AP Physics 2 – Optics Review Questions

1. The heating energy inside a microwave oven consists of electromagnetic waves.  Suppose at a particular instant in time the magnetic field inside can be described by the function B = 13 sin(51.3x), where B is in microteslas and x is position in meters.  (a)  Sketch a graph of this function and determine the position of the first “crest”.  (b) At this same position what would be the magnitude of the electric field?  (c) How are the two fields oriented relative to one another? (d) Determine the wavelength and frequency of the wave.  (e) The explanation of how it heats is that polar molecules such as water in the food oscillate back and forth in response to the electromagnetic wave – which field, electric or magnetic, would cause this and why?

2. In a popular do-it-yourself amateur scientist experiment a chocolate bar is placed in a microwave oven and zapped for about 30 seconds or so until melted “spots” form.  Measuring the distance between melted spots and then multiplying by 2 and then multiplying by 2450 million Hz is “supposed to” yield the speed of light.  The melted spots apparently get more energy (while in between it remains cool) – interference can explain this.  (a) The hot spots are nodes or antinodes?  Explain.  (b) There is only one source generating microwave energy – if there is only one wave source explain some ways that it may be possible for interference to occur.  (c) Explain why the procedure explained above should yield the speed of light.

3. The surface of a music compact disc has rows of “pits” that a laser reads when the CD is played.  Such a disk is placed on the floor of a room and a laser is pointed straight down onto its surface.  The rows of pits act like a grating (a “reflection grating”) and this produces an interference pattern on the 8.0 foot ceiling.  There are a pair of bright spots 3.0 feet on either side of a point directly above the CD. (a) Determine the location of a second pair of dimmer spots on the ceiling.  (b) Given that the rows of pits are separated by 1.5 μm what is the wavelength of the laser?  (c) At what angles would antinodal lines appear if the experiment is repeated with a red laser of frequency 480 THz?  (d) Essentially equivalent patterns could be produced with a transmission grating with how many lines per millimeter?

4. Light of wavelength 400 nm is incident on a pair of slits.  An interference pattern occurs on a screen located 2.00 m from the slits.  Bright fringes separated by 0.50 cm appear on the screen.  (a) Determine the separation of the two slits.  (b) A dark point on the screen between the 1^st and 2^nd order fringes is how much closer to one slit than the other? (c) What order line (in theory) should appear at an angle of 30.0° from the central line?  (d) A single slit of what width in this same setup would replace the 1^st order bright lines with dark lines?  Explain.

5. The following graph shows the relative intensity of light versus angle for a two-slit interference pattern.  The angle value is in radians on this graph.

(a) At what angle (in degrees) is the 2^nd order bright line?  (b) At what angle is there a nodal line along which the path difference is 2.5 wavelengths of the light?  (c) How many times wider is the separation of the slits than the wavelength of the light?  (d) Sketch a graph of intensity versus angle with all of the same parameters as this except it is produced by a diffraction grating such that the light passes through many openings.

6. Brewster’s angle is defined by the equation tan θ = n for light reflecting from material with index of refraction n.  Light reflected at this angle should be perfectly polarized (at other angles as well but to a lesser extent).  (a) Determine this angle for water, n = 1.33.  (b) Find the angle of incidence for light that is traveling initially underwater that emerges into the air at Brewster’s angle.  (c) Polarizing sunglasses are designed to eliminate this reflected light.  Explain how this helps a fisherman see underwater. 

7. Thin film interference is a phenomenon responsible for the “rainbow” colors seen in a soap bubble (not to be confused with an actual rainbow).  It occurs when light encounters a thin transparent material and undergoes partial reflections.  In order for any type of interference to occur there must be multiple waves – in this case two reflected waves.  (a) Draw a ray diagram showing a thin film (like a transparent sheet) and show the first two partial reflections that would occur for light entering at an angle.  (b) Given that the two sides of the film are parallel, show that the secondary reflection emerges from the material traveling parallel to the first reflection.  (c) Suppose a ray enters a soap film at angle of incidence 10.0° on the top surface.  Find the angle of refraction using n = 1.33.  (d)  Given the thickness of the film is 400 nm determine the total distance traveled through the material by the light that reflects off the bottom surface (and emerges parallel to the first reflection).  (e) When the secondary reflection combines with the first reflection why does the type of interference that occurs depend on the wavelength (hint – the two waves have traveled a different distance)?

8. Consider what happens when you look at yourself in a plane mirror.  (a) What type of image do you see – real or virtual?  Explain.  (b) How does the distance to the image compare to your distance from the mirror?  It is like seeing yourself from how far away?  (c) Would the thin lens/mirror equations work for a plane mirror?  If so, what is the “focal length” of a plane mirror?  Explain.

