1. No category

Physics 262 - exam#2
1. (25 pts) Answer the following questions. Justify your answers. (Use the
space provided below and the next page)
a). (9 pts) An object (an arrow) is placed as shown in front of each of the following
optical instruments. Describe the image produced in each case according to the
following three sets of identifiers: real/virtual, upright/inverted, and larger/smaller?
.
F
b). (4 pts) According to the Rayleigh’s criterion, with which color of light can a
telescope see finer detail in a distant astronomical object: red, yellow, blue, or
ultraviolet?
c). (4 pts) In a double slit diffraction experiment, the interference pattern is viewed on a
flat wall a few meters away. If the light source is changed from a green laser to a red
laser, will the total number of visible fringes on the wall become more or less?
d). (4 pts) A real object is placed in front of a convex mirror. If one moves the object
closer to the mirror, will the image move closer to or farther away from the mirror?
e). (4 pts) A thin layer of unknown oily film spreads out on a glass, the thinnest part of
the film appears dark in the resulting interference pattern. You can assume that the
thickness of the oily film at the thinnest part to be much less than one wavelength of
the light. Is the index of refraction of the oily film larger or smaller than the index of
refraction of the glass?
Answers:
a) i) virtual, upright, and larger
ii) virtual, upright, and smaller
iii) virtual, upright, and smaller
b) From the Rayleigh’s criterion, we have the minimally resolvable angular
separation between two objects given by  min  1.22

where D is the diameter of
D
a circular aperture and  is the wavelength of the light. Thus, for a telescope
with a given diameter, light with smaller wavelength will have a smaller
minimally resolvable angular separation meaning that the telescope will be able to
see finer detail using an ultraviolet light.
c) Since the locations of bright fringes in a double slit interference pattern is given
by sin   m d , the angular separation between the bright fringes will get
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Physics 262 - exam#2
further apart if one uses a longer wavelength. Since red light has a longer
wavelength than green light, it will produce less number of visible fringes on the
flat wall.
 1 1
1
d) The image distance for a convex mirror is given by      . Here, we
s'
 f s
have put in a “-“ sign explicitly for the convex mirror. For a real object (s is +),
when s decreases, 1/s gets bigger so that the magnitude of the right hand side
1 s ' of the equation will increase also. Thus, the corresponding magnitude of s’
decrease, i.e., the virtual image (note the overall negative sign on the right hand
side) will move closer to the mirror.
e) Since the thickness of the oily film at its thinnest part is less one wavelength of
the light, path difference between the two reflected lights respectively from the
top and bottom interface of the film is not an important factor in determining the
condition for interference in this case. The key determining factor will be the
relative phase shift due to reflection. Since we are observing a dark fringe at this
situation, the two reflected lights must be in destructive interference so that they
must have a relative phase shift of  . This tells us that we must have,
nair  1  noil and noil  nglass . noil is larger than nglass .
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Physics 262 - exam#2
2. (25 pts)
A thin converging lens with focal length f1  10.0cm and a thin diverging lens with focal
length f 2  20.0cm are separated by 40.0cm. An object is placed at a distance of
20.0cm to the left of the first lens. i) Find the position of the final image; ii) draw the
rays diagram for this situation; iii) what is the magnification of the final image; iv) is the
final image virtual or real?
20.0cm
f1
40.0cm
f1
converging lens
f2
f2
diverging lens
The black rays are from the original object and the blue rays are for the image formed by
the diverging lens using the intermediate image (blue arrow). The final image is indicated
by the red arrow.
For the converging lens, we have
1
1 1
  and for the diverging lens, we have
s '1 f1 s1
1
1 1
  .
s '2 f 2 s 2
By convention, f1 is positive and f2 is negative. From the first lens equation, we have
1
1 1 2 1
 

 s '1  20cm . As indicated by the rays diagram, this image will
s '1 10 20
20
be a real object for the diverging lens and s2  40cm  s '1  20cm . Putting this into the
diverging lens equation, we have
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Physics 262 - exam#2
1
1
1
1
 

 s '2  10.0cm . The final image is virtual as indicated and
s '2
20 20
10
is located 10 cm in front of the diverging lens (or 30cm behind the converging lens).
M  mcon mdiv  
s '1  s '2 
20  10 

   
  0.5 . The final image is also inverted and
s1  s2 
20  20 
smaller.
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Physics 262 - exam#2
3. (25 pts) In a double-slit diffraction experiment, two slits with
equal width a are separated by a distance d  0.12mm as shown to
the right. Take the wavelength of the light to be 500nm. A doubleslit diffraction pattern is observed on a flat screen 10m away. i) If
the 3rd order interference maximum is observed to coincide with the
1st order diffraction minimum, what is the slit width a? ii) How
many complete bright interference fringes appear within the central
diffraction envelope? iii) What is the total number of bright
interference fringes that one can see on the screen? iv) What is the
relative intensity of the second order interference fringe as compared
to the central maximum?
a
d
a
i) The 3rd order interference maximum coincides with the 1st order diffraction minimum
so that d / a  3 so that a  d / 3  0.12mm / 3  0.04mm .
ii) Including the central maximum and the remaining 2 interference maxima on either
sides, we should expect to see 5 bright interference fringes within the central diffraction
maximum.
0.12 103 m
 240 . However, since the 3rd, 6th, 9th,
 500 109 m
etc. will be missing due to diffraction minima, one should expect to see a total of 2(240240/3)+1=321 bright fringes.
iii) d sin 90  m gives mmax 
d

2
. Substituting this into
d
the equation for the intensity of a double slit diffraction, we have
iii) The second order interference fringe is located at sin  2 
 d sin   sin( a sin  2 /  ) 
I  I 0 cos (
)


  a sin  2 /  
2
2
   a  2   
 sin 

  d   
2
 I 0 cos (2 )  
   a  2   
    d  

 
 
 sin(2 a / d ) 
 I 0 1 

 2 a / d 
2
2
2
 sin(2 / 3) 
 I0 
  0.171I 0
 2 / 3 
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Physics 262 - exam#2
4. (25 pts) A light ray enters an equilateral
prism at an angle of  as shown below. The
index of refraction for the prism is 1.60.
Determine the critical angle  c with respect to
the normal so that the light ray entering from the
left face with an incident angle    c will be
totally internally reflected at P on the right face
of the prism? [Assume air is outside of the
prism.]
One should note that for this geometry, as the incident angle  increases, the incident
angle on the right face of the prism at P would get smaller. Therefore, a nearly normal
incident ray (   0 ) will be more likely to be totally internally reflected when it reaches
the right face at P. The critical angle can be calculated by going backward from P.
At P, for total internal reflection, we have
n sin   1    sin 1 (1/ n)  sin 1 (1/1.6)  38.68
Since     60 from the geometry of the prism, using Snell’s law again at the left face
of the prism, we have,
sin   n sin 
  sin 1 (n sin(60  38.68 ))
  sin 1 (0.5817)  35.6
All incident rays with incident angle   35.6 will be totally internally reflected at P.
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