SIMG-303-20033
Solution Set #6
1. An optical system consists of thin lenses L1 (f1 = 100 mm) and L2 (f2 = 200 mm)
separated by d = 800 mm. Locate and describe the image of an object that is 50 mm
high located 150 mm ”in front” of the first lens.
At least three ways to do this. Easiest is probably by “brute force”, i.e., find the image
created by the first lens and use it as the object for the second lens. The magnification
of the system is the product of the magnifications of the two lenses.
1
1
+ 0
s1 s1
=
1
f1
s01
(MT )1
µ
¶−1
µ
1
1
−
=⇒
=
=
100 mm 150 mm
µ
¶−1 µ
¶−1
3
1
2
=
=
−
300 mm 300 mm
300 mm
= 300 mm
s0
300 mm
= −2
= − 1 =−
s1
150 mm
sl1
1
1
−
f1 s1
¶−1
t = s01 + s2
800 mm = 300 mm + s2 =⇒ s2 = 500 mm
1
1
+ 0
s2 s2
=
1
f2
s02
(MT )2
¶−1
¶−1
1
1
−
=⇒
=
=
200 mm 500 mm
¶−1 µ
¶−1
µ
2
5
3
−
=
=
1000 mm 1000 mm
1000 mm
1
1000
mm = 333 mm
=
3
¡ 1000 ¢ 3
0
mm
2
s
=−
= − 2 =− 3
s2
500 mm
3
sl2
µ
1
1
−
f2 s2
µ
50mm
66.7mm
150mm
800mm
f1 = 100mm
333.3mm
f1 = 200mm
4
2
MT = (MT )1 · (MT )2 = −2 · − = +
3
3
1000
The image is located 3 mm behind the second lens and the magnification is + 43 , so
the image size is 43 · 50 mm = 200
mm = 66 23 mm. The image is upright and magnified.
3
1
A second way to do this is is to locate the principal points of the lens and measure
the object distance to the object-space principal point. Then use the thin-lens imaging
equation to find the image distance, which is measured from the image-space principal
point. The equivalent focal length of the system is:
¶−1
µ
1
1
d
f1 f2
100 mm · 200 mm
fef f =
+
−
=
=
= −40 mm
f1 f2 f1 f2
(f1 + f2 ) − d
300 mm − 800 mm
The BFD is the distance from the rear vertex to the image-space focal point. The
formula for the BFD was derived in the notes:
BF D = V0 F0 =
f2 (f1 − d)
200 mm · (100 mm − 800 mm)
=
= +280 mm
(f1 + f2 ) − d
−500 mm
The distance from the image-space principal point H0 to the image-space focal point F0
is the effective focal length, so the distance from the image-space vertex to the imagespace principal point can be found:
V0 F0 = V0 F0 − H0 F0 = 280 mm − (−40 mm) = 320 mm
The object-space focal distance (FFD) may be found from the formula that was derived
in the notes:
F F D = FV =
100 mm · (200 mm − 800 mm)
f1 (f2 − d)
=
= +120 mm
(f1 + f2 ) − d
−500 mm
The distance from the object-space vertex V to the object-space principal point H can
now be found:
= FH − FV = −40 mm − 120 mm = −160 mm
=⇒ HV = +160 mm
VH
H
F
V
V'
800mm
f1 = 100mm
F' H'
f1 = 200mm
FH = -40mm
FV = FFD = +120mm
HV = +160mm
H'F' = -40mm
V'F' = BFD = +280mm
V'H' = +320mm
The distance from the object to the object-space focal point is:
s = OH=OV+VH
= 150 mm + (−160 mm) = −10 mm
2
Since the object distance is negative, the object in the equivalent thin-lens equation is
virtual. The image distance is now found directly:
1
1
1
+ 0 =
−10 mm s
−40 mm
µ
0
0
0
s = HO =
1
1
−
−40 mm −10 mm
¶−1
=+
40
1
mm = 13 mm
3
3
The distance from the vertex to the image is obtained by substitution into:
s0 = H0 O0 =H0 V0 +V0 O0
V0 O0 = H0 O0 − H0 V0
1000
1
40
mm = 333 mm
= + mm − (−320 mm) =
3
3
3
which is the same answer as that obtained above for s02 . The image magnification is
the ratio of the distances:
MT = −
+ 40 mm
4
s0
=− 3
=+
s
−10 mm
3
which again is the same answer as that obtained using the “brute-force” method.