9. Suppose you had a lens with an “adjustable” focal length and you cast an image of an object onto a screen at a fixed image from the lens.  (a) As the object distance decreases what must happen to the focal length to keep the image focused?  (b) What change(s) to the lens would cause the focal length to change as required?  (c) Hey – you know what?  You have a lens like that – you are using it right now!  It’s in your eye!  As you look at an object that gets closer and closer your eye’s lens has to adjust to keep a real image formed and focused on the retina, which is at a fixed distance from the front of the eye.  What do you think about that?!

10. A certain pair of reading glasses (the ones I am wearing as I type this) will project an image of a recessed light in an 8.0 foot ceiling onto the floor when held directly below the bulb at a distance of 6.0 feet.  (a) Describe the type of image of the light bulb.  (b) Determine the focal length of the reading glasses.  (c) What type are the lenses in the glasses?  The side that faces my eyes is concave – what must be true of the other side (that faces outward)?  (d) The same glasses held a distance 1.6 feet from a book should produce an image where and what type and magnification?  (e) A distance of 1.6 feet is a reasonable distance at which I might read something with these glasses – why don’t I see the image described in the previous problem?  (Hint: think about where the image is and where my eyes are.)

11. The Aricebo observatory in Puerto Rico is a radio telescope consisting of a huge spherical concave reflector built into a natural bowl shaped depression in the landscape there.  The “mirror” reflects radio waves but otherwise acts like a regular optical mirror that would be found in a visual telesctope.  In December of 2017 scientists used Aricebo to measure a radio image of the asteroid 3200 Phaeton at its closest approach to Earth – a distance of 10.3 million kilometers.  The focal length of the reflector is 133 m and a radio wave detector and transmitter is located near the focal point, suspended by a support structure directly over the center.  (a) Calculate the image distance and use as many digits as your calculator will allow without rounding.  How close to the focal point is it?  (b) Phaeton comes closer to Earth than most objects in the solar system – what happens to the answer to part (a) for objects farther away?  Where should the radio wave detector be located?  (c) Given the diameter of Phaeton is 5.8 km what was the size of the image?  (d) To detect Phaeton with this device the scientists have to send radio waves at it so that the waves bounce off and return to the telescope.  The reflector is used for this purpose also.  If radio waves are generated and emanate from the focal point how does this energy travel out into space?  How is this a little like a radio flashlight?  (e) Because the reflector is so huge (300 m in diameter!) it cannot be moved and pointed at objects in the sky.  Instead the detector and transmitter are pointed in different directions on the spherical surface.  Explain how this achieves the same effect as pointing the telescope, given that it is a spherical surface.  (f) Calculate the radius of the spherical surface.

12. A fairly common “lawn decoration” is a shiny mirror-like sphere on a pedestal (how did that get started?)  Suppose that such a sphere has a diameter of 30.0 cm and a 1.8 m tall farmer is standing 3.00 m away from it, looking at it.  (a) Determine the image type, location, and size.  (b) If the farmer moves to half the distance how much bigger does he look reflected in the sphere?

13. There are lots of physics problems in the world!  Try working examples from the Giancoli book or the Open Stack online book!

Answers:

1. a. x = 3.06 cm
b. 3900 N/C
c. The electric field is perpendicular to the magnetic field.  Both fields are perpendicular to velocity.
d. 12.2 cm, 2.45 GHz
e. Polar molecules are positive on one side, negative on the other.  The electric field causes a torque and angular acceleration of the molecule.  Because this field is oscillating at any position as the wave goes by it causes the molecule to rotate back and forth.

2. a. The hot spots that melt in the chocolate bar must correspond to antinodes (or antinodal lines) occurring in the oven.  In either case waves are combining with constructive interference which increases the magnitude of the fields.  Between antinodes (or antinodal lines) there must be nodes (or nodal lines) where destructive interference is causing a reduces magnitude of the fields and hence explains the lack of melting between hot spots.
b. Even though there is only one source of waves it is possible for waves to reflect of the walls of the cavity.  The interference is the result of reflect waves encountering other waves.  Another possibility is that waves emerge from an opening that acts like a single slit, which can also result in interference due to Huygen’s principle.
c. Twice the distance between antinode should equal the wavelength assuming there is in essence a standing wave.  Multiplying wavelength times frequency equals speed for any wave phenomenon.

3. a. 7.9 feet on either side
b. 530 nm
c. 25°, 56°
d. 670 lines/mm

4. a. 0.16 mm
b. 600 nm (1.5 wavelengths difference, resulting in destructive interference)
c. 200
d. 0.16 mm – same equation now yields a location of a dark line – destructive interference explained by Huygen’s principle:  the dark location on the screen is half a wavelength closer to pairs of point sources all across the width of the slit.

5. a. 2.4°
b. 3.0°
c. The slits are separated by a distance about 47 times greater than the wavelength.
d.

