Image formation Ray tracing Calculation

Lecture 3.3.

Lenses Convex Concave

Optical instruments

Mirrors Convex Concave

Image formation

Spherical Lenses

Convex (converging) lens

Laws of refraction and reflection can be used to explain how lenses and mirrors operate

Parallel rays from the Sun passing through a convex lens

Sunlight concentrated by magnifying glass may burn hole in paper placed at focal point F

f

F F: focal point f: focal length

Example A farsighted person requires an eyeglass of strength 2.5 diopters. What is the focal length of the eyeglass lens?

D =2.5m-1 f = 1 D =1/2.5m = 0.4 m = 40cm.

F is the focal point and f is the focal length

1

f

Refractive power (or Strength) of the lens

in diopters (D) = ( f is in metres)

f

F

Image formation

(a) 4 cm, (b) 40 cm or (c) 400cm

F F

s′

h′

Image Formation (ray tracing)

Ray 1 entering lens parallel to optic axis will exit and pass through the focal point Ray 2 passing through the focal point will exit the lens and travel parallel to optic axis

Ray 3 will undergo small deviation (displacement (not shown) (thin lens)

3

Real, inverted image formed Real image (may be projected and displayed on a screen)

Object at a distance greater than the focal length from the lens

2

1

s

h

Rays are reversible

F F s

s′

h

h′

Image Formation (ray tracing)

Image • virtual • upright •magnified.

Object at a distance less than the focal length from the lens

Simple magnifier

Concave (diverging) lens

F

Dashed lines indicate the direction from which the rays appear to come

f

Rays entering lens parallel to axis appear to originate at focal point

Optic axis

Concave (diverging) lens

s

s′ F

h h′

Virtual image always produced by a concave lens

Cannot be viewed on screen since rays are diverging on the right of the lens

However can be viewed with the eye since the eye converges the rays onto the retina.

Object outside the focal length

Optic axis

Ray diagrams are useful in sketching the relationship between object and image

Relationship may also be calculated

F F

s s′

h

h′

A

B O C

D

Triangles AOB and DOC are similar

''

h hs s= − ' 'h s f

h f−= −

1 1 1's s f+ =

E

Triangles EFO and DFC are similar

Equating the expressions

' 'h sh s= −

''hh

f s f= −

−

Image Formation---calculation

' 's s fs f

−=

1 1 1's s f+ =Thin lens formula

Object distance s positive if object is in front of lens negative if object is behind lens

Image distance s′ positive if image is formed behind the lens (real) negative if is formed in front of the lens (virtual)

Focal length f •positive -- convex lens •negative --concave lens

Image ---calculation

F

s s′

h

h′ B

F

F

s s′

h

h′ B

F

'hMh

=

Magnification is defined as

M Negative : inverted image

M Positive : upright image

'sMs

= − or

Magnification

Simple magnifier

Object placed inside focal length of converging lens;

F F s

s′

h h′

Image viewed. •Virtual, •Magnified •Upright

Example An object 0.5 cm in height is placed 8 cm from a convex lens of focal length 10 cm. Determine the position, magnification, orientation and height of the image. 1 1 1

's s f+ =

1 1 1's f s= −

1 1 1' 10 8s= − s′ = 10 x 8

8-10 = - 40cm.

' 40 58

s cmMs cm

−= − = − =

h′ = M x h = 5 x 0.5cm = 2.5cm 'hM

h=

M +ve image upright

F F s

s′

h h′

Effective focal length (feff) of combination of a number of thin lenses close together

1 2

1 1 1 ......efff f f

= + +

Effective refractive power (or strength) (Seff) of combination of thin lenses close together

1 2 .....effS S S= + +

Combining Lenses

Determine the combined strength of a thin convex lens and a thin concave lens placed close together if their respective focal lengths are 10cm and -20cm.

1

f Strength S, in diopters (D) = ( f is in metres)

1 2

1 1 1eff

eff

Sf f f

= = + 1 1.1 .2effS

m m= −

10 5 5effS diopters= − =

An object is placed 45 cm from a lens of focal length -25 cm. Determine the position, magnification, and orientation of the image.

