Draw the image the object the correct location

An object O is placed at the location shown in front of a concave spherical mirror. Use ray tracing to determine the location and size of the reflected image. As you work, keep in mind the following properties of principal rays:

1. A ray parallel to the axis, after reflection, passes through the focal point F of a concave mirror or appears to come from the (virtual) focal point of a convex mirror.

Trace the path of a ray emitted from the tip of the object through the focal point of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray.

Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.

Now trace the path of a ray emitted from the tip of the object parallel to the axis of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray.

Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.

Now, using the two rays you have traced, draw the image of the object at the correct location, with the correct orientation and size.

Draw the vector ending at the intersection of the reflected rays. The location, orientation, and length of the vector will be graded.

ANSWER:

1. A ray parallel to the axis, after reflection, passes through the focal point F of a concave mirror or appears to come from the (virtual) focal point of a convex mirror.

2. A ray through (or proceeding toward) the focal point F is reflected parallel to the axis.

Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.

ANSWER:

Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.

Hint 1. Constructing the reflected ray

Correct

Part C
What type of image of the object will the convex mirror create?

Part D
Now, using the two rays you have traced, draw the image of the object at the correct location, with the correct orientation and size. Start by extending the existing virtual rays from the surface of the mirror where the reflected rays touch the mirror.

Draw the vector ending at the intersection of the virtual rays. The location, orientation, and length of the vector will be graded.

Correct

Vittorio would like to be able to see the logo on his shirt. Draw the incident and reflected rays showing the light from the logo reflecting off the mirror into his eyes. The rays should meet at the point on the mirror that needs cleaning.

Adjust the existing vectors on the diagram, for the reflected ( R ) and incident ( ) ray, starting and ending at the surface of the mirror I
respectively. The location and orientation of the vectors will be graded.

ANSWER:

(See the hints from Part A if you need help.)

Adjust the existing vectors on the diagram, for the reflected ( R ) and incident ( ) ray, starting and ending at the surface of the mirror I

Images Produced by Different Concave Mirrors Conceptual Question

The same object is placed at different distances in front of six different concave spherical mirrors. Each mirror has the focal length listed below. f

Part A

Hint 2. Definition of a real image

A real image is an image that is formed when light rays originating from an object are reflected off the mirror and intersect. If the light rays from the object intersect only when they are extended through the mirror, then the image formed is virtual.

ANSWER:

Hint 1. Ratio of focal length to object distance from focal point

You can determine the magnification of the image using the object distance, , and focal length, . The ratio of the focal length to object f

Reflection and Refraction Ranking Task

A ray of light is incident onto the interface between material 1 and material 2.

Part A

Hint 1. Distinguish between reflection and refraction

The light that reflects off of the interface does so in a similar manner to light reflecting from a mirror. The light that refracts through the interface has its path bent by the differing indices of refraction of the two materials. A phase shift occurs in only one of these two processes. Which one?

	reflection refraction

Correct

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

Hint 1. Phase change upon reflection

Correct

using the ray tracing technique. Which diagrams are accurate? Type A if you think that only diagram A is correct, type AB if you think that only diagrams A and B are correct, and so on.

Hint 3. A ray that passes through the middle of the lens

A ray that passes through the middle of a concave lens continues on its original direction with essentially no displacement after passing through the lens.

Part B

If the focal length of the concave lens is 7.50	cm	, at what distance	do	from the lens should an object be placed so that its image is formed 3.70	cm

equation.

Hint 2. The thinlens equation

	di	for a concave lens that forms an image 3.70	cm

Express your answer in centimeters.

Hint 1. Sign convention for image distances

ANSWER:


do	= 7.30

Express your answer numerically.

					m
							,
where	do	and	di	are the object distance and the image distance, respectively.

Where should the object be moved to have a larger magnification?

Hint 1. Magnification and image size.

Correct

Tactics Box 23.2 Ray Tracing for a Converging Lens

A. Draw an optical axis. Use graph paper or a ruler. Establish an appropriate scale.

B. Center the lens on the axis. Mark and label the focal points at distance on either side. f
C. Represent the object with an upright arrow at distance . It is usually best to place the base of the arrow on the axis and to draw the arrow about half the radius of the lens.

E. Extend the rays until they converge. This is the image point. Draw the rest of the image in the image plane. If the base of the object is on

The diagram below shows the situation described in the problem. The focal length of the lens is labeled ; the scale on the optical axis is in f
centimeters.

Draw the three special rays, Ray 1, Ray 2, and Ray 3, as described in the tactics box above, and label each ray accordingly. Do not draw the refracted rays.