6. a. 53.1°
b. 36.9°
c. Polarizing sunglasses have parallel long molecules in the material that are oriented to block the polarized light that reflects off of various materials.  If a fisherman is viewing a fish and the light follows the angles calculated in part (b) then his polarizing sunglasses would completely block reflected light that would otherwise “overlap” and perhaps “overwhelm” his view of the fish.

7. a.

b. First reflection is at angle θ₁, transmitted light is at angle θ₂.  Therefore the second reflection is at angle θ₂.  But θ₂ is then the resulting angle of incidence moving upward from film to air.  Therefore it emerges at the angle θ₁ by Snells’s Law.
c. 7.5°
d. 808 nm
e. Depending on the difference in the distance the crests and troughs of one wave will line up a certain way with the crests and troughs of the other wave.  (BTW – this will be constructive if the path difference is an odd multiple of a half wavelength because the first reflection from the film is inverted.)

8. a. You see a virtual image whenever you look at anything in a plane mirror.  The light producing the image appears to come from the other side of the mirror but does not actually originate or intersect on the other side of the mirror (obviously).  Such an image cannot be projected onto a screen.
b. Distance from mirror to image is equal to distance from mirror to object.  You are seeing yourself as you would appear from a distance twice as far away as you are from the mirror.
c. The thin lens/mirror equations work just fine for a plane mirror!  The catch is that the focal length is essentially infinity.  this makes the term in the equation 1/f  “disappear” by dropping to zero, which means that d_i equals negative d_o and h_i = h_o – just as expected by the type of virtual image that you see!

9. a. As object distance decreases the focal length must also decrease in order to keep the image distance the same.  At relatively small object distance the light must be bent more in order to keep the image in the same place.
b. If the index of refraction could be increased this would reduce the focal length because the light would bend more.  Or if the shape of the lens could be made more curved (more drastically convex) this would also cause more bending and reduce the focal length.
c. Pretty cool!  (The lens in your eye changes shape as it is pulled by muscles.)

10. a. Because it is projected it must be real and inverted, because the image distance is less than object distance it is reduced in size.
b. 1.5 feet
c. The lenses must be converging lenses – if one side is concave the other side must be convex and “more so”.  i.e. The convex side has a smaller radius of curvature than the concave side such that the lenses are thicker in the center than at the edges.
d. Image is 24 feet from lens (25.6 feet from the book).  It is real, inverted, and magnified 15 times.
e. I can assure you that the book is not inverted, nor 15 times bigger when viewed through my reading glasses.  The simple reason is that my eyes are not looking toward the real image – it would be about 23 feet behind me (if my head wasn’t in the way preventing it from forming).  Instead I see the image that forms upon my retina, which is affected by two lenses – that of the reading glasses and the lens in my eye.  Without the eyeglasses my old eyes cannot “bend enough” to form a focused image on the retina.  The lenses cause the light from the book to “begin converging” before entering my eye and therefore my eyes do not have to bend as much to focus on the print in the book.

11. a. The image is 1.7 μm farther from the mirror than the focal length.
b. For pretty much any object in the solar system the image distance will be essentially equal to the focal length – the farther away the closer to the focal point.  The detector and transmitter are located at the focal point.  This is true for all astronomical telescopes because of the vastness of space.
c. The image of Phaeton was only 75 μm in size at most – really tiny!  Special techniques are used to create a visual representation of the asteroid based on the radio waves that are bounced off its surface and subsequently detected.
d. This radio telescope not only receives radio waves but also sends them out!  Radio waves spreading out from equipment at the focal point reflect off the surface and travel like a beam traveling in one direction parallel to the principle axis.  This is the law of reflection “in reverse” of waves that come into the telescope from a great distance and converge on the focal point.
e. Because it is a spherical surface the principle axis can basically intersect any point on the reflector.  The principle axis would go from the center of the sphere to any point on the surface.  The equipment suspended above the mirror can be moved to effectively create a variable principle axis that does not have to point straight upward from the bottom of the mirror.
f. The radius of the surface is twice the focal length, R = 266 m.

12. a. The virtual upright image is 7.32 cm “behind” the surface of the sphere and it is 4.4 cm tall.
b. At half the distance the image is the same type and nearly in the same place but now it has height 8.6 cm, which is 1.95 times bigger.  However, from the perspective of the farmer it actually appears roughly four times bigger because not only is the image height twice as great he is also now twice as close to it.  If this doesn’t make sense to you try this simple experiment:  look at each of your index fingers with one of them twice as far away from your face than the other.  The one at half the distance looks twice as big.  (One could also point out that the nearer finger creates an image on the retina twice the size of the farther finger and your retina determines “what it looks like”!  In the case of the farmer the image on the retina of the image in the shiny sphere will be four times bigger when he moves twice as close.)