Chapter 26: Refraction, Lenses, Optical Instruments · Chapter 26: Refraction, Lenses, Optical Instruments ¥ Refraction of light, SnellÕ s law . Apparent depth ¥ Polarization of

$Page 1: Chapter 26: Refraction, Lenses, Optical Instruments · Chapter 26: Refraction, Lenses, Optical Instruments ¥ Refraction of light, SnellÕ s law . Apparent depth ¥ Polarization of$
Chapter 26: Refraction, Lenses, Optical Instruments

• Refraction of light, Snell’s law. Apparent depth

• Polarization of light on reflection

• Dispersion of light. Prisms, rainbows, sun dogs

• Formation of image by lenses, thin lens equation, magnification

• Combination of lenses

• The human eye – correction for near and farsightedness

• Optical instruments

• Omit 26.14, lens aberrations

1Friday, March 9, 2007

Refraction of light

When light travels from one transparent medium to another, it is in general

deflected from its original direction –!this is refraction.

The amount by which

the light is deflected

depends on the

refractive index of

each medium.


Refractive index – a measure

of the speed of light in a

transparent medium

Refractive index=Speed of light in vacuum

Speed of light in medium

n=c

v

Prob. 26.6: Light has wavelength 340 nm and

frequency 5.403"1014 Hz when travelling

through some substance. What substance is it?

v = c/n = f ! = (5.403"1014 Hz)"(340"10-9m)

# = 1.837"108 m/s

n = 3"108/1.837"108 = 1.633 – carbon disulphide

and v =cn

= f !


Prob. 26.8: A flat sheet of ice has a thickness of 2 cm. It is on top of a flat

sheet of quartz that has a thickness of 1.1 cm. Light strikes the ice

perpendicularly and travels through it and then through the quartz.

In the time it takes the light to travel through the two sheets, how far

would it have travelled in vacuum?

Ice: # t1 = l1/v1 = l1 n1/c as v1 = c/n1

Quartz:# t2 = l2/v2 = l2 n2/c

In time (t1 + t2), light would travel a distance L = c(t1 + t2) in vacuum.

t1 = (0.020 m) " 1.309/c = 0.02618/c

t2 = (0.011 m) " 1.544/c = 0.01698/c

So L = c(t1 + t2) = 0.0432 m = 4.32 cm

Ice

n1 = 1.309

Quartz

n2 = 1.544

l1 = 2 cm

l2 = 1.1 cm

t1 + t2 = 0.04316/c


Refraction and Reflection

Snell’s law: n1 sin!1 = n2 sin!2

Light incident from above:

If !1 = 30!

sin!2 =n1 sin!1

n2

=1! sin30"

1.33

!2 = 22.1"

“External

reflection”


Light incident from below:

sin!2 =n1 sin!1

n2

=1.33! sin30"

1

!2 = 41.7"

If !1 = 30!

Snell’s law: n1 sin!1 = n2 sin!2

“Internal

reflection”(inside medium of higher refractive index)


Prob. 26.C8: Two rays of light converge to a point on a screen.

A plane-parallel plate of glass is placed in the path of this converging

light and the glass plate is parallel to the screen.

Will the point of convergence remain on the screen?

If not, will the point move toward the glass of away from it?

n1 = 1 n2 > n1

As n2 > n1, the ray is refracted

toward the normal to the surface

when it enters the glass.

glass


The converging rays meet

farther to the right, behind the

screen.

n1 = 1 n2 > n1


Rear view mirror

Day setting:

$ 100% reflection

Night setting:

$ 10% reflection


Apparent Depth

At what angle should

the searchlight be aimed

to illuminate the chest?

Snell’s Law: n1 sin!1 = n2 sin!2

sin!1 =1.33sin31.22!

1= 0.6894

!1 = 43.6!

The searchlight is aimed above the chest – the apparent depth of the chest is

less than its actual depth.

tan!2 =2

3.3

!2 = 31.22!