3
2. A system consists of thin lenses L1 (f1 = −60 mm) and L2 (f2 =?) separated by d =
120 mm. Lens L2 is made of glass with n = 1.5 and is plano-convex with radius
r = 60 mm for the curved side. Locate and describe the image of an object that is
5 mm high located 180 mm ”in front” of the first lens.
We need to first find the focal length of L2 . The focal length is determined from the
curvatures of the surfaces and the index of refraction:
¶
µ
¶
µ
1
1
1
1
0.5
1
1
= 0.5 ·
−
=
=
= (n2 − nair )
−
f2
R1 R2
∞ −60 mm
60 mm 120 mm
f2 = 120 mm
1
1
1
+ 0 =
s1 s1
f1
¶−1
¶−1 µ
µ
1
1
1
1
0
−
s1 =
−
=
= −45 mm
f1 s1
−60 mm 180 mm
s2 = t − s01 = 120 mm − (−45 mm) = 165 mm
¶−1 µ
¶−1
µ
1
1
1
1
0
−
s2 =
−
=
= +440 mm
f2 s2
120 mm 165 mm
MT = (MT )1 · (MT )2
¶µ
¶
µ 0 ¶µ 0 ¶ µ
s2
−45 mm
440 mm
2
s1
−
= −
−
=−
=
−
s1
s2
180 mm
165 mm
3
The image is located 440 mm behind the second lens with a magnification of − 23 , so
the image size is − 10
mm = −3 13 mm
3
3. A thin lens is made of glass with index n = 1.53. In air, the lens has a focal length
f = 254 mm. What is its focal length when it is totally immersed in water (n = 1.33)?
¶
µ
1
1
1
= (n2 − n1 ) ·
−
Lensmaker’s equation
:
f
R1 R2
¶
µ
1
1
1
= (1.53 − 1.0) ·
−
If in air
:
254 mm
R1 R2
¶
µ
1
1
−
= 0.53 ·
R1 R2
¶
µ
1
1
1
1
=
−
=
=⇒
R1 R2
0.53 · 254 mm 134.62 mm
¶
µ
1
1
1
−
= (1.53 − 1.33) ·
If in water
:
f
R1 R2
1
1
= 0.20 ·
=⇒ f = 673.08 mm
f
0.53 · 254 mm
The focal length of the lens in water is considerably longer than its focal length in air
because the “relative refractivity” of the glass is much reduced in water. To see that
this result makes sense, consider what the focal length of the lens would be if immersed
in glass. The “relative refractivity” of the lens vanishes, so the focal length becomes
infinite.
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4. The surfaces of a thin equiconvex lens have equal radius of curvature: |R1 | = 150 mm.
The second surface is aluminized so that it is a mirror. Find the location of the image
of an object located 400 mm to the left of the first surface.
We can do this in three steps: refraction at the first surface, reflection at the second
surface, and refraction back at the first surface. Assume that the refractive index of
the glass is n = 1.5 :
n1 n2
1
n2 − n1
1.5 − 1
0.5
=
+ 0 =
=
=
s1
s1
R1
R1
150 mm 300 mm
¶−1 µ
¶−1
µ
0
s1
1
3
1
4
−
−
=
=
n2
300 mm 400 mm
1200 mm 1200 mm
µ
¶−1
1
=
=⇒ s01 = n2 · 1200 mm = 1800 mm
1200 mm
O'
O
V
400mm
V'
1200mm
Because the lens is thin, the distance from the first to the second surface is 0 mm. For
a mirrored surface, the image-space index n2 = −n1 :
−1.5 (−1.5 − 1.5)
1
1.5
−3
n2 n3
+ 0 =
=+
+ 0 =
=
s2
s2
−1800 mm
s2
R2
−150 mm
50 mm
µ
¶
¶
µ
−1
−1
s02
1
1
1
24
1200 mm
=
−
+
= +48 mm
=
=
1.5
50 mm −1200 mm
1200 1200
25
s02 = 1.5 · +48 mm = 72 mm
O'
72mm
O
V
5
V'
1800mm
The image is 48 mm “behind” the second surface, so s3 = −48 mm :
1.5
1
(1 − 1.5)
+ 0 =
s3
s3
−150 mm
¶−1
µ
1.5
1
400
1
−
mm ' −57.14 mm
=
=−
0
s3
300 mm 72 mm
7
This is “behind” the second surface, but since we are going in the negative direction,
the distance is “in front of” the lens. In other words, the distance is “positive”, so that
s03 = + 400
mm if measured in the same space as the object.
7
O' O
V
V'
The image is real and the magnification is
400
mm
s03
1
MT = − = − 7
=−
s1
400 mm
7
6
5. Two thin lenses of focal lengths f1 = −50 mm and f2 = +100 mm are separated by a
distance d = 50 mm. Find the focal length of the system of thin lenses and locate the
principal points.
