Physics, 6th Edition
Chapter 33. Light and Illumination
Chapter 33. Light and Illumination Light and the Electromagnetic Electromagnetic Spectrum 33-1. 33-1.
An infrar infrared ed spectro spectrophot photome ometer ter scans scans the the a!eleng a!elengths ths from from 1 to 16 µm. E"press these range in terms of the fre#uencies of the infrared rays. c
f1 =
λ1
=
3 " 1&$ m's -6
1 " 1& m
13
= 3&.& " 1& ()* f % =
c λ %
=
3 " 1&$ m's -6
16 " 1& m
= 1.$$ " 1&13 ()
+ange of fre#uencies 1.$$ " 1&13 () to 3& " 1&13 ()
33-%. 33-%.
hat hat is the the fre#uen fre#uency cy of !iole !iolett light light of a!ele a!elength ngth 1& 1& nm/ c
f =
33-3.
λ
=
3 " 1&$ m's -0
1& " 1& m
f 2.3% " 1&1 ()
*
A microa!e radiator used in measuring automoile speeds emits emits radiation radiation of fre#uency 1.% " 1&0 (). hat is the a!elength/
= λ =
33-. 33-.
c f
=
3 " 1&$ m's 1.% 1.% " 1& 1&0 ()
*
= %4& mm λ =
hat hat is the range range of fre#uen fre#uencie ciess for !isile !isile light/ light/ f1 =
c λ1
=
3 " 1&$ m's -0
1
2&& " 1& m
= .%0 " 1& ()* f % =
c λ %
5 Range of λ : 2&& nm to && nm
=
3 " 1&$ m's -0
&& " 1& m
= 2.4& " 1&1 ()
+ange of fre#uencies .%0 " 1&1 () to 2.4& " 1&1 ()
33-4 33-4..
If Plan Plan78 78ss con const stan antt h is e#ual to 6.6%6 " 1&-3 9 s, hat is the energy of light of a!elength 6&& nm/ E = hf =
hc λ
=
56.6%4 " 1&-3 9 ⋅ s53 " 1&$ m's -0
6&& " 1& m
%&4
*
E = 3.31 " 1&-10 9
Physics, 6th Edition
Chapter 33. Light and Illumination 33-6.
hat is the fre#uency of light hose energy is 4 " 1&-10 9/ f =
33-2.
E h
4.&& " 1&-10 9
=
-3
56.6%4 " 1& 9 ⋅ s
*
f 2.44 " 1&1 ()
:he fre#uency of yello-green light is 4.1 " 1&1 (). E"press the a!elength of this light in nanometers and in angstroms/ λ =
c f
=
3 " 1&$ m's 4.1 " 1&1 ()
*
λ 444 " 1&-0 m51 " 1&1& A'm*
33-$.
λ = 444 nm
λ 444& A
hat is the a!elength of light hose energy is 2 " 1&-10 9/ E =
hc λ
*
λ =
hc E
=
56.6%4 " 1&-3 9 ⋅ s53 " 1&$ m's 2 " 1&-10 9
λ = %$ nm
*
The Velocity of Light 33-0.
:he sun is appro"imately 03 million miles from the earth. (o much time is re#uired for the light emitted y the sun to reach us on earth/ t =
03 " 1&6 mi 1$6,&&& mi's
= 4&& s *
t = $.33 min
33-1&. If to e"perimenters in ;alileo8s e"periment ere separated y a distance of 4 7m, ho much time ould ha!e passed from the instant the lantern as opened until the light as oser!ed/ t =
s c
=
4&&& m $
3 " 1& m's
*
t = 12.& " 1&-6 s or
%&6
12.& µs
Physics, 6th Edition
Chapter 33. Light and Illumination
33-11. :he light reaching us from the nearest star, Alpha Centauri, re#uires .3 years to reach us. (o far is this in miles/ In 7ilometers/ s = ct 51$6,&&& mi's53.14 " 1&2 s'yr5.3& yr* s = ct 53 " 1&$ m's53.14 " 1&2 s'yr5.3& yr*
s = %.43 " 1&13 mi s =.&2 " 1&13 7m
33-1%. A spacecraft circling the moon at a distance of 3$,&&& 7m from the earth communicates y radio ith a se on earth. (o much time elapses eteen the sending and recei!ing of a signal/ $