Apparent depth – formation of image

Apparent position

of chest


Prob. 26.17/19: Find the relationship between true and apparent depth for

small angles of incidence.

!1

!2

d

d´

x

x Snell’s Law: n1 sin!1 = n2 sin!2

!1

tan!1 =x

d!

tan!2 =x

d

So, x= d!tan!1 = d tan!2

For small angles: sin!! tan!! ! radians

Therefore, n1!1 ! n2!2 and d"!1 ! d!2

So,d!

d" !2

!1" n1

n2

n1

n2

! d" # dn1

n2

Apparent depth

true depth

apparent

depth


Apparent depth (height)

Viewed from below the surface

of the water, the object appears

further above the surface than it

actually is.

The same formula holds as for

apparent depth:n

1

n2

d! " dn1

n2

As n1 > n

2, d´ > d


Prob. 26.16: A silver medallion is sealed within a transparent block of

plastic. An observer in the air, viewing the medallion from directly

above, see the medallion at an apparent depth of 1.6 cm beneath the top

surface of the block.

How far below the top surface would the medallion appear if the

observer (not wearing goggles) and the block were under water?


Prob. 26.20/18: A man in a boat is looking straight down at a fish in the

water directly beneath him. The fish is looking right back. They are the

same distance from the air-water interface.

To the man, the fish appears to be 2 m beneath his eyes.

To the fish, how far above its eyes does the man appear to be?


Chapter 26 so far...

Refractive index:

n = c/v

Snell’s law:

n1 sin%1 = n2 sin%2

Apparent depth:

d! = dn1/n2


Displacement of path of ray of light

Rotating the glass plate changes the

amount of displacement – used in some

optical instruments to line up an image

with reference lines (cross-hairs)

n1 sin!1 = n2 sin!2 = n1 sin!3

So, !3 = !1


Prob. 26.19/17:

If "1 = 30o, the glass

plate is 6 mm thick

and n2 = 1.52, what

is the displacement?

n2 = 1.52d = 6 mm

Snell: n1 sin!1 = n2 sin!2 = n3 sin!3

As n1 = n3, !1 = !3

That is, emerging ray is parallel with incident ray.

sin!2 =n1 sin!1

n2

=1! sin30"

1.52# !2 = 19.2"

x

= 30º

t = 6 mm


x

L

t = 6 mmSo, x=

t sin(!1!!2)cos!2

x=6sin(30!"19.2!)

cos19.2!

Displacement, x = 1.19 mm

!1!!2

26.19 contd

A B

C

D

x= Lsin(!1!!2)BCD:

L= t/cos!2ABD:

90º

!2 = 19.2!


For any wave: v = f !

For light: v = c/n = f !

(large n)

(small n)

Snell’s Law: a light wave hitting a boundary

The incident wave

turns into the

refracted wave and

matches onto it.

The wavefronts

crumple and tilt to

adjust to the new

wavelength # the

ray is refracted.

Rays are perpendicular

to wavefronts

So, ! =c

nf, proportional to

1n


Snell’s Law: a light wave hitting a boundary

So,c

h f= n1 sin!1 = n2 sin!2

Snell’s Law!A

B

C

D

!1 =c

n1 f= hsin"1ABC:

!2 =c

n2 f= hsin"2ACD:

n1

n2

! = c/(nf) for a light wave

90º

90º

Rays are perpendicular

to wavefronts

90º – "1

Wavefronts

Wavefronts


The normal case: both

reflected and refracted rays.

If the angle of incidence

increases, so does the angle

of refraction, until...

Total internal

reflection

Refraction at the “critical

angle for total internal

reflection” – the refracted

ray is at 90o.

n1 sin!c = n2 sin90!

sin!c = n2/n1Possible only if n2 < n

1

Internal

reflection


If the angle of incidence is greater than the critical angle:

– according to Snell’s law, the sine of the angle of refraction is greater

# than 1.0, so there is no refracted ray

– the light undergoes “total internal reflection”

Example: n1 = 1.33 (water), n

2 = 1 (air): sin "

c = 1/1.33, "

c = 48.8o

!1 > !c

n2 < n

1

26.13

n1 sin!c = n2 sin90!


