2.71 Optics
Quiz 1 Solutions
Spring ‘09
1. (a) Solution: Without correction, the focus is before the retina, leading to blurred
vision:
With correction, the image forms (focused) on the retina:
1
1
1
1
1
1
50 − 45
5
+
=
⇒
=
−
=
=
−fc + 15 50
45
−fc + 15
45 50
50 × 45
50 × 45
50 × 45
−fc + 15 =
= 450 ⇒ fc = −435 (Pc ≈ −2.3D)
5
(b) Solution: Draw again the ray-tracing diagram for the object at ∞:
x
x
x
=−
=
fc
−435
435
Draw the imaging system formed by EL above, with object = the virtual image
found by CL.
α1 = −
1
Angular magnification MA = −
450
= −9
50
⇒ α2 = −9 · α1 = −
9x
x
= − 1
435
48 3
From �ICA (see diagram on previous page),
x
= tan α2 = α2 (paraxial approximation)
EFL
x
1
= 1 ⇒ EFL = 48
3
48 3
(c) 1st method: Using the 2nd PP and EFL
1
xi = α · (EFL) = 48 α
3
2nd method: Using the imaging condition
2
x0 = αfc = −435α
The lateral magnification of EL’s imaging system (with the virtual image formed
by CL as the object and real image formed on the retina) is given by:
50
1
1
= − ⇒ xi = ML x0 = − · (−435α)
9
9
450
435
1
=
α = 48 α ⇒ Image is erect.
9
3
ML = −
(d) Solution: Use the eye’s pupil as the object and flip the optical system for proper
ray-tracing.
1
1
1
15 × 435
+ =
⇒z=−
15 z
−435
450
∴ virtual image created as shown
x�p
z
435
87
= ML = − =
=
xp
15
450
90
The myopic person’s eyes appear smaller and erect.
2. (a) Solution: First consider an on-axis point at the object plane.
NA =
a
f1
Clearly, S1 limits the angle of acceptance, so S1 is the Aperture Stop. Now
consider the chief ray from an off-axis point object.
3
FOV =
a2 ff12
f1
=
a2
f2
Clearly, S2 limits the chief ray if the point object elevation off-axis becomes suf­
ficiently large, so S2 is the Field Stop.
(b) Solution:
Entrance Pupil:
image S1 through L1
⇒ ∞ to the left
Exit Pupil:
image S1 through L2
⇒ ∞ to the right
Entrance Window: image S2 through L2, L1 ⇒ object plane
Exit Window:
⇒ collocated with S2
So the completely annotated system is:
(c) Solution: Numerical Aperture & Field of View
a1
f1
From Figure 2, S2 is imaged through L2, L1 onto the object plane. The lateral
magnification is found as follows:
From Figure 1, the angle of acceptance is NA =
� ��
a2
f
� a2 �
a2
f 1
�
�
FOV =
�
�
=
2 =
f1
f1
f2
����
limiting angle
of the chief ray
4
(half-field)
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2.71 / 2.710 Optics
Spring 2009
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