Mirrors and lenses PHY232 – Spring 2007 Jon Pumplin pumplin/PHY232 (Ppt courtesy of Remco Zegers)

mirrors and lenses

PHY232 – Spring 2007

Jon Pumplin

http://www.pa.msu.edu/~pumplin/PHY232

(Ppt courtesy of Remco Zegers)

PHY232 - Pumplin - Mirrors and lenses 2

we saw…

that light can be reflected or refracted at boundaries between material with a different index of refraction.

by shaping the surfaces of the boundaries we can make devices that can focus or otherwise alter an image.

Here we focus on mirrors and lenses for which the properties can be described well by a few equations.


the plane (=flat) mirror

in the previous chapter we already studied flat mirrors.

Distance from the object to the mirror is the object distance p

Distance from the image to the mirror is the image distance -q

in case of a flat mirror, an observer sees a virtual image, meaning that the rays do not actually come from it.

the image size (h’ ) is the same as the object size (h), meaning that the magnification h’/h=1

the image is not inverted

p -q

NOTE: a virtual imagecannot be projected on a screen but is ‘visible’ bythe eye or another opticalinstrument.


question

You are standing in front (say 1 m) of a mirror that is less high than your height. Is there a chance that you can still see your complete image?

a) yes b) no

object image


ray diagrams

to understand the properties of optical elements we use ray diagrams, in which we draw the most important elements and parameters to understand the elements

p -q

h h’


concave mirrors

C

C: center of mirror curvature

a light ray passing through the center of curvature will be reflected back upon itself because it strikes the mirror normally to the surface.

F: focal point

F

a light ray traveling parallel to the central axis of the mirrorwill be reflected to the focal point F, with FM=CM/2

The distance FM is called the focal length f.

M


concave mirrors: an object outside F

O F

step 1: draw the ray from the top of the object parallel to the central axis and its reflection (through F).

step 2: draw the ray from the top of the object through F and its reflection (parallel to the central axis)

step 3: note that a ray from the bottom of the object just reflects back.

the image of the top of the object is located where the reflected rays meet

construct the image I

I


concave mirrors: an object outside F

O FI

The image is:a) inverted (upside down)b) real (light rays pass through it)c) For this location, image is smaller than objectd) If interchange image and object, image would be bigger than object.


concave mirrors: object outside F

O FI

distance object-mirror: pdistance image-mirror: qdistance focal point-mirror: f

mirror equation: 1/p + 1/q = 1/fgiven p,f this equation can be used to calculate qmagnification: M = -q/pcan be used to calculate magnification.

• if negative: the image is inverted• if smaller than 1, object is demagnified


example

An object is placed 12 cm in front of a a concave mirror with focal length 5 cm. What are: a) the location of the imageb) the magnification

a) 1/p + 1/q = 1/f so 1/12 + 1/q = 1/5 1/q = 1/5 - 1/12 so q = 8.57 cm

b) M = -q/p = -8.57/12 = -0.71 this means that size of the image is only 71% of the the size of the object and that it is inverted.c) Image is real (whenever q>0, I.e., whenever M > 0)


concave mirrors: an object inside F

OF




I

the image of the top of the object is located where the reflected rays meet: in this you must draw virtual rays on the other side of the lens

creates a magnified image (“shaving mirror”)

the image is:a) not invertedb) virtualc) magnified


concave mirrors: an object inside F

OF Ithe image is:a) not invertedb) virtualc) magnified

The lens equation and equation for magnification are stillvalid. However, since the image is now on the otherside of the mirror, its sign should be negative


example

an object is placed 2 cm in front of a lens with a focal length of 5 cm. What are the a) image distance and b) the magnification?

a) 1/p + 1/q = 1/f so 1/2 + 1/q = 1/5 1/q = 1/5 - 1/2 so q = -3.3 cm (note the ‘-’ sign!)

b) M = -q/p = -(-3.3)/2 = +1.65 this means that size of the image is 65% larger than the size of the object and that it is not inverted (+).


demo: the virtual pig



convex mirrors: (p>|f|or p<|f| doesn’t matter)