Prob. 26.105/13: A ray of light travels from the coin to the surface of the

liquid and is refracted as it enters the air. A person sees the ray as it

skims just above the surface of the liquid. How fast is the light travelling

in the liquid?

• What are the angles of incidence, refraction? Apply Snell’s law, find

# the index of refraction of the liquid.

"

"

tan! = 5/6, ! = 39.8!n1 = ?

n2 = 1

n1 sin " = n2 sin 90º

n1 = 1/sin " = 1.56

v = c/n1 = 1.92"108 m/s

90º


Total internal reflection around the bend


Optical fibre – total internal reflection at the

walls steers the light around bends

Applications:

• Medicine – flexible optical fibres used to look inside the body.

# “Keyhole” surgery – add surgical instrument, laser beam to vaporize tissue.

• Communications – transmit telephone, radio, TV, internet signals on a

# laser beam inside a fibre optic cable – no external interference, much

# greater amount of information can be transmitted than with copper cable.

26.29


Prob. 26.-/29: The optical fibre shown consists of a core made of flint

glass surrounded by a cladding made of crown glass.

A beam of light enters the fibre from air at an angle "1 with respect to the

normal. What is "1 if the light strikes the core-cladding interface at the

critical angle "c?

n2 = 1.667

n3 = 1.523

n1 = 1 A

B

!2 = 90!"!c


Total internal reflection in a prism

!c ! sin"1(1/1.5) = 42#

Glass: n $ 1.5

45! > !c

Prisms “fold” the light

path to make the

binoculars shorter. Each

arm acts as a longer

telescope.


Polarization of light by reflection

Light reflected from a surface is in general partially polarized.

The reflected light is 100% polarized parallel to the surface when reflection

occurs at the “Brewster angle” "B

, corresponding to a 90o angle between

reflected and refracted rays.

From the triangle: sin!2 = cos!B

Snell: n1 sin!B = n2 sin!2 = n2 cos!B

So, tan!B =n2

n1

Brewster angle

!B+!2 = 90!

!B

!2

90º

!B +!2 +90! = 180!


Polarization of light by reflection

• Sunlight reflected from water – polarized horizontally. "B = 53º

• Polaroid type sun glasses reduce glare from reflected sunlight by

# filtering out horizontally polarized light.

• Digital watches – emitted light is polarized vertically (top to bottom

# in the display).

• Display turns dark if rotated by 90o when viewed through Polaroid sun

# glasses.


Dispersion by a prism

lower n, less refraction

higher n, greater refraction26.38

Violet is refracted more than red


Prob. 26.38: A ray of sunlight is passing from diamond into crown glass;

the angle of incidence is 35º.

Indices of refraction for red and blue light:

Blue: ndiamond = 2.444# # ncrownglass = 1.531

Red: ndiamond = 2.410# # ncrownglass = 1.520

Determine the angle between the refracted red and blue rays in the crown

glass.


Dispersion by rain drops – rainbows

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Primary

rainbow

Secondary

rainbow


Dispersion by a raindrop – primary rainbow

The refractive index for violet is larger than

for red.

# violet is refracted through a larger angle

# than red


Dispersion – formation of a rainbow

The colours of the rainbow come from

raindrops at different height, red from

higher up, violet from lower down.


http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Violet is refracted through a larger angle than red


Sun Dogs


Sun Dogs

Refraction by hexagonal

ice crystals

Violet is refracted through a larger angle

than red

http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/halo22.html#c3


Lenses

Focal pointA positive (converging, convex) lens

A negative (diverging, concave) lensFocal point


Formation of image by thin lenses

Parallel to axis, passes through

focal point on right

Passes through focal point on

left, emerges parallel to axis

Passes through centre of lens

in a straight line

Parallel to axis, ray traced back

to axis passes through focal

point on left

Heads toward focal point on

right, emerges parallel to axis

Passes through centre of lens

in a straight line


Formation of a real image by a converging lens

A “real” image – can be

seen on a screen placed at

the position of the image.