The focal length of the system may be found from:
¶−1
µ
1
t
1
fef f =
+
−
f1 f2 f1 f2
¶−1
µ
1
50 mm
1
+
−
=
−50 mm 100 mm (−50 mm · 100 mm)
µ
¶−1
2
1
1
=
= 0−1
+
+
−100 mm 100 mm 100 mm
The system is AFOCAL; it is a telescope.
The focal length is ∞ and all cardinal points (focal and principal points) are located at ∞.
6. The solar telescope at the Kitt Peak National Observatory has a primary mirror with
a diameter of 60 in. The image formed by this mirror is 300 ft away. If the diameter
of the sun is 864, 000 mi and its distance is 93, 000, 000 mi. Find the diameter of the
image of the sun in mm.
The image size is proportional to the ratio of the object distance to the focal length, as
shown:
864, 000 mi
x
=
93, 000, 000 mi
300 ft
864, 000 mi
x =
· 300 ft
93, 000, 000 mi
' 2.787 ft ' 850 mm
7
7. Consider an optical system composed of two thin lenses, each with focal length of
100 mm. The first lens is located at the front focal point of the second. Find the focal
length, focal points, and the principal points of the system. If the focal length of the
first lens is changed, determine the effect on these parameters of the system.
The focal length of the system is:
1
fef f
1
fef f
1
1
t
+
−
f1 f2 f1 f2
1
100 mm
1
+
−
=
100 mm 100 mm 100 mm · 100 mm
1
=
100 mm
=⇒ fef f = 100 mm
=
The focal length of the system is the same as the focal length of each lens. The imagespace focal point is located at the second lens. The object-space focal point is located at
the first lens. Since the lenses are located fef f apart, then the principal points coincide
with the lenses, as shown in the sketch.
8
If the focal length of the first lens is changed, say f1 = 150 mm, then the focal length
of the system is:
1
fef f
1
fef f
1
1
t
+
−
f1 f2 f1 f2
1
100 mm
1
+
−
=
150 mm 100 mm 150 mm · 100 mm
1
=
100 mm
=⇒ fef f = 100 mm = f2
=
Thus the focal length of the system is equal to that of the second lens, regardless of the
focal length of the first lens.
(a) (OPTIONAL BONUS) Describe any applications for this optical system.
Consider the image formed by the second lens in the second system with f2 =
100 mm of a nearby object, say at s1 = 500 mm. The image formed by the single
lens is located at the distance:
µ
¶−1 µ
¶−1 µ
¶−1
1
1
1
4
1
0
s =
−
=
=
= 125 mm
−
f2 s
100 mm 500 mm
500 mm
The magnification is
125 mm
1
s0
=−
=−
s
500 mm
4
Now add the first lens. The focal length of the system is unchanged at fef f = f2 =
100 mm. Now find the image of an object located 500 mm in front of the SECOND
lens (i.e., in the same position as before) and find its magnification. Do it brute
MT = −
9
force:
s1 = 500 mm − t = 500 mm − 100 mm = 400 mm
µ
¶−1
1
1
1
1
1
0
+
=
=⇒ s1 =
= 240 mm
−
s1 s01
f1
150 mm 400 mm
s0
240 mm
= −0.6
(MT )1 = − 1 = −
s1
400 mm
The distance to the second lens is:
s2 = t − s01
= 100 mm − 240 mm = −140 mm
¶−1 µ
µ
¶−1
1
1
1
175
1
1
0
s2 =
−
=
=
−
mm = 58 mm
f2 s2
100 mm −140 mm
3
3
175
0
mm
5
s
=+
(MT )2 = − 2 = − 3
s2
−140 mm
12
The transverse magnification of the image formed by the system is the product:
5
1
3
MT = (MT )1 · (MT )2 = − · + = −
5
12
4
The image of the object formed by the first lens alone is located 125 mm behind
the lens with a magnification of − 14 , whereas the image formed by both lenses is
located 58 13 mm behind the second lens with the same magnification. The addition of the first lens located at the front focal point of the second lens moved the
image without changing its magnification. This is the action of eyeglasses — the
optometrist locates a corrective lens at the front focal point of the eyelens to move
the image onto the retina without changing the system magnification. In the case
just considered, the “corrective” lens L1 has positive power and moved the image
“forward”, as required for a farsighted person. Had the power of L1 been negative,
the image would have moved “backward” onto the retina of a nearsighted person.
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