Lecture 14 Images Chp. 35 Opening Demo Topics –Plane mirror, Two parallel mirrors, Two plane mirrors at right angles –Spherical mirror/Plane mirror comparison.

Lecture 14 Images Chp. 35

• Opening Demo• Topics

– Plane mirror, Two parallel mirrors, Two plane mirrors at right angles

– Spherical mirror/Plane mirror comparison in forming image– Spherical refracting surfaces– Thin lenses– Optical Instruments

Magnifying glass, Microscope, Refracting telescope• Warm-up problem

Geometrical Optics:Study of reflection and refraction of light from surfaces using the ray approximation.

The ray approximation states that light travels in straight linesuntil it is reflected or refracted and then travels in straight lines again.The wavelength of light must be small compared to the size ofthe objects or else diffractive effects occur.

Reflection Refraction: Snells Law

€

n1 sinθ1 = n2 sinθ2

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θ1 = θ1'

n1

n2

Using Fermat’s Principle you can prove theabove two laws. It states that the path taken by light when traveling from one point to another is the path that takes the shortesttime compared to nearby paths.

Plane mirrors

Normal

Angle ofincidence

Angle of reflection

i = - p

Real side Virtual side

Virtual image

Problem: Two plane mirrors make an angle of 90o. How many images are there for an object placed between them?

object

eye

1

2

3

mirror

mirror

Problem: Two plan mirrors make an angle of 60o. Find all images for a point object on the bisector.

object

2

4

5,6

3

1

mirror

mirror

How tall must be a mirror be for a person to see his entire reflection in it. Let H be the height of a man and let l be the height of the mirror?

What happens if we bend the mirror?

i = - p magnification = 1

Concave mirror.Image gets magnified.Field of view is diminished

Convex mirror.Image is reduced.Field of view increased.

Thin Lenses: thickness is small compared to object distance, image distance, and radius of curvature.

Converging lens

Diverging lens

Thin Lens Equation

€

1

f=

1

p+

1

i

Lensmaker Equation

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1

f= (n −1)(

1

r1−

1

r2

)

What is the sign convention?

Sign Convention

p

Virtual side - V Real side - R

i

Light

Real object - distance p is pos on V side (Incident rays are diverging)Radius of curvature is pos on R side.Real image - distance is pos on R side.

Virtual object - distance is neg on R side Incident rays are converging)Radius of curvature is neg on the V side.Virtual image- distance is neg o the V side.

r2r1

Rules for drawing rays to locate images

•A ray initially parallel to the central axis will pass through the focal point.

•A ray that initially passes through the focal point will emerge from the lens parallel to the central axis.

•A ray that is directed towards the center of the lens will go straight through the lens undeflected.

Show Fig 35-14 figure overhead

Then go to example

24(b). Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.

pfi

111−=

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m =′ y

y= −

i

p

€

1

i=

1

5−

1

10= +

1

10

Image is real, inverted.

. .F1 F2p

Virtual side Real side

€

m = −10

10= −1

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i = +10 cm

24(e). Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

21

111

1

rrn

f

( )30

1

30

1

30

115.1

1=⎟

⎠

⎞⎜⎝

⎛−

−−=f

cmf 30=

pfi

111−=

15

1

10

1

30

11−=−=

i

cmi 15−=

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m =′ y

y= −

i

p

€

m = −−15

10= +1.5

Image is virtual, upright.

Virtual side Real side

r1. .F1 F2

pr2

27. A converging lens with a focal length of +20 cm is located 10 cm to the left of a diverging lens having a focal length of -15 cm. If an object is located 40 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. Treat each lensSeparately.

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

Ignoring the diverging lens (lens 2), the image formed by theconverging lens (lens 1) is located at a distance

€

1

i1=

1

f1

−1

p1

=1

20cm−

1

40cm. i1 = 40cm

40

This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.

30

Since m = -i1/p1= - 40/40= - 1 , the image is inverted

€

1

i2

=1

f2

−1

p2

=1

−15cm−

1

−30cm i2 = −30cm.

Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i2 < 0).

f1

f1

Lens 1 Lens 2

f2

f2

10

40

+20 -15

40

30

The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image

has the same size orientation as the object.

Optical Instruments

Magnifying lensCompound microscopeRefracting telescope