Image is inverted.

Light appears

to originate

from image

Object is outside the

focal point


Formation of a virtual image by a converging lens

A “virtual” image – cannot be formed on a

screen. Image is upright, and magnified.

Object is inside

the focal point

Light appears

to originate

from image


Formation of virtual image by

a diverging lens

Light appears

to originate

from image

Image is virtual, upright, diminished


Thin lens equation

f

!!!

!

ho

tan!=ho

do=hi

di

tan!=ho

f=

hi

di! f

So,hi

ho=di

do=di! f

fDivide by di :

1

do=1

f! 1di

That is:1

do+1

di=1

fThin lens equation Applicable to diverging

lenses too

Object distance Image distance

1

3

3

1


Linear magnification

f

!!!

!

ho

Linear magnification, m = (height of image)/(height of object)

m=hi

ho

Sign convention, image is inverted, so: m=!dido

Similar triangles:hi

ho=di

do(or from tan!)


Conventions for the thin lens equation

Virtual object to right of lens, do is negative (2 or more lenses)

Virtual image to left of lens, di is negative

Converging (positive) lens, focal length f is positive

Diverging (negative) lens, focal length f is negative

Draw ray diagrams with rays travelling from left to right.

Normal situation:

Ex8

Real object to left

of lens,

object distance,

do is positive

Real image to right

of lens,

image distance,

di is positive

Object # Lens # Image


Diverging lens, f < 0

do (> 0)

di (< 0)

f < 0

Converging lens, f > 0

do (> 0) di (> 0)

Object # Lens # Image

f > 0


Example: A 1.7 m tall person stands 2.5 m in front of a camera. The

focal length of the lens is 0.05 m.

a) Find the image distance

b) Find the magnification and the height of the image on the film.


Prob. 26.50/104: To focus a camera on objects at different distances, the

converging lens is moved toward or away from the film, so a sharp

image always falls on the film.

A camera with a lens of focal length f = 200 mm is to be focussed on an

object located at a distance of 3.5 m and then at 50 m.

Over what distance must the lens be movable?


Thin lens equation

f

!!!

!

ho


1

3

3

1

Thin lens equation:1do

+1di

=1f

Linear magnification: m =hi

ho=!di

do


Combinations of lenses – microscope

A microscope producing a virtual, inverted and magnified

final image. The eyepiece acts as a magnifying glass.

Lens 2: light appears to come

from intermediate image

1

3

1´

3´

• Find the location of the image formed by the first lens as if the second

# lens did not exist.

• Use that image as an object (source of light) for the second lens using the

# sign convention for real and virtual objects.

22


Prob. 26.59/60: Two identical diverging lenses are separated by 16 cm.

The focal length of each lens is –8 cm. An object is located 4 cm to the

left of the lens that is on the left. Determine the final image distance

relative to the lens on the right.


Thin lens equation

f

!!!

!

ho


1

3

3

1

Thin lens equation:1do

+1di

=1f

Linear magnification: m =hi

ho=!di

do

Object to left: do > 0

Image to right: di > 0

Positive lens: f > 0

(converging, convex)

Negative lens: f < 0

(diverging, concave)


Combinations of lenses – microscope

A microscope producing a virtual, inverted and magnified

final image. The eyepiece acts as a magnifying glass.

Lens 2: light appears to come

from intermediate image

1

3

1´

3´

• Find the location of the image formed by the first lens as if the second

# lens did not exist.

• Use that image as an object (source of light) for the second lens using the

# sign convention for real and virtual objects.

22


26.66: Two converging lenses (f1 = 9 cm, f2 = 6 cm) are separated by 18

cm. The lens on the left has the longer focal length. An object stands 12

cm to the left of the left-hand lens.

a) Locate the final image relative to the lens on the right.

b) Obtain the overall magnification.

c) Is the final image real or virtual, upright or inverted, larger or smaller

than the object?


