2/9/15 Oregon State University PH 212, Class #16 1

The Thin-Lens Equation

The thin lens approximation holds for a lens whose thick-ness is small compared to its focal length. With that assumption, plus some geometry and algebra, we can use this calculation for lenses: 1/d0 + 1/di = 1/f

Definitions and sign conventions:

Object distance (d0):+ when object is left of the lens (real);

Image distance (di):+ if the image is right of the lens (real);– if the image is left of the lens (virtual).

Focal length (f):+ for a converging lens.– for a diverging lens.

2/9/15 Oregon State University PH 212, Class #16 2

The Magnification Equation for Lenses

The magnification equation for lenses is:

m = hi/ho = –di/do

where ho is the object height and hi is the image height (negative if the image is inverted with respect to the positive object height).

The meaning of the sign of m:

Assuming the object is upright:

m is + if the image is upright.

m is – if the image is inverted.

2/9/15 Oregon State University PH 212, Class #16 3

1/do + 1/di = 1/f

m = hi/ho = –di/do

Practice: A lens has a focal length of –32 cm. An object is placed 19 cm in front of it (to its left).

•Calculate the image distance and magnification.•Describe the image. (Real/virtual? Upright/inverted? Enlarged/reduced?)

2/9/15 Oregon State University PH 212, Class #16 4

1/do + 1/di = 1/f

m = hi/ho = –di/do

Practice: A lens has a focal length of –32 cm. An object is placed 19 cm in front of it (to its left).

•Calculate the image distance and magnification.•Describe the image. (Real/virtual? Upright/inverted? Enlarged/reduced?)

do = 19, and f = –32, so: 1/di = 1/f – 1/do

Thus: di = (do)(f)/(do – f) = (19)(–32)/(19 - –32) = –11.9 cm

The image distance is negative, so the image is virtual, upright and reduced—indeed, the only sort of image that a diverging lens (i.e. a lens with a negative f) can produce. Verify this with the magnification eqn.: m = –di/do = –(–11.9/19) = (+)0.627

An object is placed 10.0 cm from a converging lens which has a focal length of 12.0 cm. What is the resulting magnification?

1. 6.00

2. -6.00

3. 2.00

4. -2.00

5. Not enough information.

2/9/15 5Oregon State University PH 212, Class #16

Formation of image with a Converging Lens summary:

• If object is located outside of focal point (do > f ): - Image is inverted. - Image is real. - Image is enlarged if f < do < 2f. - Image is reduced if do > 2f. - Image is on opposite side of lens from object.

- Examples: camera, eye, projector.

• If object is located inside of focal point (do < f ): - Image is upright. - Image is virtual. - Image is enlarged. - Image is on same side of lens as object.

- Example: magnifying glass.2/9/15 6Oregon State University PH 212, Class #16

Figure 23.36

2/9/15 7Oregon State University PH 212, Class #16

Figure 23.39

2/9/15 8Oregon State University PH 212, Class #16

Formation of image with a Diverging Lens summary:

- Image is upright.- Image is virtual.- Image is reduced in size.- Image is on same side of lens as object.

2/9/15 9Oregon State University PH 212, Class #16

Chapter Summary 23.4

2/9/15 10Oregon State University PH 212, Class #16

The real values of lenses and imaging, of course, are their uses:

They help to correct our view of the world (make it possible to view it clearly).

They help to enhance our view of the world (make it appear larger, more detailed).

How? The key to understanding all multiple-lens imaging systems (including your own eyes) is simple:

Light strikes the lenses in some order (first, second, third, etc.). The image produced by the first lens becomes the object for the next lens.

2/9/15 11Oregon State University PH 212, Class #16

At this point, be sure to read section 24.1 in the textbook.

Then see also the After class 16 materials for examples of how multiple lens systems work in tandem. The image produced by the first lens becomes the object that sends light to the next lens.

2/9/15 12Oregon State University PH 212, Class #16