**. ( 50 POINTS)
Below is a model for a pulsed laser used in science and medicine. The interior of the pulse can
represented by an EM ***plane* wave with frequency f = 2.50 x10 ^{14} Hz. The magnetic field _{} oscillates vertically back and forth along a line that makes a 45
degree angle with the x and y axes. The magnitude of the magnetic field is
8.00x10 ^{-9}** T . At a given instant the magnetic field makes
an angle of 45 degrees with the ***negative* x–axis. The arrow is labeled by the magnitude B of the
magnetic field . As shown by the dot centered
within the little circle,

**the wave’s propagation direction is OUT of the page.**

** **

(a) (17 points) What is the direction of
the electric field vector. Draw an arrow clearly on the x-y axes below. Draw
the tail of the arrow at the origin. What angle does the electric field make with
the horizontal? *Indicate and label* this angle in your diagram.

(b) (18 points) ) If 9020 wavelengths
are found within the pulse, the how long ( in seconds) did the pulse last?

**(c) (15 points) Suppose the pulse is transmitted from
a laboratory on the surface of the Earth toward the Moon. If the reflected pulse
(i.e., reflected off the Moon’s surface) is received at the lab 1.30 seconds
later, then what is the distance between the laboratory and the Moon’s surface?
(Assume the pulse is not affected by the
Earth’s atmosphere; the Moon has no atmosphere.) **

** ****SOLUTIONS:**

(a) The electric field vector points in the second quadrant and makes a 45
degree angle with the negative x- axis.

(b) c*t/wavelength = 9020, where wavelength = c/f. Thus, after substituting we
get:

f*t = 9020 and t = 36x10 ^{-12} seconds or 36 picoseconds.

(c) ct/2 = distance= (1/2)*(3x10 ^{8} m/s)*(1.30 s) = 1.95x10 8 m.

** ****-----------------------------------------------------------------------------------------------------------------------**

2. (50 POINTS) The upright object shown below is to the left of
a concave mirror
with a focal length of 0.25 m (not shown). The object produces a real
image which is one third the size of the
object.

** **

**(a) (30 points)
How far from the mirror is the object? **

(b) (3 points) Is the image upright or inverted (upside
down)? Explain.

(c) (13 points) How far from the mirror is the image?

**(d) (4 points) On the diagram above sketch the mirror on the
right of the object. Draw the image by
ray tracing as we did in class. Ray trace by drawing two rays from the tip of
the arrow shown, one parallel to the axis and the other intersecting the point where the mirror
crosses the axis. Label the
object distance d**_{o} and image distance d_{i}.

**SOLUTIONS: **

(a) m = -1/3 = - di/do, so that do = 3*di . Thus 1/f = 1/di + 1/do =
4/do and do = 4*f = 1.0 m.

(b) inverted since m < o.

(c) di = do/3 = 0.333 m

(d) **See class
notes or check out figure 23.14 and COMPARE with figure 23.16.**

** **

** ****-------------------------------------------------------------------------------------------------------------------**

**
**

3. (50 POINTS)
A beam of __vertically__ polarized light with __intensity__ 41 W/m^{2}
is sent leftward through three polarizing sheets. The vertically polarized wave
is shown below moving toward Sheet
1. The polarizing direction of Sheet 1 makes an angle of 30^{0} with
the __horizontal__. The polarizing
direction of Sheet 2 makes an angle
of 70^{0} with the

__vertical.__ The polarizing direction of
Sheet 3 makes an angle of 70^{0} with the __horizontal__.

(a) (17 points) What is the Intensity of
the light after it passes through Sheet 1 *before*
reaching Sheet 2.** **

(b) (17 points) What is the Intensity of the light *after*
it passes through Sheet 2 before
reaching Sheet 3?

**(c) (16 points)
What is the Intensity of the light ***after*
it passes through Sheet 3?

**SOLUTIONS: **

**(a)
****41*cos**^{2} 30 = 10.3
W/m^{2} .

(b) 10.3*cos^{2} 10 = 9.99 W/m^{2} .

**(c)
9.99*cos**^{2} 50 = 4.13 **W/m**^{2}
.

** ****----------------------------------------------------------------------------------------------------------------------**

** **

** **

** **

** **

**4. (50 points) In a university lab experiment, a
plate of glass is placed above a
container of water with the same lateral dimensions. The plate is parallel to
the water surface. The glass plate and water are separated by an air gap of
thickness t. Light of wavelength 615 nm
falls perpendicular to the plate and water. **

** **

(a) (17 points) What *minimum*
thickness t will produce
constructive interference at the
detector?

