Physics 202-Section 2G
Worksheet 11-Lenses
Formulas and Concepts
 Snell’s Law:
o When light enters a new medium (like when it’s traveling through the air and then runs
into water or glass) it will either speed up or slow down. When this happens, the light will
bend (refract).
o How fast light will move through a material is dependent on its index of refraction, n.
The index of refraction tells you the ratio between the speed of light in a vacuum (c) and
the speed of light in the new medium. Light travels at about the same speed in a vacuum
as it travels in air, so the index of refraction for air is 1.
 Movement of light to a material with a higher index of refraction means the light
will slow down. It will bend toward the normal in the new medium.
 Movement of light to a material with a lower index of refraction means the light
will speed up. In
 Mirrors can be plane (flat), concave (curved in), or convex (curved out).
 Light waves reflect off of mirrors to form images.
o An image that appears to be behind the mirror is virtual
o An image formed in front of the mirror is real
 The curvature of a mirror is defined by its radius of curvature. ½ of the radius of curvature is
called the focal length or focal distance.
1
o 2𝑅 = 𝑓
 How far away your light source/object is away from the mirror is related to how far away from the
mirror that the image appears.
o Height of the object is related to the apparent height of the image in the same way.
Sometimes images are magnified, sometimes they are reduced.
ℎ
−𝑑
o 𝑚 = ℎ𝑖 = 𝑑 𝑖
𝑜



𝑜
“m” is magnification
 Positive if the image is upright and negative if image is upside-down
 |<1| if image is reduced, |>1| is image is enlarged
o do is object distance and di is image distance
o ho is object height and hi is image height
Object and image distances are related to the focal length of the mirror
1
1
1
o 𝑓 =𝑑 +𝑑
o
𝑜
𝑖
Any distance that is in front of the mirror is positive. Any distance that is behind the mirror is
negative.
o All convex mirrors have negative focal lengths.
Convex mirrors always form reduced, virtual images.
To determine image heights and distances from a ray diagram, draw the principal rays and look for their
intersections:
1. Draw a line from the top of the object to the mirror, parallel to the axis. Reflect the ray through the
focal point.
2. Draw a line from the top of the object to the mirror through the focal point. Reflect the ray parallel
to the axis.
3. Draw a line from the top of the object to where the axis meets the mirror. Reflect the ray at the
same angle.
4. Draw a line to the mirror, connecting the top of the object with the radius of curvature. The ray will
reflect along the same line.
When necessary, extend the reflected rays behind the mirror to see where virtual images may form.
1. Draw a ray diagram to determine the location of the image formed by the object shown using
the concave mirror below.
Incident rays are blue, reflected rays are red.
The image is indicated as the large red arrow.
R
f
2. Draw a ray diagram to determine the location of the image formed by the object shown using
the convex mirror below.
f
R
3. A person is 20 cm in front of a concave mirror. Their image is 69 cm behind the mirror. Find
(a) the focal length of the mirror, as well as (b) its magnification.
Focal length: 28.16 cm. Magnification: 3.45x. Find the focal length using the formula 1/f = 1/do +1/di.
Object distance (do) is 20cm, image distance is -69cm (negative because the image is behind the mirror
and therefore virtual). Find magnification using the same numbers and plugging them into m = -di/do.
4. A mirror is placed 2.1 cm from an object. The enlarged image is located 6.2 cm behind the
mirror. (a) What kind of mirror is being used? (b) What is the focal length of the mirror? (c) What
is the magnification? (d) Is the image upright or inverted?
Cocave mirror – convex mirrors always form reduced images. This image is enlarged, so we have to
have a concave mirror.
Focal length: 3.176 cm. 1/f = 1/do + 1/di. Do is 2.1 cm, and di is -6.2 cm (negative because it’s behind
the mirror and therefore virtual).
Magnification: 2.95x. m = -di/do.
Image is upright. If the image were inverted, magnification would be negative. Magnification is positive
when image is upright.
5. If you want to create a 2.57 m tall virtual image of a 36.0 cm tall object, (a) what kind of mirror
should you use? (b) If the object is placed 3.00 m from the mirror, what radius of curvature
should the mirror have? (c) How far from the mirror will the image be formed?
Concave mirror: we’re trying to get an enlarged image, and only concave mirrors can do that.
R = 6.98 m. We know that radius of curvature is two times the focal length. So we really need to find the
focal length. We’re give image height (2.57m) and object height (0.36 m), and object distance (3 m), so
we can find image distance using the formula –di/do = hi/ho. Doing that, we get -21.4617 m for image
distance. Now we can find f using the formula 1/f = 1/di + 1/do. Focal length f is 3.49 m, so we double it to
6.98 to get radius of curvature.
21.417 m behind the mirror. We solved for this in the last part of this question. We get image distance to
be -21.417. Because it is negative, we know it is behind the mirror.