OPTI 415 Homework 2
1. Is this person near-sighted or far-sighted? Justify your answer.
The lens is a positive lens since the side of his face appears to move outwards when
looking through the glasses as compared to outside the glasses. The drawing below
illustrates this effect. Positive powered lenses correct for far-sightedness.
2. Using raytracing code, find the locations of the cardinal points in the following system.
Where are the entrance and exit pupils located? If the physical diameter of the aperture
stop is 13.4 mm, what are the diameters of the entrance and exit pupil? What is the FNumber of the system?
Radius (mm)
Thickness (mm)
Index
24.607
5.080
1.517
-36.347
1.600
1.620
212.138
12.300
1.000
∞ (Stop)
21.700
1.000
-14.123
1.520
1.517
-38.904
4.800
1.620
-25.814
37.934
1.000
I used Zemax with a wavelength of 587.6 nm and BK7 (n = 1.517) and F2 (n = 1.620) for
the materials. The locations of the cardinal points are below, as well as F/# and pupil
properties.
3. You need to design a red light camera. If the working distance is 15 m and the full field of
view in the horizontal direction is 40°. The image should look something like the image
below. Answer the following questions:
a) How wide is the scene in the object plane?
The FFOV is given as 40°, so the HFOV is 20°. The width of the scene will be given by
2 × 15m × tan(HFOV ) = 10.9m
b) If the camera sensor has dimensions of 23.6 x 15.8 mm, what is the magnification of the
system?
From (a), the object width of 10.9m needs to be mapped to a width of 23.6 mm on the
sensor, so the magnification is
m=
image size
0.0236
=−
= −0.0022
object size
10.9
Note the minus sign since we expect the image to be upside down on the sensor.
c) What is the focal length of the lens that is needed?
We can write the Gaussian imaging formula as
z=
1− m
− 0.0022 × (− 15m )
f ⇒ f =
= 32.9mm
m
1.0022
d) A character on the license plate is 75 mm tall. We would like this height to correspond to
25 pixels to be able to read the character. What are the number of pixels (assume square
pixels) in the horizontal and vertical directions?
75 mm will map to an absolute height of 75mm × −0.0022 = 165µm on the sensor, so 25
pixels corresponding to this height means the pixels are 165µm 25 = 6.6µm square. The
sensor size is 23.6 x 15.8 mm, so the pixel dimensions are 23.6 0.0066 = 3576 pix in the
horizontal direction and 15.8 0.0066 = 2394 pix in the vertical direction.
e) Roughly, how many megapixels are in the sensor? Is this a reasonable number for
currently available technology?
The sensor has 3576 × 2394 = 8,560,944 pix ≅ 8Mpix . These sensors are readily available.