(b) (17 points) Suppose the air gap
between the plate and water has the
minimum thickness t you computed in part (a). Suppose you *raise*
the glass above the water surface and *increase*
the air gap thickness until the signal at the detector *first* becomes ZERO. What is the new thickness t’ ?

** (c) (16 points)
Suppose the air gap between the plate
and water has the thickness t’ you computed in part (b). Suppose you ***further
raise* the glass above the water surface and increase the air gap thickness. What is the *next *thickness t” that will
produce *constructive interference?*

**SOLUTIONS:**

(a) 2t = wavelength /2 so t =154 nm

(b) 2t = wavelength so t = 308 nm

(c) 2t = (3/2)*wavelength so t = 461 nm. **
**

**---------------------------------------------------------------------------------------------------------------------------**

** **

**5. ( 51 POINTS) A magnifying glass has focal length f
= 8.5 cm. The magnifying glass is used to read print (the object) placed a distance
d**_{o} = 7.5 cm from the
lens. See figures 1 and 2. Note that N =
near point *distance* = 25 cm.

(a) (3 points) Is the image of the news
print virtual or real ? Explain.

**(b) (20 points) What is the image position d**_{i} ?

(c) (8 Points) Using the definition of M, compute the angular magnification M = θ’/θ. See figures
1 and 2.

(d) (10 points) What would be the angular magnification M if this lens forms a
virtual image at the person’s near
point, assumed to be 25 cm away from the lens?

(e) (10 points) What would be the
angular magnification M if the eye is
fully relaxed and the lens forms a
virtual image essentially at d_{i} = -_{} ?

**SOLUTIONS:**

**(a)virtual
image from the diagram**

(b) 1/8.5 = 1/7.5 + 1/di so di = -64 cm

(c) M = (h/do)/(h/N) = N/do = 3.3 x

(d) N/ f + 1 = 3.9 x

(e) N/f = 2.9 x

----------------------------------------------------------------------------------------------------------------

**6. ( 20 points) A United Nations
observer on Planet Earth (1) remotely
views two vehicles in space. A small rocket (3) has been fired from a large rocket (2). Rocket 3’s
velocity relative to rocket
2 is V**_{32} = 0.600c. Rocket 2’s
velocity relative to the Earth is V_{21} = 0.600c. See Figure 1.

(a) (13 points) Find the velocity V_{31 }of rocket 3 relative to the Earth.

(b) (2 points) Explain why the answer
to part (a) should be less than the
speed of light c.

(c) (5 points) See Figure 2 on the next
page. Suppose rocket 2 sends out a pulse
of light (3) . In this case V_{32 } =c. Assume
Rocket 2’s velocity relative to the Earth is V_{21} = 0.600c . Using
the method and formulas you used in part (a) , compute the velocity V_{31 }of
the light pulse relative to the
Earth. Does your answer

** make sense? Explain.**

SOLUTIONS:

(a) V_{31} = (V_{32} + V_{21})/(1 + V_{32}*V_{21}/c^{2})
= 1.2*c/1.36 = 0.882 *c

(b) Objects with mass cannot exceed c since the relativistic momentum
discussed in Ch. 26 would become imaginary.

(c) V_{31} = (V_{32} + V_{21})/(1 + V_{32}*V_{21}/c^{2})
= 1.6*c/1.6 = c. The speed of light is the same in all
reference frames !!

**--------------------------------------------------------------------------------------------------------------------------**

** **

**7. (12 points) When a
beam of UV (ultra violet) radiation,
wavelength 285 nm, falls on a metal surface, the maximum
kinetic energy of emitted electrons is KE = 5.20x10 **^{– 19}
J. Note: h = 6.626x10 ^{-34} J·
s.

What is the maximum kinetic energy KE’ (
in Joules) , when a beam of *visible*
light, wavelength 632. 8 nm, falls on the same metal surface?

**SOLUTION**

h*f = = h*c/wavelength = 5.20X10 ^{-19}
J + W_{o} , where wavelength = 285 x**10**^{–
9}
m.

W_{o} = 1.77x10 ^{-19}.

**Now let** ** h*f' =
h*c/wavelength' = KE' + W**_{o } = KE'
+ 1.77x10 ^{-19} , where wavelength now is 632.8 nm = 632.8
x**10**^{– 9}
m. Thus: KE' = 1.37x**10 **^{-19} J.