Prob. 26.64: A coin is located 20 cm to the left of a converging lens (f =

16 cm). A second, identical, lens is placed to the right of the first lens,

such that the image formed by the combination has the same size and

orientation as the original coin. Find the separation between the lenses.


The Human Eye

Biomedical Applications of Introductory

Physics, Tuszynski & Dixon

Most of the

refraction occurs at

the cornea

n = 1.34

n = 1.41-1.45

n = 1.33-1.34

n = 1.38

Sharpest image,

best colour

discrimination


The human eye

The eye focuses an image onto the retina by adjusting the focal length of

the eye lens. This is known as accommodation.

Closest distance at which eye can focus: “near point”

Normal value: N = 25 cm

Eye lens compressed,

focal length decreased

Greatest distance at which eye can focus: “far point”

Normal value: infinity

Eye lens has its

longest focal lengthCiliary muscle,

relaxed


Near and far points

Near point: closest distance at which unaided eye can focus,

normal value, N = 25 cm

Far point: greatest distance at which unaided eye can focus,

normal value infinity

Accommodation: the ability of the eye to adjust its focal length to focus on

objects at different distances.

Nearsighted eye: far point less than infinity " distant objects are blurred

Farsighted eye: near point greater than 25 cm " objects close by are

blurred


Nearsightedness

The eye lens forms an image of a distant object in front of the

retina # blurred image on the retina

Correction is with a diverging lens that moves the image back

onto the retina. The corrective lens forms a virtual image in front

of the eye that is close enough for the eye to focus on.

Objects in focus


Farsightedness

The eye lens forms an image of a nearby object behind

the retina # blurred image on the retina

Correct with a converging lens that forms a virtual

image far enough away for the eye to focus on.

Objects in focus


Correction of near and farsightedness

Use a corrective lens to form a virtual image at a distance at which

the eye can focus.

Nearsighted:

• The corrective lens forms an image of a distant object at the person’s far

point, or closer.

Farsighted:

• The corrective lens forms an image of an object at the person’s near

point, or further.

Power of a lens:

• Power is 1/f, focal length in metres, power in diopters.

# Example, f = –10 cm, power = 1/(– 0.1) = –10 diopters.


Prob. 26.107/71: A nearsighted person cannot read a sign that is more

than 5.2 m from his eyes. He wears contact lenses that do not correct his

vision completely, but do allow him to read signs located up to distances

of 12 m from his eyes.

What is the focal length of the contacts?


Prob. 26.67: A farsighted person has a near point that is 67 cm from

her eyes. She wears eyeglasses that are designed to enable her to read a

newspaper held at a distance of 25 cm from her eyes.

Find the focal length of the eyeglasses, assuming that they are worn –

a) 2.2 cm from the eyes,

b) 3.3 cm from the eyes.


Angular size, magnification

Angular size of the object: !=ho

N

(small angle, in radians, object

viewed by unaided eye at near

point, N)

Angular magnification, M:

M =!!

!=ho

do" N

ho=N

do

(near point, closest eye can focus at)

The magnifying glass lets the user view the object closer than the near point

Angular size of image formed

by the magnifying glass:

!! =ho

do


Angular magnification: M =!!

!=N

do

2) Final image is at the near point, so di = – N

Magnifying Glass

Then, M =N

do=f +N

f= 1+

N

f$ maximum usable magnification

Two cases:

1) Final image is at infinity: so do = f

Then: M =N

do=N

f$ minimum magnification

1

do=1

f! 1

!N =f +N

fNThin lens equation:

!1do

=1f! 1!!

"

di = – !


Magnification Markings

Lenses are sometimes marked with the magnification they produce

when an image is formed at infinity.

For example, “10"”.

