1. No category

LIGHT SAMPLE TEST PROBLEMS
Name_____________________
1. An object 5.00 cm tall is placed 30.0 cm from a mirror with a 10.0 cm focal length. (a) Make a ray
diagram showing the object image relationships. (b) Where will the image be formed? (c) How tall will the
image be. (d) What are the characteristics of the image? (e) calculate the magnification.
(b)
hi
d
hi
15.0cm
=! i
=!
!hi = -2.50cm
ho
do
5.00cm
30.0cm
h
!2.50cm
(e) !!m = i ! m =
= -0.500x
ho
5.00cm
(c) !!
1 1 1
1 1
1
=
+ !
!
=
f d o di
f d o di
1
1
1
!
=
di = 15.0cm
10.0 30.0 di
(d) real,inverted, reduced
2. If the object in #1 is moved until it is 7.50 cm from the mirror, find (a) the location, (b) size
(c) magnification. and (d) describe the image formed.
(a)
1 1 1
=
+ !
f d o di
(b) !!
hi
d
=! i
ho
do
1 1
1
!
=
f d o di
1
1
1
!
=
10.0 7.50 di
hi
!30.0cm
=!
5.00cm
7.50cm
!hi = +20.0cm
di = -30.0cm
(c) !!m =
+20.0cm
=! 2.67x
7.50cm
(d) virtual, erect, enl arg ed
3. An object 5.00 cm tall is placed 30.0 cm from a convex mirror with a 10.0 cm focal length. (a) Make
a ray diagram showing the object image relationships. (b) Where will the image be formed? (c) How tall
will the image be. (d) What are the characteristics of the image? (e) calculate the magnification.
1 1 1
1 1
1
=
+ !
!
=
f d o di
f d o di
1
1
1
!
=
di = -7.50cm
!10.0 30.0 di
(b)
hi
d
hi
!7.50cm
=! i
=!
ho
do
5.00cm
30.0cm
h
!1.25
(e) !!m = i = !
= -0.250x
ho
5.00
(c) !!
!hi = -1.25cm !!!!(d)real,inverted, reduced
4. Calculate the radius of curvature of mirror #1 and the convex mirror in #3. In what way are they the
same? How do they differ?
R = 2 f = 2(30.0cm) = 60.0cm
R = 2 f = 2(!10.0cm) = !20.0cm
concave!has!a!real!(+)! focus,!!!convex!is!virtual!(!)! focus
5. A ray of light travels from air into water at an angle of 40.0o from the normal. The index of refraction of
water is 1.33. Find (a) the angle of refraction and (b) the speed of light in water.
(a) n1 sin !1 = n1 sin !1 1.00 sin 40.0 o = 1.33sin ! r !!!! 28.9 o from normal
(b)
n=
c
v
v=
c 3 x 10 8 m s
=
= 2.26 x 10 8 m s
n
1.33
6. The focal length of the lens in a box camera is 10.0 cm. The fixed distance between the lens and the film
is 11.0 cm. if an object is to be clearly focused on the film, how far must it be from the lens?
1 1 1
=
+ !
f d o di
1 1
1
1
1
1
! =
!!!!
!
=
f di d o
10.0 11.0 do
do = 110cm
12. An object 8.0 cm tall is placed 20 cm from a converging lens. A real image is formed 10 cm from the
lens. What is (a) the focal length of the lens, (b) the size of the image?
1 1 1
1
1
=
+ =
+ !
f do di 20 10
f = 6.67cm
hi
d
=! i
ho
do
hi
10
= ! !!!!hi = -4.0cm !
8.0cm
20
7. (a) When do concave mirrors form (a) real enlarged images? (b) real reduced images? (c) enlarged
virtual images? (d) When do convex mirrors form (1) real images (2) enlarged images?
(a) when object is between f and R (b) when object is beyond R (c) when object is less than f
(d) (1) never (2) never
8.A scuba diver is seen by another diver who is in a raft at the surface of the water.
(a) If the diver is 15.0 meters under the water, how far under the surface of the water does he appear to the
other diver? (b) If the diver under the water shines a light 20.0o away from the normal, at what angle will it
leave the water? (c) At what angle with the normal would the light be unable to leave the water? What is
this angle called. (d) Describe what happens to the light at angles beyond this angle.
h n
15.0m 1.33 '
(a) ' = 1
=
h = 11.3m (b) n1 sin !1 = n2 sin ! 2 1.33sin 20.0 o = 1.00 sin ! r ! r = 27.1o
h n2
h'
1.00
(c) 1.33sin ! c = 1.00 sin 90.0 o ! c = 48.8 o = critical angle!!(d) total internal reflection
9. An object 5.00 cm tall is placed 30.0 cm from a converging lens with a 20.0 cm focal length.
(a) Make a labeled ray diagram showing the object image relationships.
(b) Where will the image be formed? (c) How tall will the image be. (d) what is the magnification?
(e) What are the characteristics of the image?
