Exam 3 Solutions -- VERSION B

Conceptual Questions (5 points each)

Questions 1 -- 3 refer to the following situation.

$\parbox{2in}{A driver, traveling to the east, approaches a red light ($f= 4.688\ \times\ 10^{14}$\ Hz). }$

1. What is the wavelength of the red light?

   (d) 640 nm

   $$\lambda f = c \quad\Rightarrow\quad \lambda = \frac{c}{f} = \frac... ...s\ 10^{14}\ {\rm s^{-1}}} = 6.4\ \times\ 10^{-7} \ {\rm m} = 640\ {\rm nm}\ \ .$$

2. What would the driver's speed be, if he perceived the light as green ( $f= 5.555\ \times\ 10^{14}$ Hz)?

   (b) $5.548\ \times\ 10^7\ {\rm m/s}$

   The driver perceives a Doppler shifted light ray, with frequency

   $$f' = f\ \left(\ 1 - \frac{u}{c}\ \right)\ \ ,$$
   where $u$ is the relative speed between the driver and the stoplight. Thus
   $$u = -c\ \left(\ \frac{f'}{f} - 1\ \right) = - 3.0\ \times\ 10^{8}... ...\left(\ \frac{5.555}{4.688} - 1\ \right) = -5.548\ \times\ 10^7\ {\rm m/s}\ \ ,$$
   and the driver is approaching the light with speed $5.548\ \times\ 10^7\ {\rm m/s}$ (very fast indeed!).

3. An electromagnetic wave, traveling from the stoplight to the driver, has an electric field which oscillates up, then down, with an amplitude of 18 N/C. What is the magnetic field of the electromagnetic wave?

   (e) It has amplitude $6\ \times\ 10^{-8}$ T, and oscillates north then south, in phase with the electric field.

   $$B= \frac{E}{c} = \frac{18\ {\rm N/C}}{ 3.0\ \times\ 10^{8}\ {\rm m/s}} = 6\ \times\ 10^{-8}\ {\rm T}\ \ .$$

   Thus the magnetic field has amplitude $6\ \times\ 10^{-8}$ T. Its direction of oscillation is related to that of $\vec{E}$ and the direction of propagation by the right hand rule: $\vec{E}\ \times\ \vec{B}$ points in the direction of propagation. Since $\vec{E}$ oscillates up (then down), and the propagation velocity is westward (toward the car), $\vec{B}$ oscillates north (then south).

   
   

   Questions 4 -- 6 refer to the following situation:

   
   

   An object is placed in front of a concave spherical mirror as shown below. The three principal rays 1, 2, and 3 leave the top of the object and, after reflection, converge at a point on the top of the image. Ray 1 is parallel to the principal axis, ray 2 passes through the focal point F, and ray 3 passes through the center of curvature C.

   

4. Which ray(s) will pass through F after reflection?

   (a) 1 only

   Rays initially parallel to the principal axis all are focused at $F$ after reflection; this is the definition of the focal point. The other rays don't reflect back through F: ray 2 emerges parallel to, and below, the principal axis, while ray 3 is reflected straight back on itself.

5. Which ray(s) will reflect straight back on itself(themselves)?

   (c) 3 only

   Ray 3 approaches the mirror along a radius of the spherical mirror; thus it is normal to the mirror, with angle of incidence $0^o$. It is thus reflected straight back on itself, with angle of reflection $0^o$. The other 2 rays, as discussed above, are not reflected straight back on themselves.

6. Which one of the following groups of terms best describes the image?

   (d) real, inverted, reduced

   See the completed ray diagram above. The image is manifestly inverted and reduced; it is real because the reflected rays actually pass through the image.

7. Which one of the following statements concerning a virtual image produced by a mirror is true?

   (d) A virtual image is always upright relative to the object.

   The magnification equation, $m = h_i/h_o = -d_i/d_o$, confirms that a virtual image (negative $d_i$) is always upright (positive $h_i$) relative to the object. Thus choice (c) never occurs (a virtual image cannot be inverted with respect to the object). Choice (e) also never occurs, as only real images can be photographed or projected onto screens, since only real images have actual light rays delivered to the image to expose film or scatter from a screen. Finally, a convex mirror produces virtual images which are smaller than the object, while a concave mirror, with object inside the focal point, produces a virtual image larger than the object. Thus neither (a) nor (b) is true.

