November 21, 2008
PHY2054 Discussion-Fall ‘08
Quiz 9 (Chapter 23.1-23.5)
Name:
UFID:
*1. (5pts) A person walks into a room that has, on opposite walls, two plane mirrors producing
multiple images. Find the distance from the person to the third image seen in the left-hand mirror
when the person is 1.50 m away from the mirror on the left wall and 2.50 m away from the mirror on
the right wall.
The image for a flat mirror is as far behind the mirror as the object is in front. Thus the 1st image in
the left-hand mirror is 1.50 m behind the mirror.
The 1st image in the right-hand mirror is 2.50 m behind the mirror, which is 6.50 m in front of the
left-hand mirror. Thus the 2nd image in the left hand mirror is 6.50 m behind the mirror.
The 1st image in the left-hand mirror forms the 2nd image in the right-hand mirror, which is 5.50 m
behind the mirror. Since this image is 9.50 m from the left-hand mirror. The 3rd image in the left-hand
mirror is 9.50 m behind the mirror, which is 11.0 m away from the person.
**2. (5pts) A ball is dropped from rest 2.00 m directly above the vertex of a concave mirror having a
radius of 0.500 m and lying in a horizontal plane. At what times is the size of the image double the
size of the object?
If the image is real, the magnification is negative. Thus we have
M = -2 = -q/p ⇒ q = 2p
We apply the mirror equation and solve it for the object distance.
1/p+1/2p = 2/R ⇒ p = 3R/4 = 3×0.5/4 = 0.375 m
Using one of the kinematics equations, the time is calculated as
Δy = -(1/2)gt2 ⇒ t = √(-2Δy/g) = √(-2×(0.375-2)/9.8) = 0.576 s
If the image is virtual, the magnification is positive. Thus we have
M = 2 = -q/p ⇒ q = -2p
1/p+1/(-2p) = 2/R ⇒ p = f/4 = 0.5/4 = 0.125 m
t = √(-2Δy/g) = √(-2×(0.25-2)/9.8) = 0.618 s
**3. (5pts) A glass plate with a refractive index of 1.65 rests on the bottom of an aquarium tank. The
plate is 12.0 cm thick and covered with water (n = 1.33). The top surface of the plate is 15.0 cm
below the water surface. What is the apparent thickness of the plate as viewed from above the
water?
Light rays from the top of the plate are refracted at the water surface. The image is located at
qt = -(na/nw)pt = -(1/1.33)×15 = -11.28 cm or 11.28 cm below the water surface.
Light rays from the bottom surface of the plate are refracted twice, once at the top of the plate and
once at the water surface. The location of the image formed by the top surface of the plate is
q1b = -(nw/ng)p1b = -(1.33/1.65)×12 = -9.67 cm or 9.67 cm below the top surface of the plate.
This image, which is 24.67 cm below the water surface, serves as an object for the refraction at the
water surface. The final image is formed at
q2b = -(na/nw)p2b = -(1/1.33)×24.67 = -18.55 cm
The thickness is given by
t’ = |qt-qb| = |-11.28-(-18.55) | = 7.27 cm (=t/ng = 12/1.65)
***4. (5pts) A parallel beam of light enters a transparent hemisphere perpendicular to the flat face.
The radius of the hemisphere is 8.00 cm and the index of refraction is 1.50. A mirror is in contact
with the hemisphere at the vertex and it is perpendicular to the principal axis. Determine the point at
which the beam is focused. (Assume paraxial rays-that is, assume that all rays are located close to
the principal axis.)
The beam is refracted at the spherical surface when it goes out of
the hemisphere. Taking object distance to be infinity, we have
n/p1+1/q1 = (1-n)/R ⇒ 1.5/∞+1/q = (1-1.5)/(-8) ⇒ q1 = 16.0
cm
Then it reflects back from the flat mirror. The image is formed at
q2 = -p2 = -(-16) = 16.0 cm in front of the mirror.
The beam enters the hemispherical lens again, thus we have
1/p3+n/q3 = (n-1)/R ⇒ 1/(-16)+1.5/q3 = (1.5-1)/8 ⇒ q3 = 12.0 cm to the left of the mirror or 4.00
cm left of the flat surface of the lens.
Finally it refracts at the flat surface of the lens. The final image is located at
q4 = -(1/n)p4 = -(1/1.5)(-4) = 2.67 cm to the left of the flat surface of the hemisphere.