(b)
f
ho
hi
O
do
(c) !!
hi
d
=! i
ho
do
1 1 1
1 1
1
=
+ !
!
=
f d o di
f d o di
1
1
1
!
=
di = 60.0cm
20.0 30.0 di
di
hi
60.0cm
=!
5.00cm
30.0cm
!hi = -10.0cm
(d) !!m =
(e) real, inverted, enlarged
hi
d
60.0
=! i! m=!
= -2.00x
ho
do
30.0
10. An object 5.00 cm tall is placed 30.0 cm from a diverging lens with a 20.0 cm focal length.
(a) Make a labeled ray diagram showing the object image relationships.
(b) Where will the image be formed?
(c) How tall will the image be. (d) what is the magnification?
(e) What are the characteristics of the image?
1 1 1
1 1
1
=
+ !
!
=
f d o di
f d o di
1
1
1
!
=
di = -12.0cm
!20.0 30.0 di
do
(b)
(c) !!
hi
d
=! i
ho
do
(d) !!m =
hi
!12.0cm
=!
5.00cm
30.0cm
!hi = +2.00cm
hi
d
!12.0
=! i! m=!
= +0.400x
ho
do
30.0
(e) virtual, erect, reduced
ho
hi
f
di
O
11. Describe the lens type and object location required to form a ... (a) real enlarged image
(b) real reduced image (e) virtual enlarged image (d) virtual reduced image
(a) converging lens... between F and 2F (b) converging lens... beyond 2F
(c) converging lens.. less than a foca llength (d) diverging lens ... always makes this image type
12. Monochromatic light ( λ = 700 nm) is shined through two small slits . A 2nd order line appears 36.5
cm to the left of the bright central line on a screen opposite the slits. If the distance from the center of the
slits to the second-order image is 74.5 cm, calculate the distance between the slits.
tan ! =
x
L
! = tan "1
36.5
= 26.10 o
74.5
m# = d sin !
d=
m# 2(700nm)
1µ m
=
x
= 3.18 µ m
o
sin ! sin 26.10 1, 000nm
13. A diffraction grating has 5720 lines/cm and is located 1.00 m away from a gas discharge tube. When a
student looks through the grating she sees a bright spectral line 32.8 cm to the right (and left too) of the
discharge tube. Calculate the wavelength of this spectral line in nanometers and Angstroms.
32.8
= 18.159 o
m# = d sin !
100
1cm
sin18.159 o
d sin ! 5, 720lines
10 7 nm
#=
=
x
= 545nm
m
1
1cm
tan ! =
x
L
! = tan "1
o
! = 5.45x10 3 A
14. Compare the camera and human eye in structures and function. LOOK in your book or at your notes
B. E =
C.
hc
hc
1 eV
=
x
= 41, 000 = 41keV
"10
! 3.0 x 10 m 1.60 x 10 "19 J
nx sin ! c = nr sin 90 o
D. ! =
nc sin 41o = 1.00 sin 90 o
nx = 1.52
h
h
=
= 3.33 x 10 -29 m
6
m
mv 1 x 10 kg(20 s )
E. m! = d sin "
d=
1
N
sin " =
m!
d
$ m! '
)
" = sin #1 &
&% 1 N )(
$
'
#9
$ m! '
&
)
1
(750
x
10
m)
o
) = sin #1 &
"1 = sin #1 &
) = 17.5
1
0.010m
&% N )(
&
)
&%
)(
4, 000
F. m! = d sin " max ima : m = 1, 2, 3
$
'
#9
&
)
3
(750
x
10
m)
o
" 3 = sin #1 &
) = 64.2
0.010m
&
)
4, 000
%&
()
min ima : m =
1 3 5
, ,
2 2 2
90cm
!
tan ! =
!=
3.0mm
900mm
! = 0.19099 o
3.0mm
d sin "
0.30m sin(0.199099 o )
=!=
= 0.000667mm = 667nm
3
m
2
G. v =
x
t 1 8 rev
t 1 8 rev =
x
70, 400m
=
= 0.00023467s
v 3.00 x 10 8 ms
1
rev
8
!=
= 533 rev s
0.00023467s
H. m! = d sin "
d sin "
2.33 x 10 #8 m sin15 o 1 nm
!=
=!=
x #9 = 6.0nm
m
1
10 m
I. Bremsstrahlung and characteristic X ! rays
LOOK AT YOUR NOTES TO ANSWER THESE QUESTIONS
J. K! and K " show quantization of energy
c=
2d
t 1 rev
8
35.2km
m=0
m=0
3
!
2
note : A + B = 180 o
40
o
A
65 o
!D
!1 = 25.37 o ! 2 = 39.63o
B
ni sin ! i = nr sin ! r
1.00 sin 40 o = 1.5 sin ! r
!1 = 25.37 o
A + B = 180 o
B = 180 o ! 65 o = 115 o
! B = 180 o " 115 o " 25.37 o = 39.63o
ni sin ! i = nr sin ! r
1.50 sin 39.63o = 1.00 sin ! D
! D = 73.09 o = 73.1o