: n

n=1.33

c = 3.00x108 m/s

Unknown: v

(1

(2

medium 2

medium 1

normal

60o

(2

water

air

Sketch:

n1 = 1.00 n2 = 1.33
(1 = 60o
Unknown: (2

(part 1) n1=1.33 n2=1.52
(1=40o
Unknown: (2

Sketch:

40o

water

air

glass

(1

(2

(2

(part 2) n1=1.52 n2=1.00
(1=34.2o
Unknown: (2

41o

25o

Sketch:

glass

air

(2=41o n2=1.00 (1=25o
Unknown: n1

reflected ray

90o

(c2

high

low

normal

[pic]

n1 = 1.33 n2 = 1.00
(2 = 90o
Unknown: (c

(c

[pic]

90o

water

air

Sketch:

C

C

B

B

A

A

[pic]

incident ray

45(

45(

f1

60o

60o

45o

60o

Normal

Refracted ray

Incident ray

height image (hi)

ray 3

ray 1

primary focal point

secondary focal point

object distance (do)

shallow

deep

image distance (di)

optic axis

image

Ray 2

f1

object

ray 2

f2

fast

[pic]

height object (ho)

shallow

deep

slow

Ray 1

f2

fast

slow

b) The reflected ray and wavefronts are shown below. Draw in the incident and refracted wavefronts, the normal, the incident and refracted rays.

fast

slow

slow

fast

slow

fast

air: n=1.00

n=1.60

glass

diamond

water

air: n=1.00

n=2.42

n=1.33

o

Answers: erect and virtual di = -1.67 cm x 4 = -6.7 cm hi = .67 cm x 10 = 6.7 cm [pic]

ray 3

ray 1

ray 2

i

o

1 cm = 4 cm

f2

f1

Answers: erect and virtual di = -5.0 cm x 4 = -20. cm hi = 2.0 cm x 4 = 8.0 cm
[pic]

Since the rays will never focus on the right side, they are traced back behind with dashed lines to find the image. Behind the lens yields a negative image distance. Ray 2 comes from f2 instead of going through it.

ray 1

1 cm = 5 cm

f1

Ray 3

i

1 cm = 4 cm

[pic]

do is positive (+) for real objects (objects behind the lens), negative (-) for virtual objects. di is positive (+) for real images, negative (-) for virtual images. f is positive (+) for convex lenses, negative (-) for concave lenses.

Knowns: do = 50 cm f = 20 cm ho = 15 cm
Unknowns:
di hi

Solution: (part a)
[pic]

Equation:(part a)
[pic]

Solution: (part b)
[pic]

Equation: (part b)
[pic]

Solution: (part a)
[pic]

Knowns: do = 5.00 m = 500. cm f = 35.0 mm = 3.50 cm ho = 185 cm
Unknowns:
di hi

Solution: (part b)
[pic]

Solution:
[pic]

Knowns: do = 30.0 cm di = -10.0 cm
Unknown: f

glass

glass

glass

glass

glass

air

water

top of bulb

top of bulb focused here

top of bulb

mask blocks top half of lens

f

f

f

f

f

f

f

f

f

f

Image forms in front of the retina.