Consider Refraction at Spherical Surfaces:

Starting point for the development of lens equations

Vast majority of quality lenses that are used today have segments containing spherical shapes. The aim is to use refraction at surfaces to simultaneously image a large number of object points which may emit at different wavelengths.

Point V (Vertex)

SVsO )object distance(

VPsi )image distance(

i - Angle of incidence

t - Angle of refraction

r - Angle of reflection

The ray SA emitted from point S will strike the surface at A, refract towards the normal, resulting in the ray AP in the second medium (n2) and strike the point P.

All rays emerging from point S and striking the surface at the same angle i will be refracted and converge at the same point P.

Let’s return to Fermat’s Principal

io lnlnOPLwithOPLdx

d210

Using the Law of cosines:

cos)cos(

cos2

cos2

cos2

2/122

2/122

222

with

RsRRsRland

RsRRsRl

RsRRsRl

iii

ooo

ooo

2/122

2

2/1221

21

cos2

cos2

,

RsRRsRn

RsRRsRnOPL

lnlnOPLwithTherefore

ii

oo

io

Note: Si, So, R are all positive variables here. Now, we can let d(OPL)/d = 0 to determine the path of least time.

Then the derivative becomes

0sin)(2cos22

1

sin2cos22

1)(

2/1222

2/1221

RsRRsRRsRn

RsRRsRRsRnd

OPLd

iii

ooo

We can express this result in terms of the original variables lo and li:

o

o

i

i

io

i

i

o

o

l

sn

l

sn

Rl

n

l

n

andl

RsRn

l

RsRn

1221

21

1

0sinsin

However, if the point A on the surface changes, then the new ray will not intercept the optical axis at point P.

Assume new small vaules of the radial angle so that cos 1, lo so, and li si .

R

nn

s

n

s

nthen

io

1221

This is known as the first-order theory, and involves a paraxial approximation. The field of Gaussian Optics utilizes this approach .

Note that we could have also started with Snell’s law: n1sin 1= n2sin2 and used sin .

Again, subscripts o and i refer to object and image locations, respectively.

Using spherical (convex) surfaces for imaging and focusing

i) Spherical waves from the object focus refracted into plane waves.

Suppose that a point at fo is imaged at a point very far away (i.e., si = ).

so fo = object focal length

Object focus

Rnn

nf

R

nn

f

n

R

nnn

s

nthen

oo

o

12

1121

1221

Suppose now that plane waves (parallel rays) are incident from a point emitting light from a point very far away (i.e., so = ).

ii) Plane waves refracted into spherical waves.

Rnn

nsf

R

nn

s

n

R

nn

s

nnswhen

iii

io

12

2122

1221

Diverging rays revealing a virtual image point using concave spherical surfaces.

Virtual image point

Parallel rays impinging on a concave surface. The refracted rays diverge and appear to emanate from the virtual focal point Fi. The image is therefore virtual since rays are diverging from it.

R < 0

fi < 0

si < 0

Signs of variables are important.

Rnn

nsf ii

12

2

A virtual object point resulting from converging rays. Rays converging from the left strike the concave surface and are refracted such that they are parallel to the optical axis. An object is virtual when the rays converge toward it.

so < 0 here.

012

1121

1221

Rnn

nsf

R

nn

s

n

R

nn

s

n

s

nswhen

ooo

ioi

The combination of various surfaces of thin lenses will determine the signs of the corresponding spherical radii.

S

)a(

)b(

)c(

As the object distance so is gradually reduced, the conjugate image point P gradually changes from real to virtual.

The point P’ indicates the position of the virtual image point that would be observed if we were standing in the glass medium looking towards S.

We will use virtual image points to locate conjugate image points.

In the paraxial approximation:

R

nn

s

n

s

nfrom

R

nn

s

n

s

n

io

ml

i

l

o

m 1221

111

The 2nd surface “sees” rays coming towards it from the P’ (virtual image point) which becomes the 2nd object point for the 2nd surface.

Therefore

dssssssdss ioiiooio 12112212 ,,,

)A(

)B(Thus, at the 2nd surface:221 R

nn

s

n

ds

n lm

i

m

i

l

Add Equations (A) & (B)

112121

11

ii

lml

i

m

o

m

sds

dn

RRnn

s

n

s

n

Let d 0 (this is the thin lens approximation) and nm 1:

)(11

111

21

cRR

nss l

io

and is known as the thin-lens equation, or the Lens maker’s formula ,

in which so1 = so and si2 = si, V1 V2, and d 0. Also note that

oos

iis

fsfsio

lim,lim

For a thin lens (c) fi = fo = f and

fss

RRn

f

io

l

111

111

1

21

Convex f > 0

Concave f < 0

Also, known as the Gaussian lens formula

Location of focal lengths for converging and diverging lenses

1m

llm n

nn 1

m

llm n

nn

If a lens is immersed in a medium

with

21

111

1

RRn

f lm

m

llm n

nn

f2f

2ffObject

Real image

Convex thin lens

Simplest example showing symmetry in which so = 2f si = 2f

Concave, f < 0, image is upright and virtual, |si| < |f|

f

fObject

Virtual image

2

3

1siso

Note that a ray passing through the center is drawn as a straight line.

