Lenses

• Knowing how curved mirrors reflect light can help you understand how lenses affect light rays.

• A lens is a curved piece of transparent material. Light refracts as it passes through a lens, causing the light rays to bend.

Basic Lens Shapes

•A converging lens is a lens that is thickest in the middle and causes incident parallel rays to converge at a single point.

•A diverging lens is a lens that is thinnest in the middle and causes incident parallel rays to spread apart after refraction.

Diverging (Concave) Lenses

• A concave lens is thinner and flatter in the middle than around the edges.

• Light passing through the thicker, more curved areas of the lens will bend more than light that passes through the flatter, thinner area in the middle.

• Rays of light are spread out (diverged) after passing through the lens.

Optical Center

Secondary Principal Focus (F’)

Converging (Convex) Lenses

• A convex lens is thinker in the middle than around the edges.

• This thicker middle causes the refracting light rays to come together (converge)

Optical Center

Secondary Principal Focus (F’)

Three Rules for a Converging Lens

1. A ray parallel to the principal axis is refracted through the principal focus.

2. A ray through the secondary principal focus is refracted parallel to the principal axis.

3. A ray through the optical center continues straight through without being refracted.

Using Ray Diagrams with Converging Lenses

• Draw the ray parallel to the principal axis. Draw the refracted ray so that it passes through the principal focus.

• Draw a ray from the top of the object through the middle of the lens. This ray is undeviated. Where the rays meet, that is where the image is.

• You can also draw the incident ray passing through F’ on the way to the lens. The result will be a parallel ray to the principal axis.

Locating Images – Object beyond 2F’

•Image is • Smaller• Inverted• Between F and 2F• Real

Locating Images – Object at 2F’

•Image is • Same size• Inverted• At 2F• Real

Locating Images – Object between 2F’ and F’

•Image is • Larger• Inverted• Beyond 2F• Real

Locating Images – Object AT F’

•Image is • No clear image due to

emergent rays being parallel

Locating Images – Between F’ and Lens

•Image is • Larger• Upright• Behind the lens• Virtual

Three Rules for a Diverging Lens

1. A ray parallel to the principal axis is refracted as if it had come through the principal focus.

2. A ray that appears to pass through the secondary principal focus is refracted parallel to the principal axis.

3. A ray through the optical center continues straight through without being refracted.

•A diverging lens always produces the same image no matter where the object is.

• Smaller• Upright• On the same side as the object• Virtual

The Lens Equation

• There are two ways to determine the characteristics of images formed by lenses: with ray diagrams or with algebra!

Lens Terminology• do = distance from the

object to the optical center

• di = distance from the image to the optical center

• ho = height of object• hi = height of image• f is the focal length of the

lens; distance from the optical center (f’ is the same length as f)

Thin Lens Equation

•Object distance (do) is always positive•Image distance (di) is positive for real images and negative for virtual images.•F is positive for a converging lens and negative for a diverging lens

Sample Question

• A converging lens has a focal length of 34. A candle is located 96 cm from the lens. What type of image is formed and where is it located?

•do = 96 cm•f= 34 cm•2f = 68 cm

•The object is further than 2f. According to the rules, the image should be smaller, inverted, between f’ and 2f’, and it is real.

• 1/f = 1/34 = 0.0294

• 1/do = 1/96 = 0.0104

• 1/di = 1/f – 1/do 0.0294 – 0.0104

0.0190

• 1/di = 1/0.0190 = 52.6 cm

Example 2• A pencil is located 18 cm from the lens. The

image is real and located 24 cm away from the optical center. What is the focal length of the lens?

• 1/do = 1/18 = 0.0556• 1/di = 1/24 = 0.0417• 1/f = 1/do + 1/di

0.0556 + 0.04170.0973

• 1/f = 1/0.0973 =10.2 cm

Example 3• A diverging lens (image is upright and virtual)

has a focal length of 25 cm. A virtual marble has a distance of 12 cm in front of the lens. What is the distance of the object?

• 1/f = -1/25 = -0.0400 (- because of virtual)• 1/di = -1/12 = -0.0833• 1/do = 1/f - 1/di

-0.0400 – (-0.0833)0.0433

• 1/do = 1/0.0433 =23.09 cm

The Magnification Equation

• When you compare the size of an image with the size of an object, you are determining the magnification of the lens.

•Object height and image heights are positive when measured upright and negative when measured downward

•Magnification is positive for an upright image and negative for an inverted image.

•Magnification is dimensionless.

Sample Question

• A toy has a height of 8.4 cm. It stands in front of a converging lens. The image is real, inverted, and has a height of 23 cm. What isthe magnification?

• ho = 8.4 cm• hi = - 23 cm (inverted)• M = hi/ho• M = -23 / 8.4• M = -2.7

Sample Question 2

• A brick is placed 7.2 cm from the optical center. An upright, virtual image with a magnification of 2.3 is noticed. What is the di?

• do = 7.2 cm• M = 2.3 cm• M = -di/do• di = -M x do• di = -2.3 x 7.2• di = -16.56 cm (image is on the same side as the object,

16.6 cm from the optical lens.