4 . ( 15 point s )
A ray of red light in air (7 = 650 nm ) is incident on a semicircularblock of clear plastic (n = 1.51for this light),
as shown above. The ray strikes the block at its center of curvature at an angle of incidence of 27".
(a) Part of the incident ray is reflected and part is refracted at the first interface.
i. Determine the angle of reflection at the first interface.Draw and label the reflected ray on the
diagram above.
ii. Determine the angle of refraction at the first interface. Draw and label the refracted ray on the
diagram above.
iii. Determine the speedof the light in the plastic block.
iv. Determine the wavelength of the light in the plastic block.
(b) The sourceof red light is replacedwith one that producesblue light ( )" - 450 nm ), for which the plastic
has a greater index of refraction than for the red light. Qualitatively describe what happensto the
reflected and refracted rays.
(c) The semicircular block is removed and the blue light is directed perpendicularly through a double slit and
onto a screen.The distancebetween the slits is 0.15 mm. The slits are 7.4 m from the screen.
i. On the diagram of the screenbelow, sketch the pattern of light that you should expect to see.
i
Midpoint
bstweenslits
ii. Calculatethe distancebetweentwo adjacentbrightfringes.
5 . ( 10 point s )
A thin converging lens L of focal length 10.0 cm is used as a simple magnifier
to examine an object O that is
placed 6.0 cm from the lens.
(a) on the figure below, draw a ray diagram showing at least
two incident rays and the position and size of the
lmage fbrmed.
(b)
i.
Indicate whether the image is real or virtual.
-
Real
Virtual
ii. Justify your answer.
(c) Calculate the distanceof the image from the center of the lens. (Do
NoT simply measureyour ray diagram.)
(d) The objectis now moved3.0 cm to the right,as shownabove.
How doesthe heightof the new image
comparewith thatof the previousimage?
It is larger.
Justify your answer.
It is smaller.
It is the same size.
(a)
(i) 2 points
For drawinga reflectedray at approximatelythe sameangleto the normalas
the incidentray
For clearlyindicatingthatthis is the reflectedray
( ii) 4 points
Snell'slaw is usedto find the angleof refraction
n, sin0t = frzsin9,
For correctlysubstitutingvaluesinto Snell'slaw
l . 0 s i n 2 ' 7=o 1 . 5 1 s i n 0 ,
= 0.30
sin0, = sin27"11.51
For the correctvalueof the angle
0z = 17.5"
For drawinga ray at approximatelythe correctangle
For clearly indicatingthat this is the refractedray
(iii) I point
The speedin the block canbe determined
usingthe definitionof indexof refraction.
u - cln = (:.oox lo8 ^ls)ft.st
For the correct answer
u - L99 x 108 m/s
(a) (continued)
(i v) 2 points
For a statement
For
statr
that the frequencyis the samein the two materials,or an equation
that is an applicationof thatfact
that
t ..;ufl - . =
uair
u,ai,
lai'
trur,
= ?plutti.
Aptar..
\
I
uo,.r,,ri'.',, I
^
Zplastic = -;/plastic
oR
lPhstic=
I _
T
)
"air
=tiffiff(650nm)
=
zprastrc
zplastic
oR lphstic=#
For the
For
the correct
co
answerwith units
= 431nm OR 430 nm
=
2plastic
2plastic
(b)
2 points
The following
follov
pointswereonly awardedif rayswereshownor describedin part (a)
For indicating
indica
thatthe angleof reflectiondoesnot change
For indicating
indica
thatthe angleof refractionbecomessmaller
(c)
(i) 2 points
Examplein which the dark linesin the drawnpatternrepresent
the brightbandsof blue light
I
I
I
Iilill
llllil
i
Midpoint
befir'ecnslie
For indicatinga centralpeakin the pattern
For havingapproximately
evenspacingbetweenmaxima
A sketchof the intensitygraphwasalsoacceptable
(ii) 2 points
For usingan appropriate
formula(or combinationof formulas)andcorrectlysubstituting
For example
m)"L
Y=
'-m
d
Xt^t t=
(r)(+so
x lo-ern){r.a
*)
0. 15 x 10-3 m
For the correctanswer
x-4.2mm
3 points
Example:
(a)
For eachcorrectray from the objectusedto locatethe image,1 point wasawarded,to a
maximumof 2 points
For the correctsizeandorientationof the image
(b)
(i)
I point
For an indicationthatthe imaseis virtual
(ii)
1 point
For a correctjustification
Example:The raysemergingfrom the lensdid not actuallyconvergeat the imagebut
only appearto havedoneso.
This point wasonly awardedif the point for part(bxi) wasawarded.
(c)
2 points
For correctuseof the equationrelatingimagedistanceto objectdistanceandfocal
length,with correctsubstitutions
l+1_1
'ti ro
f
-'
J;--
fs.
so-,f
_ _ ( 1 0 . 0c m ) ( 6 . 0c m )
"t
6 . 0 c m - 1 0 . 0c m
For the correctanswerwith the correctsign
s; = -15 cm
(d) 3 points
Example:
For an indicationthat the heightof the new imageis larger
For a correctjustification(only whenthe previouspointhasbeenawarded)
Justificationapproach1:
The rayspassingthroughthe lensbecomelessdivergingfrom eachotherdueto the
geometry.Their extensions
meetfurtherawayfrom the lensmakingthe image
larger.
Justificationapproach2: Constructa new ray diagram
For eachcorrectray from the objectusedto locatethe image,1 point was awarded,to a
maximumof 2 points.The studentmustexplicitlyreferto the diagramasthe
justificationfor the largerheightin orderfor the diagramto be considered.
Justificationapproach3: Calculatethe changein magnification
For a correctcalculationof the imagelocationafterthe objectis moved
111
si so J
- (lorocmX2'Qcm)
=-eocm
",,t =-F,- ^
so f
9 . 0c m - 1 0 . 0c m
For a correctcalculationof the magnificationbothprior to movingthe objectandafter
the objectis moved,leadingto the conclusionthatthe new imageis larger
M ', = - s i
M . 't = - t '
so
so
---15cm -*2.5
ocm
-90 cm
=
= *[0
9cm
The magnificationincreases
so the heightof the new imageis larger.