Part B

Now, draw the refracted segments of the three special rays considered previously. Use the labels	Ray1r Ray2r	, and	Ray3r

ANSWER:

	at approximately 2	cm
	at approximately 2
	at approximately 4
	at approximately 4
	at approximately 6

Learning Goal:

To practice Tactics Box 23.3 Ray Tracing for a Diverging Lens.

4. Draw the three "special rays" from the tip of the arrow. Use a straightedge.

1. A ray parallel to the axis (Ray 1) diverges along a line through the near focal point. 2. A ray along a line toward the far focal point (Ray 2) emerges parallel to the axis. 3. A ray through the center of the lens (Ray 3) does not bend.

Part A

The diagram below shows the situation described in the problem. The focal length of the lens is labeled ; the scale on the optical axis is in f
centimeters.

Correct

Part B

	Ray1r Ray2r	, and	Ray3r

ANSWER:

at approximately 3.5
at approximately 3.5		cm on the same side of the lens
at approximately 5	cm on the opposite side of the lens
at approximately 5

To practice Tactics Box 23.4 Ray tracing for a spherical mirror.

The procedure known as ray tracing is a pictorial method of understanding image formation when lenses or mirrors are used. It consists in locating the image by the use of just three "special rays." The following Tactics Box explains this procedure for the case of a concave mirror.

1. A ray parallel to the axis (Ray 1) reflects through (concave) or away from (convex) the focal point.

2. An incoming ray passing through (concave) or away from (convex) the focal point (Ray 2) emerges parallel to the axis.

ANSWER:

	Ray1r	for the reflected segment of Ray 1,	Ray2r	for

ANSWER:


	at about 5.5	cm
	at about 7.5	cm
	at about 13	cm

Learning Goal:

To understand how the properties of a convex lens and the distance between the object and the lens affect the image.

Feel free to play around with the simulation. When you are done, begin part A.

Part A

The screen is then at the focal plane. As the lamp is moved closer to the lens, the distance between the focal plane and the lens _____.

ANSWER:

Correct

When the lamp is closer to the lens, the rays going into the lens are diverging more quickly than when the lamp is further away. Thus, after going through the lens, they aren't converging as quickly, so it takes a longer distance for them to focus.

Do the rays that go through the lens always converge on the right side of the lens (forming a real image), regardless of the position of the lamp?

ANSWER:

Correct

When the object is at the focal point, the rays neither diverge nor converge, but are perfectly parallel. Conversely, if the object were at infinity, the rays going from the object to the lens would be parallel, and after passing through the lens, they would converge at the focal point.

Part E

The focal length __________ when the diameter of the lens is increased.

ANSWER:

Correct

The focal length does not depend on the lens's diameter; a greater diameter simply allows more light to be focused.

The focal length of the lens is ____________, and the focal plane is ___________ from the lens.

Hint 1. How to approach the problem


30	cm		/ 120
120		cm		/ 120
60	cm		/ 60
60	cm		/ 180		cm
60	cm		/ 120		cm

Now, move the lamp so that it is 90	cm	from the center of the lens.


1			+	1	=	1
s				s′		f		if is twice the focal length of the lens
If decreases, then	s′	must increase (as you found in Part A). Notice that, as you found in Part F,					s′= s	if is twice the focal length of the lens

(1/120 + 1/120 = 1/60).

greater than the focal length of the lens)?

ANSWER:

Correct
The image gets very big as the object approaches the focal point.

						, where	h′
then	h′	is negative.				, where	h′
Place the pencil 90			cm	from the lens. What is the magnification of the image (be sure the curvature radius is still 0.6

How far from the lens does the pencil need to be for the magnification	M	to be 1?

Notice that, as you saw earlier for this lens, when the object is 120 cm away, the image is also 120 cm , so the ratio of the distances and equal to one. In fact, as you might have guessed by now, the magnification can also be expressed as M = −s′	s′

One of the contests at the school carnival is to throw a spear at an underwater target lying flat on the bottom of a pool. The water is 1.20 m deep. You're standing on a small stool that places your eyes 3.20 m above the bottom of the pool. As you look at the target, your gaze is 30 below horizontal. At what∘

angle below horizontal should you throw the spear in order to hit the target?

35.5 ∘

Correct

Express your answer as an integer.

ANSWER:

	(x;y)	coordinates of each image?

Express your answer using two significant figures. Give your answer in the form (x;y). If there is more than one answer, separate them by commas. Give your answer in meters.

	mm/s ) of the sprinter's image at the instant the sprinter is 15

ANSWER: