Physics E1bx April 7 – April 14, 2015

Physics E1bx: Assignments for April 7 – April 14 Homework #7: Light Rays; Reflection, Refraction; Lenses Due Tuesday, April 14, at 6:00PM

This assignment must be turned in by 6:00PM on Tuesday, April 14. Late homework will not be accepted. Please write your answers to these questions on a separate sheet of paper with your name and your section TF’s name written at the top. Turn in your homework to the mailbox marked with your section TF’s name in the row of mailboxes outside of Sci Ctr 108.

You are encouraged to work with your classmates on these assignments, but please write the names of all your study group members on your homework.

After completing this homework assignment, you should be able to… • understand what it means for an image to be real, virtual, upright, inverted, reduced, and enlarged • use the three principal rays (perpendicular, parallel, focal) to perform ray tracing for planar, concave, and convex mirrors • use the mirror and magnification equations • identify whether a lens is converging or diverging • use the three principal rays (central, parallel, focal) to perform ray tracing for converging and diverging lenses • use the thin lens and magnification equations • perform ray tracing for a combination of lenses • apply the thin lens equation and magnification equation to combination of lenses

1 Physics E1bx April 7 – April 14, 2015 Here are summaries of this module’s important concepts to help you complete this homework:

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1. Lens and mirror (2 pts). A bright object is 20 cm from a thin lens with focal length f = 12 cm. 25 cm behind the lens is a convex spherical mirror with radius of curvature r = 20 cm, as in the figure below (not to scale).

a) Draw a ray diagram for the lens only and locate the image formed by the lens. Calculate the image distance. Where is the image relative to the mirror?

The ray diagram: parallel ray (blue), focal ray (green), central ray (red)

1 1 1 1 1 1 = − = − = di f do 12 cm 20 cm 30 cm

The image is 30 cm behind (to the right of) the lens, which means it is 5 cm to the right of the mirror surface. € b) Taking the mirror into account, calculate the position of the final image formed by the mirror. Is the final image real or virtual? Calculate the magnification. Is it upright or inverted relative to the original object? Larger or smaller?

The image produced by the lens becomes the object for the mirror. Since the object is located behind the mirror, the object distance is negative. (This is an example of a virtual object.) Plugging in the negative object distance into the mirror equation: 1 1 1 1 1 1 = − = − = di f do −10 cm −5 cm 10 cm The final image is 10 cm in front of the mirror, which is 15 cm behind the lens. Since the image is in front of the mirror, the light rays actually do converge at the image location, which makes it a real image. € 5 Physics E1bx April 7 – April 14, 2015

We can calculate the magnification from the mirror alone from the magnification equation:

diM 10 cm mM = − = − = +2.0 , doM −5 cm which means that the mirror image is twice as big (and not inverted) relative to the mirror object. But the mirror object was the image produced by the lens, which also had been magnified:

diL 30 cm € mL = − = − = −1.5, doL 20 cm meaning that the lens produced an image which was 1.5 times as big and inverted relative to the original object. The total magnification is given by the product of the two, which means that the final image is inverted and 3 times as big as the original object. € We could also draw a ray diagram for the mirror, but it’s a little strange because the object is behind the mirror, so light isn’t actually coming from it. (This was not asked of you, but you might be curious to know how it’s done.) Instead, we have to imagine the principal rays as incoming rays that are aimed at the object, and get reflected in the usual way. The parallel ray (blue) comes in parallel aimed at the virtual object, and reflects on a line through the focal point. The focal ray (green) comes in directed at both the virtual object and the focal point, but reflects parallel to the axis. And the perpendicular ray (red), comes in directed at the virtual object and the center of curvature, strikes the mirror at normal incidence, and returns exactly along the line it came in. as you can see from the diagram below, the rays do intersect to form a real image 10 cm in front of the mirror.

2. Optometry practice (2 pts). Your friends Abelard and Bernadette are both farsighted, and both wear glasses to correct their vision. Abelard has an unaided near point of 120 cm, and Bernadette has an unaided near point of 60 cm. a) Assume that both wear their glasses so that the lens is 2 cm in front of their eyes. What focal length would you prescribe for each of your friends to correct their farsightedness?

For farsightedness, the goal of corrective lenses is to take an object at the desired near point (25 cm from the eye, usually) and produce an image at the patient’s unaided near point. That image then becomes the object for the second lens in the “system” (the patient’s eye).

Let’s start with Abelard. An object 25 cm from his eye is only 23 cm from the lens, so do = +23 cm. When light from that object passes through the lens, we want it to form an image 120 cm from his eye, which is 118 cm from the lens. (i.e., on the same side as the object), which corresponds to a di = –118 cm. Plugging these into the thin lens equation, we can solve for f:

6 Physics E1bx April 7 – April 14, 2015 1 1 1 0.035 = + = fA 23 cm −118 cm cm so fA = 29 cm.

For Bernadette, we do the exact same thing, except that we want the image produced by the glasses to be 60 cm from her eye instead of 120. That means di = –58 cm, so: 1 1 1 0.026 = + = fB 23 cm −58 cm cm or fB = 38 cm. b) Abelard forgets his glasses one day and borrows Bernadette’s. What will be his near point when wearing the wrong glasses?