Ideal behavior of 2 sets of parallel rays; all sets of parallel rays are focused on one focal plane.

For case (b) below

1,1

111111

fswheref

fs

sfs

sfsfss

oo

oi

oiio

0, ii sfs

Tracing a few key rays through a positive and negative lens

yo

S2

S1

Consider the Newtonian form of the lens equations.

From the geometry of similar triangles:

fxfxfssfy

y

s

s

ioioi

o

i

o

111111,

22

1

fxxfxx

ff

fxx

fxf

fxfxf

ffx

ioio

io

o

io

o

Newtonian Form:

xo > 0 if the object is to the left of Fo .

xi > 0 if the image is to the right of Fi .

The result is that the object and image must be on the opposite sides of their respective focal points.

Define Transverse (or Lateral) Magnification:

f

x

x

f

xfx

xxff

fx

fxf

fx

fx

s

s

y

yM

i

o

oo

oo

o

o

o

i

o

i

o

iT

1//2

2f f

f 2f

Image forming behavior of a thin positive lens.

MT > 0 Erect image and MT < 0 Inverted image. All real images for a thin lens will be inverted.

f2f

2ffSimplest example 2f-2f conjugate imaging gives

1f

f

x

fM

oT

Define Longitudinal Magnification, ML

022

22

To

Lo

io

iL M

x

fM

x

fxand

dx

dxM

This implies that a positive dxo corresponds to a negative dxi and vice versa. In other words, a finger pointing toward the lens is imaged pointing away from it as shown on the next slide.

The number-2 ray entering the lens parallel to the central axis limits the image height.

The transverse magnification (MT) is different from the longitudinal magnification (ML).

Image orientation for a thin lens:

)a (The effect of placing a second lens L2 within the focal length of a positive lens L1. (b) when L2 is positive, its presence adds convergence to the bundle of rays. (c) When L2 is negative, it adds divergence to the bundle of rays.

Two thin lenses separated by a distance smaller than either focal length.

Note that d < si1, so that the object for Lens 2 (L2) is virtual.

Note the additional convergence caused by L2 so that the final image is closer to the object. The addition of ray 4 enables the final image to be located graphically.

Fig. 5.30 Two thin lenses separated by a distance greater than the sum of their focal lengths. Because the intermediate image is real, you could start with point Pi’ and treat it as if it were a real object point for L2. Therefore, a ray from Pi’ through Fo2 would arrive at P1.

Note that d > si1, so that the object for Lens 2 (L2) is real.

11

112

11

1122

21

21

22

222

222

2

2

12

11

111

111

,111

)(0

)(0

,111

fsfs

fd

fsfsf

df

fsd

fsd

fs

fss

sfs

reals

virtualssds

fs

fss

sfs

o

o

o

o

i

i

o

oi

oi

o

o

io

o

oi

oi

For the compound lens system, so1 is the object distance and si2 is the image distance.

The total transverse magnification (MT) is given by

1111

21

2

2

1

121 fsfsd

sf

s

s

s

sMMM

oo

i

o

i

o

iTTT

For this two lens system, let’s determine the front focal length (ffl) f1 and the back focal length (bfl) f2.

Let si2 then this gives so2 f2.

so2 = d – si1 = f2 si1 = d – f2 but

21

211

211112

2

11111

ffd

fdfsffl

fdfsfS i

i

soiso

From the previous slide, we calculated si2. Therefore, if so1 we get,

21

12

12

12

12

12

1222

111

,0

fff

fff

fffflbfldfor

ffd

fdf

ffd

ffdfsbfl

ef

ef

i

fef = “effective focal length”

Suppose that we have in general a system of N lenses whose thicknesses are small and each lens is placed in contact with its neighbor.

1 2 3……… NThen, in the thin lens approximation:

Nef fffff

1...

1111

321

Fig. 5.31 A positive and negative thin lens combination for a system having a large spacing between the lenses. Parallel rays impinging on the first lens enable the position of the bfl.

Example A Example B

Example A: Two identical converging (convex) lenses have f1 = f2 = +15 cm and separated by d = 6 cm. so1 = 10 cm. Find the position and magnification of the final image.

111

111

fss io

si1 = -30 cm at (O’) which is virtual and erect

Then so2 = |si1| + d = 30 cm + 6 cm = 36 cm

222

111

fss io

si2 = i’ = +26 cm at I’ Thus, the image is real and inverted.

The magnification is given by

17.236

26

10

30

2

2

1

121

o

i

o

iTTT s

s

s

sMMM

Thus, an object of height yo1 = 1 cm has an image height of yi2 = -2.17cm

Example B: f1= +12 cm, f2 = -32 cm, d = 22 cm

An object is placed 18 cm to the left of the first lens (so1 = 18 cm). Find the location and magnification of the final image.

111

111

fss io

si1 = +36 cm in back of the second lens, and thus creates a virtual object for the second lens.

so2 = -|36 cm – 22 cm| = -14 cm

222

111

fss io

si2 = i’ = +25 cm; The magnification is given by

57.314

25

18

36

2

2

1

121

o

i

o

iTTT s

s

s

sMMM

Thus, if yo1 = 1 cm this gives yi2 = -3.57 cm

Image is real and Inverted