Chapter 25/26 - Part B Mirrors & Lenses Flashcards
Lamp is placed 10cm in front of concave mirror, that forms image of lamp placed 3m from mirror.
a) What is the radius of curvature?
b) Lamp is 5mm high, how tall is the image? What is the magnification?
a) Here you will use the mirror equation!
1/f = 1/d0 + 1/di
do = 10cm (positive because real object (in front of mirror)
di = 300cm (positive because image is on screen in front of mirror, meaning its a real image)
focal length = positive because concave mirror
Applying mirror equation you get f = 9.68cm
R = 2f soooo R = 19.4cm
b) Use magnification equation
-di/do
= -300/10 = -30M
Negative sign means image is inverted
hi = Mho
(-30)(5mm) = -150mm
Negative height means inverted image
Sign Convention of Radius of Curvature/Focal Length
Positive for concave
Negative for convex
What kind of image do convex mirrors and diverging lenses (concave) always produce
Smaller, upright, virtual
Diverging/Convex SUV
Da,Ca,Sa
What is the relationship between concave/convex mirrors and converging/diverging lenses?
Converging Lens: Convex but behaves like concave mirror
Diverging Lens: Concave but behaves like a convex mirror
Santa stands 0.750m from ornament. Diameter of ornament is 7.20cm. His height is estimated at 1.6m. Where and how tall is the image of Santa? What is his orientation (upright or inverted)
do = 0.750 (real object so positive)
R = 7.20/2 = -0.0360
f = -0.0360/2
Negative, since it’s a convex mirror
h0 = 1.6m (real object so positive)
Use information, do and f to solve for di
-0.0176
Negative, so virtual image
Magnification
-di/do 0.0235
Positive so image is upright, less than one so image is smaller
Image height
hi = M * ho
0.0235 * 1.6
3.76 cm
A dentist using a concave mirror with magnification 2 when mirror is 1.25cm from a tooth. What must the radius of curvature be?
1)
Solve for magnification first
m = -di/do
di=-2*1.25 = -2.50
2)
Use mirror equation to solve for f using di = -2.50 and do = 1.25
3) R = 2f
so R = 5cm
What is the sign convention for di for a lens?
+ (real image opposite side of object)
- (virtual image on object side
How do you determine what type of lens it is?
How do you determine real or virtual image?
Use the information to determine the focal length
Determine the sign of di (negative di means virtual)
How do virtual images work for mirrors and lenses?
For mirrors, virtual image is behind mirror.
For lenses, virtual image is in front of lens
In a lens, what is the object and virtual side? How do determine sign of R?
Incoming light is the virtual side.
Determine C1 and if C1 is on the same side as the outgoing, then R1 is positive. If it is on opposite sides then it is negative
Figure shows a jalapeño
seed O1
that is placed in front of
two thin symmetrical coaxial lenses
1 and 2, with focal lengths f1
= +24 cm
and f2
= +9 cm, respectively, and with
lens separation L = 10 cm. The seed is
6.0 cm from lens 1.
Where does the system of two lenses
produce an image of the seed?
Step 1
f1 = +24
d01 = -6 (since object is on “virtual” side where light came from)
Using lensmaker equation you get -8cm
Image is 8cm to the left of Lens 1
Step 2
Image is 8cm to the left of lens 1, lens 2 is 10cm to the right of lens 1
so 10 + 8 = 18cm
thus d02 = -18cm
Now apply lens formula for lens 2
f2 = 9cm
d02 = -18cm
di2 = 18cm
Image is 18cm to the right of lens 2
focal length of objective lens: 15mm
focal length of eyepiece lens: 25mm
distance between lenses = 61mm
object distance for objective lens d01 = 24.1mm
STEP 1
Find image formed by objective lens.
f0 = 15 and d01 = 24.1
di1 = 39.7
STEP 2
do2 = 61mm - 39.7 = 21.3
Distance between object and eyepiece
Now use lens formula
fe = 25.5
do2 = 21.3
di2 = 130
The wavelength of the red light from a helium–neon laser is 633 nm in
air, but 474 nm in the jellylike fluid inside your eyeball, called the
vitreous humor. Calculate the index of refraction of the vitreous humor
n = lair/l
=633/474
=1.34
You (height of your eyes above the water, 1.75 m) are standing 2.00 m
from the edge of a 2.50-m-deep swimming pool. You notice that you can
barely see your cell phone, which went missing a few minutes before, on
the bottom of the pool. How far from the side of the pool is your cell
phone? nwater
= 1.333
Draw a line to the normal
Find the angle of reflection based on trigonometry
48.81 degrees
Apply Snells Law
1.00sin48.81 = 1.333sinx
x = 34.37
Use trig again to find distance of cell phone
d= 2.50tan34.37
= 1.71m
When solving for a convex mirror, what do you ensure to do in your mirror equation?
Make the focal length NEGATIVE
A small tropical fish is at the center of a water-filled spherical fishbowl 28.0 cm in
diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The
effect of the thin walls of the bowl may be ignored
Use the equation
n2/di - n1/do = n2-n1/R
Treat it like a concave mirror, so the focal length is positive
1/di - 1.33/13 = 1-1.33/14
di = 14cm on same side as fish
What is the distance between successive wave fronts?
One wavelength
A plate glass window (n = 1.5) has a thickness of 4.0 x 10-3 m. How long does it take light to
pass perpendicularly through the plate
t = d/v
v = c/n
t = dn/c
Answer should be 2 x 10^-11seconds
A flat sheet of ice has a thickness of 2.0 cm. It is on top of a flat sheet of crystalline quartz
that has a thickness of 1.1 cm. Light strikes the ice perpendicularly and travels through it and
then through the quartz. In the time it takes the light to travel through the two sheets, how
far (in centimeters) would it have traveled in a vacuum?
Find the tiem it takes to travel through ice and then quartz
t = d/v
v = c/n
t = dn/c
soooo total t = dn/c + dn + c
That gives total time
Now in a vacuum, n = 1 so d = ct
using time from before
The crystalline lens of the human eye is a double-convex lens made of material having
an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 mm, which also
varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the
radii of curvature of this lens. (b) If an object 16 cm tall is placed 30.0 cm from the eye lens, where would
the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note:
The results obtained here are not strictly accurate, because the lens is embedded in fluids having refractive
indexes different from that of air.)
Use the lensmaker equation for a)
Note that R1 = R becuase converging lens and R2 = -R
So you get
1/f = (1.44-1)(2/R)
Solve for R and you get 7mm
The cornea behaves as a thin lens of focal length approximately 1.8 cm,
although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front
surface is convex, with a radius of curvature of 5.0 mm. (a) If this focal length is in air, what is the radius of
curvature of the back side of the cornea?
Use the lens maker equation
1/0.018 = (1.38-1)(1/0.005 - 1/R2)
R2 = 1.86cm
