Test 3 Practice Problems
Ch 21
1) Xc = 1/ ( 2 pi f C) = 1 / [2 pi (2000 Hz) (10 µf) ] = 8.0 Ω
XL = 2 pi f L = 2 pi (2000 Hz) 10 mH = 126 Ω
2) Vrms = 0.707 Vp = 0.707 (12 V) = 8.5 V
Ip = I rms / 0.707 = 8.0 A / 0.707 = 11.3 A
Ch 22
1:
To solve the problem we use Snell’s law twice: at the first air-water interface:
nair Sin (80)  nair Sin ( )  (1) Sin (80 )  (1.33) Sin ( )  Sin ( ) 
Sin (80 )
 0.7405    Sin 1 (0.7405) 
1.33
47.8º
Now, according to the geometry of the problem the incident angle on the second interface water-air is
  90    42.2o
And using Snell’s law for a second time:
nwater Sin( )  nwater Sin(  )
 (1.33)Sin(42.2 )  (1)Sin( )  Sin( )  0.8939    Sin 1 (0.8939)  63º
Ch 23
1) The image is real, inverted and larger.
2) The problem says that the distance from the object (in this case you) to the mirror is 4.0 meters and the
magnification is 0.35, so:
do  4.0 and m  0.35 ,
but m  
d
di
, so 0.35   i  di  1.4
4.0
do
and now we can find the focal length:
1
1 1
1
1
1

  

 0.25  0.714  0.464  f  -2.15m
f do di
f 4.0  1.4
1
 0.4m . The problem says that the distance to the object is
3) A power of 2.5D means a focal length f 
2.5 D
do  0.3m , so we can calculate the distance to the image:
1
1 1
1
1
1
1

 

 
 2.5  3.33  0.833  d i  -1.2m
f do di
0.4m 0.3m d i
di
Since di is negative the image is virtual (it is to the left of the lens), so we cannot project it on a screen or
photographic film.
d
 1.2m
The magnification is m   i  
 4
do
0.3m
Since the magnification is positive the image is upright.
Ch 24 – Wave Nature of Light
1) Double Slit
A parallel beam of light from a He-Ne laser, with a wavelength of 656 nm, falls on two very narrow slits
that are 0.050 mm apart. How far apart are the fringes in the center of the pattern on a screen 4.2 m away?
For constructive interference, the path difference is a multiple of the wavelength. The location on the screen is given
by x  l tan  , For small angles, we have sin   tan   x l . For adjacent fringes, m  1.
x
 ml
d sin   m  d  m  x 

l
d
633  109 m   3.8m 

l
x  m
 1
 0.035m  3.5cm
d
 6.8 105 m 
x  m
l
d
2) Single Slit
 (1)
(656 x109 m)( 4.2m)
 0.055m
(0.050 x103 m)
In an experiment you shine red light of wavelength 533 nm on a slit, generating a central diffraction peak
of 2θ = 8.4o as shown in the figure. How wide is the slit?
The first dark fringe happens when:   D sin  so: D 

sin 

533nm
 7.28 µm
sin 4.2
3) Diffraction Grating
A diffraction grating has lines separated by 3.5 μm. Calculate the angle of diffraction of the first order
fringe for green light (λ = 530 nm) and red light (λ = 650 nm).
The equation that describes diffraction gratings is:
dSin  m
If we are looking for the first fringe m=1, so:

  Sin 1  
d
For the given wavelengths we get:
 530nm 
  8.7º
(a) Green light θ  Sin 1 
 3.5 μm 
 650nm 
  10.7º
(b) Red light θ  Sin 1 
 3.5 μm 
4) Diffraction Grating
A diffraction grating is used to determine wavelengths in a spectrometer. Calculate the angular separation
(in degrees) between the two main lines of sodium (589.0 nm and 589.6 nm) if the grating has a density of
15,000 lines per inch. [1 inch = 2.54 cm]
The density of lines implies a distance between slits of:
d
1inch
2.54 x10 2 m

 1.6933x10-6 m
15000lines 15000lines
To get the angular separation we will find the angle for each wavelength and calculate the difference:

d sin(  )      sin 1  
d 
 589.0 x10 9 m 
  20.355º
1  sin 1 
6 
 1.6933x10 
 589.6 x10 9 m 
  20.376º
 2  sin 
6 
 1.6933x10 
1
So the difference is 0.022º
5) Polarization
Find how much intensity of a beam of un-polarized light will go through three polarizers, where the first
and second are rotated 37o with respect to each other and the second and third are also rotated 37o with
respect to each other.
If you start with unpolarized light of intensity Io, after the first polarizer you will have polarized light of
1
intensity I o , the second polarizer is rotated by 37o, so there will be additional losses, the third polarizer is also
2
rotated by 37o, and the final intensity will be:
I FINAL 
1
Io [Cos 4 (37 o )]  0.203 Io  20.3% of Io
2