93% Transmission over a km!

Discussion question: A fish swims below the surface of the water. The observer sees the fish at

A. A greater depth than it really is

B. The same depthC. A smaller depth than it really

is.

Imaging Assumptions “Thin Lens” approximation Diameter of lens/mirror is much

larger than the wavelength of light–This lets us do ray approximations–We’ll discuss what happens if this

isn’t true later Aberations

Making Images with Lenses and Mirrors

Conventions for Imaging Calculations

The Principal or Optical Axis– It is a line drawn from + infinity to –

infinity which passes through the center of the lens/mirror and is normal to the surface of the lens/mirror at its center.

Conventions for Imaging Calculations

p=distance from the object to lens or mirror along principle axis– p is usually positive...– but p is sometime negative (when I have a

“virtual object” as in the case for the second lens below).

p

-p

Conventions for Imaging Calculations

q = distance from the lens or mirror to the image– Measured using a “ruler” which is parallel to the

optical axis.

q

Conventions for Imaging Calculations

q = distance from the lens or mirror to the image– Measured using a “ruler” which is parallel to the

optical axis.– q is positive if the image is on the side of the

lens or mirror where we expect the light to go. In this case we say that the image is real.

Conventions for Imaging Calculations

q = distance from the lens or mirror to the image– q is negative if the image forms on the side of

the lens or mirror where the light doesn’t really go. In this case we say that the image is virtual.

-q

Is the image you see when you look in the mirror real or

virtual?A. RealB. VirtualC. YesD. Why?E. 42

Images Formed by a Flat Mirror

Is the image in a flat mirror real or virtual?

Is q for a flat mirror negative or positive?

Conventions for Imaging Calculations

M = the lateral magnification of the image– M=image height / object height– M is negative if the image is inverted– For a flat mirror, M is always 1

f = the focal length of a mirror or lens– If f is positive, it is the distance from the

mirror/lens that collimated light passing through it is focused to.

– If f is negative, it is the distance away from the mirror/lens of a spot that light appears to be diverging from when collimated light passes through it.

Negative and Positive f Mirrors

f

-f

Finding the Focal Length of a Concave Mirror

F=R/2

θ

θ

θ

Finding the Focal Length of a Convex Mirror

F=-R/2

Let us be of good cheer as we go about our lives. Although we live in increasingly perilous times, the Lord loves us and is mindful of us. He is always on our side as we do what is right. He will help us in time of need. Difficulties come into our lives, problems we do not anticipate and which we would never choose. None of us is immune. The purpose of mortality is to learn and to grow to be more like our Father, and it is often during the difficult times that we learn the most, as painful as the lessons may be. Our lives can also be filled with joy as we follow the teachings of the gospel of Jesus Christ. 

-- President Thomas S. Monson, General Conference, October 2012

R

Finding Image Location and Magnification by Drawing Lines

Finding Image Location and Magnification with Equations

The Mirror/Lens Equation,Magnification

How far from a converging mirror should I put an object if the mirror has a radius of curvature of 1 meter and I want the image to be the same size as the object?

A : less than 0.25 mB : 0.25 m to 0.75 mC : 0.75 m to 1.25 mD : 1.25 m to 3 mE : more than 3 m

I look at my image in a round Christmas ornament. Which of the following will be true?

A : The image will be upright (not inverted) and real.

B : The image will be upright and virtual.C : The image will be inverted and real.D : The image will be inverted and virtual.E : The ornament will shatter.