Chapter 30 Lenses

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# Chapter 30 Lenses - PowerPoint PPT Presentation

Chapter 30 Lenses. refraction The bending of the path of light refraction occurs as light passes across the boundary between two media A one-word synonym for refraction is "bending”. Refraction.

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## Chapter 30 Lenses

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### Chapter 30Lenses

refraction
• The bending of the path of light
• refraction
• occurs as light passes across the boundary between two media
• A one-word synonym for
• refraction is "bending”
Refraction
• The refraction occurs at the boundary and iscaused by a change in the speedof the light wave upon crossing the boundary
• The speed of a light wave is dependent upon theoptical densityof the material through which it moves
Lens
• A lens is an optical device which transmits and refracts light
• There will be two main types

Converging

Diverging

Converging Lens
• a lens which converges rays of light which are traveling parallel to its principal axis
• Converging lenses can be identified by their shape
• they are thicker across their middle and thinner at their upper and lower edges
Converging Lenses
• When parallel rays pass through a converging lens, they converge on a point and then diverge as they go on their way.
• Focal point (F)
• point, through which the rays all pass, F.
Converging Lenses
• Turn the lens around and it behaves the same way, therefore a focal point is on both sides of the lens.
• The focal length is positive for a converging lens.
Converging Lens Parts
• Focal Point (F)
• a point of convergence of light or a point from which it diverges
• Focal Length (f)
• the distance from a lens to its focus
• Focus
• a point of convergence of light
• a fixed reference point on the concave side
Converging Lens Parts
• Center of curvature (2 f)
• the center of the circle of curvature
• Optical axis
• A line that passes through the vertical of a lens
Converging Lens Parts
• Principal axis
• a line that passes through the center of curvature of a lens so that light is neither reflected nor refracted
Converging Lens Parts
• Object distance (do)
• the distance from the optical axis to the object
• Image distance (di)
• the distance from the optical axis to the image
Image Formation
• Three principal raysare necessary for image formation
• The point where the three rays meet is where the image is formed
Converging Lens Image Formation

Instead, we will continue the incident ray to the optical axis of the lens and refract the light at that point.

To simplify the construction of ray diagrams, we will avoid refracting each light ray twice - upon entering and emerging from the lens.

2F

F

F

2F

Convex Lens

Lenses

Object: f < do < 2f

pa

Image Characteristics

Real

Inverted

di > 2f

hi > ho

Any incident ray parallel to the principal axis will refract through the focus (on the other side of the lens).

Any incident ray passing through the (near side) focus will refract parallel to the principal axis

Any incident ray passing through the center of the lens will pass straight through the lens.

2F

F

F

2F

Convex Lens

Lenses

Object: do > 2f

pa

Image Characteristics

Real

Inverted

f < di < 2f

hi < ho

Any incident ray parallel to the principal axis will refract through the focus (on the other side of the lens).

Any incident ray passing through the (near side) focus will refract parallel to the principal axis

Any incident ray passing through the center of the lens will pass straight through the lens.

2F

F

F

2F

Convex Lens

Lenses

Object: do = f

pa

Image Characteristics

no image

Any incident ray parallel to the principal axis will refract through the focus (on the other side of the lens).

Any incident ray passing through the (near side) focus will refract parallel to the principal axis

Any incident ray passing through the center of the lens will pass straight through the lens.

2F

F

F

2F

Convex Lens

Lenses

Object: do = 2f

pa

Image Characteristics

Real

Inverted

di = 2f

hi = ho

Any incident ray parallel to the principal axis will refract through the focus (on the other side of the lens).

Any incident ray passing through the (near side) focus will refract parallel to the principal axis

Any incident ray passing through the center of the lens will pass straight through the lens.

2F

F

F

2F

Convex Lens

Lenses

Object: do < f

pa

Image Characteristics

Virtual

Upright

di < 0

hi > ho

Any incident ray parallel to the principal axis will refract through the focus (on the other side of the lens).