   
   

   Questions 8 -- 10 refer to the following situation:

   
   

   $\parbox{3.8in}{A beam of light that consists of a mixture of red, gr... ...ated in the table. An observer is located to the right of the prism as shown. }$

   color	n
   red	1.43
   green	1.40
   violet	1.37

   
   

   

8. At what angle $\theta$ will the observer see green light?

   (b) $82^o$

   Snell's law tells us

   $$n_{\rm prism} \ \sin (45^o) = n_{\rm air} \ \sin\theta \quad\Righ... ... prism}}{n_{\rm air}}\ \ \sin (45^o)\ \right) = \sin^{-1}\ (0.9899) = 82^o\ \ .$$

9. Which physical phenomenon is illustrated by the fact that the observer will not see all three colors of light?

   (d)total internal reflection

   We do not see all three colors because some have their angle of incidence, $45^o$, greater than the critical angle $\theta_c$ for total internal reflection. For the three colors, $\theta_c$ is

   $$\theta_c = \sin^{-1} (1/n) = \left\{\ \begin{array}{ll}44.4^... ...6^o&\mbox{green}\\ 46.9^o&\mbox{violet}\end{array}\right.\ \ .$$
   Thus red light is incident at greater than its critical angle, and is totally internally reflected; while green and violet light are incident at less than the critical angle, and so send a refracted ray out to the observer.

10. Which physical phenomenon is illustrated by the fact that the emerging rays are spread into component colors of the beam?

    (c) dispersion

    The emerging rays are spread into their component colors because they are bent differently by the prism, because their refractive indices in the prism differ, and this effect is dispersion. While diffraction will also separate colors (not at the central peak, but at the side peaks), this is due not to varying $n$, but to the fact that constructive and destructive interference occurs at different locations for light of different wavelengths. That is not the cause of the separation of colors here.

    
    

    Questions 11 and 12 refer to corrective lenses for the human eye.
    

11. In a scene from a movie, a nearsighted character removes his eyeglasses and uses them to focus the nearly parallel rays of the sun to start a fire. What is physically wrong with this scene?

    (b) The eyeglasses have diverging lenses and cannot be used to focus parallel rays.

    A nearsighted eye converges light too strongly, in front of the retina, so it is corrected by a diverging lens. Thus (a) is wrong. Parallel rays can be focused to a point, either by a convex lens or a concave mirror, both of which have real focal points where the parallel rays actually meet after refracting/reflecting. When these parallel light rays are focused, the energy density that was diffuse is concentrated at a point, and it can be enough (felt as heat) to start a fire. Thus (c) and (e) are wrong. However, to start a fire, the light rays have to actually focus, bringing actual energy to the point. This happens only for a real image or focal point, so (d) is wrong and divergent lenses cannot start the fire because they do not actually focus the parallel rays at the focal point.

12. Without his contact lenses, Mike can focus from 0.80 m to infinity. What refractive power of the lenses does he require for normal reading (0.25 m from the eyes)?

    (c) 2.75 diopters

    Mike is farsighted, and the corrective lens must take an object at the desired near point ( $d_o = 0.25$ m), and create a virtual image at his actual near point ( $d_i = -0.80$ m). Thus the corrective lens must have

    $$RP = \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{0.25\ {\rm m}} - \frac{1}{0.80\ {\rm m}} = 2.75\ {\rm diopters}\ \ .$$
    Note that the sign of $d_i$ is negative because the image is virtual, and that the lens is converging (has positive focal length), as expected to correct a farsighted eye which doesn't converge light strongly enough.

Quantitative Problems (Point Value as Marked)

1. (20 points)
   Vertically polarized light, with an electric field magnitude $E_o= 660 \ {\rm N/C}$ and intensity $S_o = 1200\ {\rm W/m^2}$, is incident on a series of 2 polaroid filters. The first has a vertical polarization axis; the second, a polarization axis at $30^o$ with respect to the vertical.

   

   1. What is the intensity $S$ of light leaving the second filter?
   2. What is the electric field $\vec{E}$ leaving the second filter? Give magnitude and direction.
   3. How do your answers to (a) and (b) change, if the order of filters is reversed?

   
   
   
   
   
   
   

   1. At each stage, the intensity is reduced by a factor of $\cos^2\theta$, where $\theta$ is the angle between the polarization axis of the incident polarized light ray and the filter's polarizing axis.