s 53$,&&& 7m51&&& m'7m 3.1& " 1& m*
t =
" 1&1& m 3 " 1&$ m's
*
t =
3.$ " 1&$ m 3 " 1&$ m's
= 1.%$ s
t = 133 s
33-13. An spacecraft sends a signal that re#uires %& min to reach the earth. (o far aay is the spacecraft from earth/ s ct 53 " 1&$ m's5%& min56& s'min*
s 3.6& " 1&11 m
Light Rays and Shadows 33-1. :he shado formed on a screen m aay from a point source of light is 6& cm tall. hat is the height of the o
=
h
6& cm
&& cm 1m
h =
5&& cm56& cm 1&& cm
*
h = %.& m
%&2
3m
h
Physics, 6th Edition
Chapter 33. Light and Illumination
33-14. A point source of light is placed 14 cm from an upright 6-cm ruler. Calculate the length of the shado formed y the ruler on a all & cm from the ruler. h 44 cm
6 cm
=
14 cm
*
h
6 cm
h = %%.& cm 14 cm
& cm
33-16. (o far must an $&-mm-diameter plate e placed in front of a point source of light if it is to form a shado && mm in diameter at a distance of % m from the light source/ x $& mm
=
%&&& mm && mm
*
x = && mm
$& mm
&& mm
x %&&& mm
33-12. A source of light & mm in diameter shines through a pinhole in the tip of a cardoard o" % m from the source. hat is the diameter of the image formed on the ottom of the o" if 6& mm
the height of the o" is 6& mm/ y 6& mm
=
& mm %&&& mm
& mm
y
*
y = 1.%& mm %&&& mm
=33-1$. A lamp is co!ered ith a o", and a %&-mm-long narro slit is cut in the o" so that light shines through. An o
the slit. 5>imilar ∆8s a %& mm
=
u 14&& + a b %&
=
4&& + a 3& mm
=
%& a
4&& − b 3&
*
14&& mm
a *
u 14&& + 1&&&
=
b
%& 1&&&
* b = %&& mm* Now,
30 mm
20 mm
a = 1&&& mm
*
c
u = 4& mm p
14&& − b
%&$
=
%& b
from which p = 13& mm
u
p
Physics, 6th Edition
Chapter 33. Light and Illumination
Illumination of Surfaces 33-10. hat is the solid angle sutended at the center of a 3.%&-m-diameter sphere y a &.4-m% area on its surface/
Ω=
A R %
=
&.4 m% 51.6 m%
*
Ω &.104 sr
33-%&. A solid angle of &.&$& sr is sutended at the center of a 0.&& cm diameter sphere y surface area A on the sphere. hat is this area/ A = Ω R2 = 5&.&$ sr5&.&0 m%*
A = 6.$ " 1&- m%
33-%1. An $? " 11 cm sheet of metal is illuminated y a source of light located 1.3 m directly ao!e it. hat is the luminous flu" falling on the metal if the source has an intensity of %&& cd. hat is the total luminous flu" emitted y the light source/ A = 5&.&$4 m5&.11 m*
Ω=
A R
%
=
0.34 " 1&−3 m% 51.3 m
%
A = 0.34 " 1&-3 m% 1.3 m
= 4.43 " 1&-3sr
= ! Ω = 5%&& cd54.43 " 1&-3 sr* "otal lux = #π ! = π5%&& cd*
= 1.11 lm "otal lux = %41& lm
33-%%. A &- monochromatic source of yello-green light 5444 nm illuminates a &.4-m% surface at a distance of 1.& m. hat is the luminous intensity of the source and ho many lumens fall on the surface/ = 56$& lm'5& %2,%&& lm* ! =
! =
R % A
=
5%2, %&& lm51 m% &.4 m %
%&0
*
Ω
* Ω=
A R %
! 4,&& cd
Physics, 6th Edition
Chapter 33. Light and Illumination
33-%3. hat is the illumination produced y a %&&-cd source on a small surface .& m aay/ E =
!