1 1 1's s f+ = 1 1 1

's f s= −

M +ve image upright

Example

1 S’

1 -25 cm

1 45 cm = - S’ = -16.1 cm

M = - S’ S M = - -16.1 cm

45 cm = 0.36

s

s′ F

h h′ Optic

axis

Mirrors Flat Mirror Concave Mirror Convex Mirror

Curved mirrors analogous to lenses Ray tracing and thin lens equation also valid. real and virtual images are also formed

Upright, virtual image

Object and image distance equal

Flat Mirror

d d object image

Spherical Mirrors

Spherical mirror

C

R

Hollow sphere

Spherical mirror is a section hollow sphere

Principal or optic axis

Radius of curvature R = 2f

Curved (Spherical) Mirrors

F f

f F

Concave mirror (converging)

Convex mirror (diverging)

Thin lens formula may be used to determine object and image distances and focal lengths etc

Real image : inverted (h′ negative), positive image distance s′ (RHS of lens, LHS of mirror) Virtual image: upright (h′ positive), negative image distance s′ (LHS of lens, RHS of mirror)

Positive focal length

Negative focal length

For all single lenses and mirrors:

Mirrors Concave shaving/makeup mirrors

C F

Image is virtual, upright and enlarged.

Object placed at distance < f from mirror

Question: if object is placed at the focal point, where is image located?

Application: searchlight

Mirrors Example

An object is positioned 5 cm in front of a concave mirror of focal length 10 cm. Determine the location of its image and its characteristics.

1 1 1's s f+ =

s = 5 cm f = 10 cm

1 1 1's f s= −

1 1 1' 10 5s cm cm= −

1 1 2 1' 10 10 10s cm cm cm= − = − S’ = -10cm

Characteristics. •Image virtual •Located behind mirror

'sMs

= −

105

2

cmMcm

M

−= −

=

Optical instruments

System may have many optical elements (example, lenses and mirrors)

Microscopes, telescopes, cameras etc

Thin lens formula or ray tracing may be used to analyse behaviour of such systems

Simple compound microscope two convex lenses

Fo

Fe

h′′

eyepiece

Fe

Objective lens

h′

Fo

Object (height h)

Final image

image formed by objective lens is inside focal length of eyepiece lens.

Simple refracting telescope

Fe

Objective lens eyepiece

Fo,

Optical instruments

Virtual image at infinity, Magnified and inverted

Objective lens forms image (real, inverted) at focal point F0 which is also the focal point Fe of the eyepiece; virtual image is then formed at infinity.

0

e

fMf

= −

f0 fe

Optical instruments

Effective focal length of the objective in the Hubble telescope is 57.8 m. What focal length eyepiece is required to give a magnification of -8.0 x 103.

0

e

fMf

= −0

effM

= −

33

57.8 7.23 108.0 10e

mf m−= − = ××

Eyepiece lens

Objective, concave mirror

Flat mirror

Simple reflecting telescope

Example

Optical instruments

Medical loupes Galilean Design objective

eyepiece Operating site

Typically Magnification m ≈ 2.5 → 4.5 Optical design allows

observer focus at infinity thereby relieving eyestrain

Typical working distance 28-38 cm

Endoscope for medical investigations— inserted through small incision or orifice to inspect and facilitate operation on interior parts of the body

flexible shaft includes: light source to illuminate area, image channel to view area under investigation, air or water conduit to clear debris, instrument conduit

Endoscope

Typical endoscope eyepiece

Transmits light, air , water

Flexible shaft

Instrument entry

Optical instruments Optical fibre: Total Internal Reflection

Camera Optical instruments

CCD array

Real, inverted image formed on CCD array

Lens translated to change image distance to adjust for different object distances.

aperture

Example Microscope as shown in the previous figure has the following characteristics objective lens focal length = 0.7 cm eyepiece lens focal length = 4.5 cm separation between lenses = 21 cm. Object size = 0.01 cm. object to objective lens distance = 0.73 cm. Calculate total magnification produced and size of the final image.

1 1 1's s f+ = 1 1 1

's f s= −

image distance s′ due to objective lens

1 1 1' 0.7 0.73s= −

s′ = 17 cm

Magnification M0 = -s′/s = -17/0.73 = -23.3

size of the image = h′ =M0h = (-23.3)(0.01)cm = -0.233 cm.

M0 = h′/h Magnification negative because image is inverted

Magnification Me = -s′/s = -(-36/4) = 9

Size of the final image = h′ ′ =Meh′ = (9)( -0.233) = -2.1cm.

Total magnification MT = (-2.1cm)/(0.01 cm) = -210

Total magnification MT = M0 x Me = (-23.3) x 9 = - 210

1 1 1's s f+ =

1 1 1's f s= − 1 1 1

' 4.5 4s= −

Image produced by objective is situated at a distance (21-17) cm = 4 cm from the eyepiece.

s′ = -36 cm

s′ negative: image is on same side of eyepiece as object