Download AP Physics B – Geometric Optics – FR 2 Answer Key SECTION C

AP Physics B – Geometric Optics – FR 2 Answer Key
SECTION C – Geometric Optics
#1 (1974-B3)
a)
b) 1/f = 1/do + 1/di … 1/3 = 1/4 + 1/ di … di = 12 cm
c)
M = – di / do …
M = –12/4 = –3 … M = hi / ho … –3 = hi / 1cm … hi = – 3cm
#2 (1976-B6)
a)
1/f = 1/do + 1/di … 1/8 = 1/6 + 1/di … di = –24 cm, left of lens, virtual
b) M = – di / do …
M = – (–24) / 6 = +4 … M = hi / ho … 4 = hi / 1 cm … hi = 4 cm, upright
c)
d) Magnifying glass, or telescope / microscope.
#3 (1978-B5)
a)
b) 1/f = 1/do + 1/di … 1/10 = 1/20 + 1/ di … di = 15 cm
c)
M = – di / do …
M = –15/30 = –0.5 … M = hi / ho … –0.5 = hi / 6cm … hi = – 3cm
d) Reflection Ray PQ is shown in the diagram as the topmost ray. For this ray, the incoming θ = reflection θ.
#4 (1979-B6)
a)
Note: The angles given are tricks. They are not measured from the normal, we must use the angle from normal.
Air  Plastic
ni sin θi = nr sin θr,
(1) sin (53) = nr sin(37)
nr = 1.33
n = c /v
1.33 = 3x108 / v
v = 2.26 x 108 m/s
b) Check the critical angle Plastic  Air. ni sin θc = nr sin (90) …
(1.33) sin θc = (1) … θc = 48.75°
The incoming angle (53°) is larger than the critical angle, so total
internal reflection will occur.
c) This looks like a converging lens, but in lens problems, the lenses always have a higher index than the
surrounding material. Since this ‘air lens’ has a smaller index of
refraction it will behave the opposite as a normal lens would. From
a simple refraction standpoint, there will be no refraction entering
the air since the ray is perpendicular to the surface. Then on exiting,
as you move less–more dense the ray bends towards the normal.
#5 (1980-B4)
e) Based on 1/f = 1/do + 1/di
Rearrange for y as function of x (make f – since concave)
1/di = (–1/f) + (–1/do) ..
the focal point term is negative and constant so …
As do increases the 1/do term decreases adding less
and less – values to the right side of the equation. When do
gets very large the right side becomes constant.
#6 (1981-B5)
a)
b) This is a real image. The image is on the other side of the lens. It is ‘projectable’ into real space.
The math below also proves that its real
c) 1/f = 1/do + 1/di … 1/6 = 1/18 + 1/di … di = 9 cm
d)
#7 (1982-B6)
a) 1/f = 1/do + 1/di … 1/f = 1/ (3f/2) + 1/di … di = 3f
b)
c) The new lens makes a larger magnification. M = – di / do  with the same do, di must have increased. Then
based on 1/f = 1/do + 1/di, to get a larger di with the same do the focal point must be larger. You can test this with
sample numbers if you don’t see it at first.
#8 (1984-B5)
a) c = f λ … 3x108 = f (6x10–7) … f = 5x1014 Hz.
b) n1 λ1 = n2 λ2
… (1) (6x10–7) = (1.25) = 4.8x10–7 m = 480 nm
c) The light is traveling to progressively more dense materials, so undergoes ½ λ phase shifts at both boundaries S 1
and S2 essentially canceling out this phase shift (from flips). To get minimum intensity (destructive), the total phase
shift from traveling in the film should be ½ λfilm so the film thickness should ¼ λfilm = ¼ 480 nm = 120 nm
d) Based on the same analysis above. To get maximum intensity (constructive) the total phase shift from traveling
in the film should be 1 λfilm, so the film thickness should be ½ λfilm = ½ 480 nm = 240 nm
# 9 (1987-B5)
a) Law of reflection, angle of incidence = angle of reflection. θ 3=30°
b) ni sin θi = nr sin θr, … (1.6) sin 30 = 1 sin θ2 … θ2 = 53.1°
c) n = c /v
d) First c=fair λair
(1.6) = 3x108 / v
n1 λ1 = n2 λ2
v = 1.875x108 m/s
… nair λair = nglass λglass … nair (c/fair) = ng λg … λg = 3.125x10–7 m
e) Determine critical angle, glass to air.
ni sin θc = nr sin 90
Any angle larger than 38.68° will cause total reflection.
(1.6) sin θc = (1)
θc = 38.68°
#10 (1988-B5)
a)
b) ni sin θi = nr sin θr, … 1.5 sin (37) = 1 sin θr
c) ni sin θi = nr sin θr, … ni sin (37) = 1
θr = 65° … therefore angle α shown above is 28°.
ni = 1.66, any n higher than this causes total internal.
d)
This ray would not totally reflect because by putting it in water we have reduced
the difference between the n’s meaning less bending and the limit of total internal
has not be reached.
e) ni sin θi = nr sin θr, … (1.66) sin (37) = (1.33) sin θr … θr = 48.7° – 37° = 11.7°
#11 (1990-B6)
a) The reflected angle is the same as the angle of incidence which can be found using the geometry.
Tan θ = o/a = 2/3 … θ = 34°.
b) ni sin θi = nr sin θr … (1.33) sin (34°) = (1) θr … θr = 48°.
c) We need to find the critical angle first … ni sin θc = nr sin 90 … (1.33) sin θc = 1 … θc = 49°.
Keeping the light source at the same horizontal distance
away (2m) we have to raise up the light source
(decreasing the 3m depth) until the we reach a point
New location of light
where the new depth and the 2m horizontal
source to increase θi to
side of that triangle reach a 49° angle
the needed 49°
Using geometry. Tan θ = o/a
tan(49) = 2 / ynew  ynew = 1.8 m
d) Light travels from less n, to more n, to less n. A 180° ( ½ λ) phase shift will happen at the air–oil boundary only.
e)
Based on the ½ λ phase shift air–oil, we need to create and additional ½ λ, or 1½ λ, or 2 ½ λ … phase shift from
traveling in the oil to create maximum (constructive) intensities. To do this, oil thicknesses of
t = 1/4 λoil , 3/4 λoil , 5/4 λoil … can be used. For an oil thickness of 100 nm (as given), these correspond to
wavelength in the oil of 400 nm, 133.33 nm, 80 nm …
Now we have to convert these oil wavelengths to air wavelengths using
n1 λ1 = n2 λ2 …
nair λair = nfilm λfilm … (1)( λair) = (1.5)(400) … λair = 600 nm. … repeating for the other λ’s and we get
possible air wavelengths of: 600 nm, 200 nm, 120 nm …
Since the visible spectrum in air ranges between 400–700 nm, only the 600 nm λ qualifies.
#12 (1992-B6)
a)
b) i) 1/f = 1/do + 1/di … 1/15 = 1/45 + 1/di … di = 22.5 cm.
ii) M = – di / do … M = –(22.5)/45 = –0.5 … M = hi / ho … –0.5 = hi / 8 cm … hi = – 4 cm
c) As describe in MC section question 18, this would cause the image to dim only.
d)
#13 (1997-B5)
a) The object is located in front of the lens (to the left). The image is located behind the lens (to the right). That
means this is a real image, which can only be created by a converging lens.
b) 1/f = 1/do + 1/di … 1/f = 1/30 + 1/90 … f = 22.5 mm
c)
d) Real, Larger, Inverted.
e)
e)