This means that,

"

"

with N = 25 cm, the normal near point.

M =N

f= 10,

So, f =N

10=

2510

= 2.5 cm


Prob. 26.112: A stamp collector is viewing a stamp with a magnifying

glass held next to her eye. Her near point is 25 cm from her eye.

a) What is the refractive power of a magnifying glass that has an angular

magnification of 6 when the image of the stamp is located at the near

point?

b) What is the angular magnification when the image of the stamp is 45

cm from the eye?


Astronomical Telescope

Formation of intermediate image

by the objective.

The eyepiece acts as a magnifying

glass to produce a magnified final

image.

!=hi

di1! hi

fo

!! =" hi

do2#"hi

fe

M =!!

!"#hi

fe$ fo

hi=# fo

fe (exact when object and final image are at infinity)

di1

do2

(small angles)

fo fe


Astronomical Telescope

di1

do2

Object at infinity:

• first image at focal point of objective

# di1

= fo

Final image at infinity:

• first image at focal point of eyepiece as well

# do2

= fe

# distance between lenses is:

" L = fo + f

e

and

#M = ! di1

do2= !fo

feexactly


Prob. 26.92/90: An astronomical telescope has an angular

magnification of –132 and uses an objective with a refractive power

of 1.5 diopters.

What is the refractive power of the eyepiece?

Angular magnification,

Refractive power,

so,

Therefore, Pe = 132 # Po = 132 # 1.5 = 198 diopters.

M = !fo

fe= !132

P =1f

M = !Pe

Po= !132

fe =1Pe

= 0.00505 m = 5.1 mm


Opera Glasses

Like an astronomical telescope but with an eyepiece that is a diverging

(negative) lens.

The distance between the lenses is still:

L = fo + f

e and the angular magnification is

M = – fo/f

e

But, as fe < 0:

• the length is less than an astronomical telescope of the same

# magnification

• the image is the right way up (M > 0)


Compound Microscope

!!

With:

• object just outside the focal point of the objective, so do1

$ fo

• first image at focal point of eyepiece (# final image at infinity)

Distance between lenses = L

fo f

e

Magnification: compare angular size of final image, "%,

to angular size, ", of object at near point viewed with

the naked eye.

Angular magnification, M =!!

!"#(L# fe)N

fo feN = near point


ho1

hi1 !!do2 ! fe

!! =hi1

do2" hi1

fe

So, !! " #!L# fe

fo fe

"$ho1

With the object at the near point and no microscope: !=ho1

N

Compound Microscope

di1

Final image

do1 ! fo

Therefore, M =!!

!"#

!L# fe

fo fe

"$N

hi1 = m1ho1 =!!

di1

d01

""ho1 #!

!L! fe

fo

""ho1

fo

fe


Prob. 26.86/98: A microscope for viewing blood cells has an objective

with a focal length of 0.5 cm and an eyepiece with a focal length of

2.5 cm.

The distance between the two lenses is 14 cm.

If a blood cell subtends an angle of 2.1 " 10-5 rad when viewed with

the naked eye at a near point of 25 cm, what angle does it subtend

when viewed through the microscope?


Summary of Chapter 26

• Snell’s Law: ## # # n1 sin "1 = n2 sin "2, # v = c/n

• Apparent depth: ## # d! = d n1/n2

• Total internal reflection: #n1 sin "c = n2, "" "c = critical angle for total

" " " " " " " " " " " " internal reflection

• Lens equation: # # # 1/do + 1/di = 1/f

• Linear magnification: # m = – di/do

" Two lenses:# # # # m = m1 m2

• Angular magnification:# M = "!/"

• Magnifying glass:# # M = N/f # # (image at infinity, N = near point)

# # # # # # M = N/f + 1# (image at near point)


Documents

Chapter 26: Refraction, Lenses, Optical Instruments · Chapter 26: Refraction, Lenses, Optical Instruments ¥ Refraction of light, SnellÕ s law . Apparent depth ¥ Polarization of