Any incident ray passing through the (near side) focus will refract parallel to the principal axis

Any incident ray passing through the center of the lens will pass straight through the lens.

The Mathematics of Lenses
• a ray diagram can only determine the approximate location and size of the image
• it will not provide numerical information about image distance and image size
The Mathematics of Lenses
• a ray diagram can only determine the approximate location and size of the image
• it will not provide numerical information about image distance and image size
• it is necessary to use the Lens Equationand the Magnification Equation
Lens Equation
• object distance (do)
• image distance (di)
• focal length (f)
Magnification Equation
• image height (hi)
• object height (ho)
• Magnification (M)
Combination equation
• You can combine the Lens Equationand the Magnification Equation
Sample Problem
• A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance (di ) and the image size(hi ).
Sample Problem (part 1)
• Step 1 Given:
• ho= 4.00 cm, do= 45.7 cm, f =15.2 cm, di = ?
• Step 2 Formula:
• 1/f = 1/do + 1/di
• Step 3 Plug in:
• 1/(15.2 cm) = 1/(45.7 cm) + 1/di
• Step 4 Calculate
• 0.0658 cm-1 = 0.0219 cm-1 + 1/di
• 0.0439 cm-1 = 1/di
• di= 22.8 cm
Sample Problem (part 2)
• Step 1 Given:
• ho= 4.00 cm, do= 45.7 cm, f =15.2 cm, di = 22.8, hi = ?
• Step 2 Formula:
• hi/ho = - di/do
• Step 3 Plug in:
• hi /(4.00 cm) = - (22.8 cm)/(45.7 cm)
• Step 4 Calculate
• hi = - (4.00 cm) • (22.8 cm)/(45.7 cm)
• hi = -1.99 cm the negative sign means it is inverted
Real images
• An object further than the focal length away from the lens forms a convergent real image.
Diverging Lens
• When rays pass through a diverging lens, they diverge from a point on the other side of the lens.
• This point is called the focal point F. there is a focal point on both sides of the lens.
• The focal length is negative for a diverging lens.
Diverging Lens Diagram

The image formed will be virtual and at the focal point in front of the lens

Diverging lenses have a negative focal length because the parallel rays entering the lens diverge

Ray Diagrams

Converging Lens

Diverging Lens

Real Image Formation
• Real images are like the one in the diagram below.
• The light actually passes through the image. The image will be projected on a screen placed at the image position.
• The image distance di is positive for a real image.
Virtual Image Formation
• Virtual images are like the ones in the diagram below.
• The light does not pass through the image. The image can not be projected on a screen placed at the image position.
• The image distance di is negative for a virtual image.
Real image vs. Virtual image
• What is the difference?
Lens Equation
• a ray diagram can only approximate the location and size of the image
• The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f)
Magnification Equation
• relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho)
Sample Problem
• A 6.0-cm tall light bulb is placed a distance of 35.3 cm from a double convex lens having a focal length of 10.2 cm. Determine the image distance

Given: ho = 6.0 cm

do = 35.3 cm

f = 10.2 cm

di= ?

M = ?

1/f = 1/do + 1/di

1/(10.2 cm) = 1/(35.3 cm) + 1/di

0.0 cm-1 = 0.0 cm-1 + 1/di

0.0 cm-1 = 1/di

di = cm

Sample Problem
• A 8.0-cm tall light bulb is placed a distance of 65.2 cm from a double convex lens having a focal length of 25.2 cm. Determine the image size

hi/ho = - di/do

Given: ho = 8.0 cm

do = 65.2 cm

f = 25.2 cm

di= ?

M = ?

hi/(8.0 cm) = - ( cm)/(65.2 cm)

hi = - (cm) * (cm)/(cm)

hi = - cm

Human eye

The light that enters the eye is refracted (bent) and then it reflects off objects in the eye. It then passes through several layers that focus and magnify the image

The eye lens curves depending on the image in focus. For closer images it increases its curve and for images farther away it decreases its curve

Study Well !!!!