=
R %
%&& cd 5 m%
*
E = 1%.4 l"
33-%. A lamp % m from a small surface produces an illumination of 1&& l" on the surface. hat is the intensity of the source/ ! = ER2 = 51&& l"5% m%*
! && cd
33-%4. A tale top 1 m ide and % m long is located .& m from a lamp. If & lm of flu" fall on this surface, hat is the illumination E of the surface/ E
=
A
.& lm =
51 m65% m6
* E = %& l"
33-%6. hat should e the location of the lamp in Prolem 33-%4 in order to produce tice the illumination/ 5 !llumination $aries in$ersely with s%uare of &istance'( 2
2 2
2 2
E ) R) = E 2 R = *2E ) (R +
R% =
R1% %
=
5 m% %
*
R2 = %.$3 m
=33-%2. A point source of light is placed at the center of a sphere 2& mm in diameter. A hole is cut in the surface of the sphere, alloing the flu" to pass through a solid angle of &.1% sr. hat is the diameter of the opening/
@ + '% 34 mm B
A = Ω R2 = 5&.1% sr534 mm%* A = 12 mm%
A =
π ,
%
*
,=
A π
=
512 mm% π
%1&
*
, = 13.2 mm
Physics, 6th Edition
Chapter 33. Light and Illumination
Challenge Prolems 33-%$. hen light of a!elength 44& nm passes from air into a thin glass plate and out again into the air, the fre#uency remains constant, ut the speed through the glass is reduced to % " %&$ m's. hat is the a!elength inside the glass/ 5 f is same for both( f =
$air λair
$ glass
=
λ glass
λ glass
*
=
$glass λ air $air
=
5% " 1&$ m's544& nm 3 " 1&$ m's
+
λ glass= 362 nm
33-%0. A 3&-cd standard light source is compared ith a lamp of un7non intensity using a grease-spot photometer 5refer to ig. 33-%1. :he to light sources are placed 1 m apart, and the grease spot is mo!ed toard the standard light. hen the grease spot is %4 cm from the standard light source, the illumination is e#ual on oth sides. Compute the un7non intensity/ @ "he illumination E is the same for each. B ! x r x%
=
!s rs%
* ! x =
! s r x% r s%
=
53& cd524 cm% 5%4 cm%
*
! x
3& cd %4 cm
! x = %2& cd
24 cm
33-3&. here should the grease spot in Prolem 33-%0 e placed for the illumination y the un7non light source to e e"actly tice the illumination of the standard source/ E x = % E s * .4 51- x
%
=
! x % x
r 1 x
%
*
=
% ! s % s
r
*
%2& cd 51- x
%
=
.4 x % = 51 − x% *
3.1% x = 1*
x=
1 3.1%
%53& cd x
%
1m
*
3& cd
%.1% x = 1 − x*
= &.3%& m*
%11
x
%2& cd 1 m D x
x = 3%.& cm from stan&ar&
! x
Physics, 6th Edition
Chapter 33. Light and Illumination
33-31. :he illumination of a gi!en surface is $& l" hen it is 3 m aay from the light source. At hat distance ill the illumination e %& l"/ 5 Recall that ! = ER is constant ( % 1
% %
E1 R = E% R *
E1 R1%
+% =
5$& l"53 m %
=
E %
5%& l"
*
R2 = 6&& m
33-3%. A light is suspended 0 m ao!e a street and pro!ides an illumination of 34 l" at a point directly elo it. etermine the luminous intensity of the light. E =
! R
* ! = ER % = 536 l"50 m% *
%
! = %0%& cd
=33-33. A 6&- monochromatic source of yello-green light 5444 nm illuminates a &.6 m% surface at a distance of 1.& m. hat is the solid angle sutended at the source/ hat is the luminous intensity of the source/
Ω=
= 56$& lm'56& &,$&& lm*
! =
Ω
=
5&,$&& lm &.6& sr
*
A R %
=
&.6 m% 51 m%
= &.6&& sr
! 6$,&&& cd
=33-3. At hat distance from a all ill a 34-cd lamp pro!ide the same illumination as an $&-cd lamp located .& m from the all/ @ E ) = E 2 B !1 r1%
=
!% r%%
*
r % =
! % r 1% ! 1
=
534 cd5 m % $& cd
*
r 2 %.64 m
=33-34. (o much must a small lamp e loered to doule the illumination on an o
cm directly under it/
R% =
R1% %
=
5$& cm% %
*
R2 = 46.6 cm*
%1%
y $& cm D 46.6 cm %3. cm
Physics, 6th Edition
Chapter 33. Light and Illumination
=33-36. Compute the illumination of a gi!en surface 1& cm from a 2-cd light source if the normal to the surface ma7es an angle of 3$& ith the flu". E =
! cos θ R %
=
52 cd5cos 3$& 51.& m%
*
E = %0.$ l"
=33-32. A circular tale top is located m elo and 3 m to the left of a lamp that emits 1$&& lm. hat illumination is pro!ided on the surface of the tale/ hat is the area of the tale top if 3 lm of flu" falls on its surface/ 3m
tan θ =
θ 36.0&
m !
E =
! cosθ R
=
%
1$&& lm
=
.π
= #π !
=
.π
513 lm'sr cos 36.0& 54 m
%
*
*
!
=
+4 m
m
3m
1.3 lm'sr
E = .4$ l"*
θ
A =
E
&.644 m%
=33-3$. hat angle θ eteen the flu" and a line dran normal to a surface ill cause the illumination of that surface to e reduced y one-half hen the distance to the surface has not changed/ E1 =
!1 % 1
R
*
E% =
>ustitution yields
! cosθ R%%
*
E1 = % E% *
2! cosθ = !
