Download Discussion Question 15C

Discussion Question 15C
P212, Week 15
Convex Mirrors
A convex mirror is mounted in a loading dock in order to provide a wide field of view. This
convex mirror is spherical, with an 80 cm radius of curvature. A simple camera is placed in
front of this mirror to monitor the activity in the loading dock. The camera consists of a single
converging lens of focal length 3.75 cm, placed 4.0 cm in front of a CCD array. A truck enters
the loading dock and stops at a position 22 m in front of the mirror.
4 cm
x
Rmirror = 80 cm
flens = 3.75 cm
22 m
Camera
(a) Consider the image of the truck formed by the convex mirror. Where is this image
located relative to the mirror?
First, use your knowledge of spherical mirrors to determine the focal length f of the mirror.
What sign should you use for f ?
R
= 40 cm , but the mirror is diverging ∴ f = -40 cm
2
1
1
1
+ =
→ s ' = −39.28 cm
2200 s ' −40
f =
(b) Could a person standing in front of the convex mirror see an image of the truck? If so,
what sort of image would it be? (Upright or inverted? Magnified or diminished?)
Since s' < 0, the image is in back of the mirror and thus visible to a person in front
−39.28
s'
of the mirror. m = - = −
= 1.8% hence image is upright and much smaller.
s
2200
4 cm
x
Rmirror = 80 cm
flens = 3.75 cm
22 m
Camera
(c) Where should the camera be placed to provide a focused image of the truck? Give the
distance x between the convex mirror and the camera lens.
Sign conventions again. As in the first discussion question, the image created by the first
optical element (the mirror) becomes the object for the second element (the lens). Just
remember that all directions and sign conventions in optics are based on the direction in
which light travels …
The image for the mirror is 39.286 behind the mirror. Hence object
for the lens is s = 39.286 + x and s ' = 4 . Hence :
1
1
1
+
=
→ x = 20.714 cm
4 39.286 + x 3.75
(d) If one half of the camera lens were covered up, what would happen to the image
recorded by the camera? Would half of it disappear?
Covering up ½ of the lens cuts half the light passing through the lens. Hence the image’s
intensity is cut in half.
2