We want to know the object distance at which the glasses he’s wearing would produce an image at his unaided near point. If he’s wearing the right glasses, this is 25 cm from his eye, or an object distance of 23 cm from the lenses. But when he’s wearing the wrong glasses, we just need to solve the thin lens equation for do: 1 1 1 1 1 = + = + fB do di do −118 cm 1 1 1 = + do 38 cm 118 cm

do = 29 cm The near point is measured from the eye, so his near point while wearing Bernadette’s glasses is 2 cm longer than this, or 31 cm.

3. Mirror, mirror (2 pts). You peer into a curved mirror at a distance of 1 m. The image of your face in the mirror is inverted and half the size of your face. a) Where is the image located? Is it real or virtual?

The magnification is -0.5 and the object distance is +1 m. Calculating the image distance:

di m = − ⇒ di = −mdo = +0.5 m. do We can conclude that the image is 0.5 cm in front of the mirror. Because the image distance is positive, the image must be real. a, cont.) What is the mirror’s radius of curvature? Is it concave or convex?

We can use the mirror equation to solve for the focal length: 1 1 1 1 1 1 = + = + = f do di 1 m 0.5 m 33.3 cm The mirror’s focal length is 33.3 cm, which means the radius of curvature is r = 2 f = +66.6 cm . A positive r means the center of curvature is in front of the mirror, so the mirror is concave.

7 Physics E1bx April 7 – April 14, 2015 b) Draw a ray diagram for the image of your face in the mirror. Your diagram should include the mirror, object, image, focal point, and three principal rays. You should use a straightedge, but the diagram does not have to be exactly to scale.

The diagram is shown below. The parallel ray is in blue, focal ray in green, and perpendicular ray in red.

8 Physics E1bx April 7 – April 14, 2015 4. Putting Everything Together (Exam-Type Question): Compound Microscope (4 pts)

Two thin convex lenses and an object are arranged as shown below. Two rays from the tip of the object are drawn in order to determine the location of the image produced by lens 1. Lens 2 is placed so that one of its focal points coincides with the location of the image produced by lens 1.

a) Treating the image produced by lens 1 as an object for lens 2, draw two principal rays from the tip of this image that pass through lens 2. (Note that one of the principal rays cannot be drawn in this case.)

Treating the image produced by lens 1 as the object for lens 2, the two principal rays drawn are the central ray (blue) and the parallel ray (purple). These rays produce a virtual image (shown in red), as shown in the diagram above. We cannot draw the focal ray in this case because the image produced by lens 1 is already at the focal point for lens 2.

b) Where is the tip of the image seen by the observer located? Explain. (Hint: From where do the rays on the right side of lens 2 appear to have come?)

The tip of the image seen by the observer is located to the left of lens 1. The observer sees an enlarged, inverted, virtual image of the object. Lens 1 produces a real, enlarged, inverted image

9 Physics E1bx April 7 – April 14, 2015 located at the focal point of lens 2. Lens 2 then uses this image as its object and creates an even more enlarged, inverted, virtual image that is located behind the original object.

c) Which would appear larger: the image seen by the observer (with both lenses present) or the object itself (with the lenses removed)? Explain.

The image seen by the observer (with both lenses present) appears larger than the actual object itself. This is because each of the lenses produces an enlarged image of its object. Lens 1 magnifies the object according to �!! ℎ!! �! = − = − �!! ℎ!! where d1i and d1o are the image distance and object distance for lens 1, respectively, and h1i and h1o are the image height and object height for lens 1, respectively. As seen from the ray tracing,

�!! > �!! so the magnification of lens 1 is greater than 1. Then,

ℎ!! > ℎ!! which means that height of the image produced by lens 1 is larger than the height of the object.

Lens 2 similarly magnifies its object (the image produced by lens 1) according to �!! ℎ!! �! = − = − �!! ℎ!! Because

�!! > �!! the magnification of lens 2 is greater than 1, so

ℎ!! > ℎ!!. This means that height of the image produced by lens 2 is larger than the height of the object. Since the object for lens 2 is the image produced by lens 1,

��� > ��� = ��� > ���

The image seen by the observer is larger than the object itself. Note that in this explanation, we were considering the magnitudes of all variables.

d) The diagram in this problem illustrates a compound microscope. Lens 1, called the objective, is placed near the object of interest. Lens 2, known as the eyepiece, is placed so that one of its focal points coincides with the image produced by the objective.

In order to improve the angular magnification of the microscope shown above, would you replace the eyepiece (lens 2) with another lens that has a smaller focal length or a larger focal length? Explain your reasoning.

The total magnification of the microscope is equal to the product of the individual magnifications: �!"!#$ = �!"#�!"!#$!%! The magnification due to the eyepiece can be written as �!! �!"!#$!%! = − �!!

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where d2i and d2o are the image and object distances for lens 2, respectively. Since the image for lens 2 is placed at its focal point,

�!! = �! and �!! �!"!#$!%! = − . �! So, decreasing the focal length of lens 2 will increase the angular magnification.

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