      Here, the light is initially vertically polarized, so $\theta$ for the first filter is 0 and the intensity is unchanged. $\theta$ for the second filter is $30^o$, since light leaves the first filter still vertically polarized. Thus

      $$S= S_o\ \cos^2 (0^o)\ \cos^2 (30^o) = 1200\ {\rm W/m^2}\ (0.75) = 900 \ {\rm W/m^2}\ \ .$$

   2. At each stage, the incident $\vec{E}$ is reduced to its component $E_\parallel$ parallel to the filter, which is emitted pointing along the filter's polarization axis. Again, in the first stage, the filter's and light ray's polarization axes are aligned, so $E_\parallel = E$ and the entire $\vec{E}$ is transmitted by the first filter, still pointing up. The second filter passes
      $$E_\parallel = E\cos (30^o) = E_o\ (0.866) = 570\ {\rm N/C}\ \ ,$$
      pointing along the polarization axis of filter 2, $30^o$ from the vertical.

      Many students obtained the magnitude of $E$ leaving the second filter from the equation

      $$S = c\epsilon_oE^2\ \ ,$$
      relating the light ray's intensity to its electric field. This gives
      $$E = \sqrt{\ \frac{S}{c\epsilon_o}\ } = \sqrt{\ \frac{900 \ {\rm W... ...\rm m/s}) (8.85\ \times\ 10^{-12}\ {\rm C^2/(nm^2)})}\ } = 580 \ {\rm N/C}\ \ .$$
      The difference between this result and the one above is merely due to roundoff errors (we have only one significant figure anyway.)

   3. If the order of the filters is reversed, the vertically polarized light encounters first a filter with polarizing axis at $30^o$ from the vertical; which projects $\vec{E}$ to $E_\parallel$, pointing $30^o$ from the vertical (along the first filter's polarization axis). This projected $\vec{E}$ then continues to the second, vertical filter, which again reduces $\vec{E}$ by passing its vertical component, pointing up.

      Quantitatively, the angle between the incident polarization and the filter's polarizing axis is $30^o$ each time, so

      $$S= S_o\ \cos^2 (30^o)\ \cos^2 (30^o) = 1200\ {\rm W/m^2}\ (0.75)^2 = 680 \ {\rm W/m^2}\ \ .$$
      $$E= E_o\ \cos (30^o)\ \cos (30^o) = 660\ {\rm N/C}\ (0.75)= 500 \ {\rm N/C}\ \ .$$

      The $\vec{E}$ field points in the direction of the polarizing axis of the last filter, or up. Again, the magnitude of $\vec{E}$ could also be found by

      $$E = \sqrt{\ \frac{S}{c\epsilon_o}\ } = \sqrt{\ \frac{680 \ {\rm W... ...\rm m/s}) (8.85\ \times\ 10^{-12}\ {\rm C^2/(nm^2)})}\ } = 510 \ {\rm N/C}\ \ ,$$
      which differs from the above only by roundoff error.

2. (20 points)
   An object placed 6 cm in front of a lens produces an image with magnification $m= +0.25$.
   1. Note below whether the image is upright or inverted, and real or virtual ; and whether the lens is converging or diverging.
   2. What is the focal length of the lens? Remember signs!
   3. Draw a ray diagram for this situation. Be sure to show object and lens, labeling the lens' focal point, and showing the 3 principal rays and their determination of the image.

   
   
   
   
   
   
   

   1. Note that $m = h_i/h_o = -d_i/d_o$ tells us, for positive $m$, that the image is upright ($h_i$ positive) and virtual ($d_i$ negative). Since the virtual image is smaller than the object, it must be obtained by tracing back diverging rays from the lens, so the lens must be divergent (as you could see from parts (b) and (c) below).

   2. The object distance is $d_o = +6$ cm; the image distance can be found from the magnification equation
      $$d_i = -m\, d_o = - (0.25)\ (6\ {\rm cm}) = - 1.5\ {\rm cm}\ \ .$$
      Thus the focal length is, by the lens equation,
      $$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{6\ {\rm cm... ...\ {\rm cm}} = -0.5\ {\rm (cm)^{-1}}\quad \Rightarrow\quad f = -2\ {\rm cm}\ \ .$$
      Note that the negative focal length means that the lens is diverging.

   3. The ray diagram is above. Note for this diverging lens, the first principal ray starts parallel from the top of the image, then hits the lens and diverges as if emanating from the focal point on the left. The second principal ray heads straight for the focal point on the right, and is bent parallel by the lens, and the third principal ray goes straight through the center of the lens. The image is obtained by extrapolating each of the 3 principal rays emerging on the right side of the lens straight back, to where they intersect, on the left of the lens, at the top of the image. Note that the image is inside the focal point.