!1 = ! % and R) = R2
an& cos θ &.4
%13
or
θ 6&&
Physics, 6th Edition
Chapter 33. Light and Illumination
=33-30. In ichelson8s measurements of the speed of light, as shon in ig. 33-11, he otained a !alue of %.002 " 1&$ m's. If the total light path as 34 7m, hat as the rotational fre#uency of the eight-sided mirror/ "he time for light to reappear from e&ge ) to e&ge 2 is: t =
s
34, &&& m
=
c
$
3 " 1& m's
= 1.162 " 1&- s
"he time for one re$olution is -t: " = $51.162 " 1&- s f =
1 "
=
1 $51.162 " 1&- s
*
f = 1&21 ()
=33-&. All of the light from a spotlight is collected and focused on a screen of area &.3& m%. hat must e the luminous intensity of the light in order that an illumination of 4&& l" e achie!ed/ E =
! A
%
! = EA = 54&& l"5&.3& m = 14& cd
*
! = 14& cd
=33-1. A 3&&-cd light is suspended 4 m ao!e the left edge of a tale. ind the illumination of a small piece of paper located a hori)ontal distance of %.4 m from the edge of the tale/ R =
5%.4 m% + 54 m% = 4.40 m + θ
tan θ =
E =
! cos θ R %
%.4 m 4m
=
*
θ =
%6.6& %.4 m
53&& cd5cos %6.6& 54.40 m%
*
%1
E = $.40 l"
4m
Physics, 6th Edition
Chapter 33. Light and Illumination
Critical Thin!ing Prolems 33-%. A certain radio station roadcasts at a fre#uency of 114& 7()* A red eam of light has a fre#uency of .2& " 1&1 () * and an ultra!iolet ray has a fre#uency of %. " 1&16 (). hich has the highest a!elength/ hich has the greatest energy/ hat are the a!elengths of each electromagnetic a!e/ @ Recall that h 6.6%4 " 1&-3 9.B λ =
λ =
λ =
c f c f c f
=
=
=
3 " 1&$ m's 6
*
λ = %61 m
E = hf = 2.6% " 1&-%$ 9
*
λ = 630
E = hf = 3.11 " 1&-10 9
*
λ = 1%.4
1.14& " 1& () 3 " 1&$ m's 1
.2& " 1& () 3 " 1&$ m's 16
%.& " 1& ()
nm
nm
E = hf = 1.40 " 1&-12 9
"he ra&io wa$e has highest wa$elength+ "he ultra$iolet ray has highest energy' =33-3. An un7non light source A located $& cm from a screen produces the same illumination as a standard 3&-cd light source at point . located 3& cm from the screen. hat is the luminous intensity of the un7non light source/ ! x r x%
=
%
!s rs%
* ! x =
! s r x r s%
=
53& cd5$& cm% 53& cm%
*
! x = %13 cd
=33-. :he illumination of a surface 3.& m directly elo a light source is %& l". ind the intensity of the light source/ At hat distance elo the light source ill the illumination e douled/ Is the luminous flu" also douled at this location/ ! = ER2 = 5%& l"53.& m% *
I %31 cd R2
2
2
E 2 = 2E ) an& E ) R) = E 2 R2 = *2E ) (R2
2
R) R% =
R1% %
=
53.& m% %
*
R2 = %.& m
%14
∆
=0
Physics, 6th Edition
Chapter 33. Light and Illumination
=33-4. :he illumination of an isotropic source is E A at a point A on a tale 3& cm directly elo the source. At hat hori)ontal distance from A on the tale top ill the illumination e reduced y one half/ E A =
! R A%
! % A
R
=
*
E. =
% ! cosθ R.%
*
! cosθ % .
R + %A + %F
% So that: 5cosθ =
E A = % E. *
*
+ F
! A = ! .
θ
+ A 3& cm
=
1 % cosθ
1 % cosθ
tan θ =
*
+A
= cosθ +F
but
5cos θ 3 =
*
x 3& cm
1 %
cos θ = 3 &.4 = &.20* and θ 32.4&
*
x = 53& cm tan 32.4& *
*
x
x = %3.& cm
=33-6. In i)eau8s e"periment to calculate the speed of light, the plane mirror as located at a distance of $63& m. (e used a heel containing 2%& teeth 5and !oids. E!ery time the rotational speed of the heel as increased y %.% re!'s, the light came through to his eye. hat !alue did he otain for the speed of light/ An increase of f %.% () is nee&e& for light to reappear from E&ge ) to E&ge 2' "hus/ the time for one re$olution of entire wheel is: " =
1 f
=
1 %.% ()
= &.&13 s
Now/ the time for one tooth an& one $oi& between ) an& 2 is: t=
&.&13 s 2%& teeth c =
= 4.2 x1&−4 s*
%5$63& m -4
4.2 " 1& s
*
"ime for one roun& trip'
c = 3.&1 " 1&$ m's
%16
