1 32 Optical Images image formation reflection & refraction mirror & lens equations Human...

1

32 Optical Images

• image formation

• reflection & refraction

• mirror & lens equations

• Human eye

• Spherical aberration

• Chromatic aberration

2

image formation

• real image: rays converge to a pointEx. sun’s rays focused by magnifier

• virtual image: apparent source of light divergence Ex. image seen in mirror

3

Plane Mirror Image

• distances measured from mirror & axis

• image height hi = object height ho

• object distance do = image distance di

44

concave mirror images

55

convex mirror images

• always “virtual”, “upright”, “diminished”

66

lens images

77

Lens & Mirror Equation

• relate do, di & f.

• lateral magnification (LM) = hi/ho.

you must learn the sign conventions to use these formulas

fdd io

111

o

i

o

i

d

d

h

hLM

8

imbedded object

• image distance < object distance

99

lens power

• power = 1/(focal-length(meters))

• unit: [D, diopters, 1/m]

• used for corrective lenses

• Ex. near-sighted person P = -5.0Df = 1/P = 1/(-5) = -0.2m = -20cm.

10

human eye• average index of refraction ~ 1.4

11

Far Sighted Eye

• correction requires converging lens

12

near sighted eye

• correction requires diverging lens

13

Spherical Aberration

• Spherical aberration: light striking near edge focus at different points than light striking near center

14

ChromaticAberration

• Chromatic aberration: different colors focus at different points

15

Summary

• Image formation by lenses & mirrors

• Real/virtual, orientation, magnification.

• Near & far-sighted problem & correction

• Spherical & chromatic aberration

16

2. A document is sealed in a glass cube (n=1.5) at a depth of 10cm from one surface. Calculate s’ and m.

5.100.11

10

5.1

s

00.1)10(1

)67.6(5.1

2

1

sn

snm

Object is in n1=1.5. Viewer is in air n2=1.00. Radius of surface is r = infinity.

1/s’ = -0.15 s’ = -6.67cm.

.

17

OI

18

Question: Reverse the designations in the example above. Object is now the fish, so n1 = 1.33 (fish location). Let the fish be 10cm from bowl surface. Where is the image of the fish formed that the cat will see?

15

33.100.100.1

10

33.1

s

197.1)10(00.1

)00.9(33.1

2

1

sn

snm

Answer: s = 10cm , s’ = ?, r = -15cm, n1 = 1.33, n2 = 1.00.

0.133 + 1/s’ = 0.022 1/s’ = -0.111 s’ = -9.00cm

19

Calculating Focal Length for a Thin Lens

cmrr

nf

1215

1

10

1)15.1(

11)1(

1

21

Example: A double convex, thin glass lens with n = 1.5 has radii of curvature of 10 and 15cm. Find its focal length.

20

21

22

r

nn

s

n

s

n 1221

15

00.133.133.1

10

1

s

s, cat’s nose

n1 holds object, here equal to 1.00

0.1 +1.33/s’= 0.022

1.33/s’ = -0.078

s’ = -17.1cm

sn

snm

2

1

29.1)10(33.1

)1.17(00.1

m

23

24

25