O F


the image of the top of the object is located where the reflected rays meet

construct the image I (examples: garden ball, rearview mirrors that warn “objects are closer than they appear”)

I


F is now located on the otherside of the mirror


convex mirrors

O FIF is now located on the otherside of the mirror

the image is:a) not invertedb) virtualc) demagnified

The lens/mirror equation and equation for magnification are still valid. However, since the image and focal point are now on the other side of the mirror, their signs should be negative


example

an object with a height of 3 cm is placed 6 cm in front of a convex mirror, with f=-3 cm. What are a) the image distance and b) the magnification?

answer a): 1/p+1/q=1/f with p=6 cm f=-3 cm1/6 + 1/q = -1/3 so q=-2 cm b) M=-q/p=-(-2)/6=1/3the image is only 33% of the height of the object.


Mirrors: an overview

mirror equation 1/p + 1/q = 1/f F = R/2 where R is the radius of the mirror magnification: M = -q/p

type p? image image direction

M q f

concave p>f real inverted |M|>0 M - + +

concave p<f virtual not inverted

|M|>1 M + - +

convex any p virtual not inverted

|M|<1 M + - -


Lenses Lenses function by refracting light at their

surfaces Their action depends on

radii of the curvatures of both surfacesthe refractive index of the lens

converging (positive lenses) have positive focal length and are always thickest in the center

diverging (negative lenses) have negative focal length and are thickest at the edges

+

- used indrawings


lensmakers equation

R1

R2

f: focal length of lensn: refractive index of lensR1 radius of front surfaceR2 radius of back surface

1 2

R2 is negative if the center of the circle is on the left of curvature 2 of the lensR1 is positive if the center of the circle is on the right of curvature 1 of the lens

object

if the lens is not in air then (nlens-nmedium)


example

Given R1=10 cm and R2=5 cm, what is the focal length? The lens is made of glass (n=1.5)R1

R2

1 2

R1 is on the right of the curvature, so positive +10 cmR2 is on the left of the curvature, so negative –5 cmn=1.51/f=0.5(0.1-(-0.2))=0.15 f=+6.67 cm

object


example 2

Given R1=5 cm and R2=10 cm, what is the focal length? The lens is made of glass (n=1.5)R2

R1

1 2object

R1 is on the left of curvature 1 so R1=-5 cmR2 is on the right of curvature 2 so R2=+10 cmn=1.51/f=0.5(-0.2-0.1)=-0.15 f=-6.67 cm


converging lens p>f

O F

+

F

1) A ray parallel to the central axis will be bend through the focal point

2) A ray through the center of the lens will continue unperturbed

3) A ray through the focal point of the lens will be bend parallel to the central axis

I

4) the image is located at the crossing of the above 3 rays (you need just 2 of them).

A real inverted image is created. The magnificationdepends on p: |M| can be <1, 1 or >1


lens equation

O F

+

F

I

The equation that connects object distance p, imagedistance q and focal length f is (just like for mirrors):1/p + 1/q = 1/fSimilarly for the magnification:M=-q/p

q is positive if the image is on the opposite side of the lens as the objectNOTE THAT THIS IS DIFFERENT THAN THE CASE FOR MIRRORS. In either case, q >0 means the image is on the side where the light from the lens or mirror is.


example

an object is put 20 cm in front of a positive lens, with focal length of 12 cm. a) What is the image distance q? b) What is the magnification?

a) P = 20 cm, f = 12 cm use 1/p + 1/q = 1/f1/20 + 1/q = 1/12 solve for q gives q=30 cm

b) M = -q/p = -30/20 = -1.5

The image is inverted (M negative).The image is magnified (|M|>1)


converging lens p<f

OF

+

F

A virtual non-inverted image is created. Magnification >1

1) A ray parallel to the central axis will be bend through the focal point

3) A ray through the focal point of the lens will be bend parallel to the central axis



I


example

an object is put 2 cm in front of a positive lens, with focal length of 3 cm. a) What is the image distance q? b) What is the magnification?

a) p=2 cm, f=3 cm use 1/p + 1/q = 1/f1/2 + 1/q = 1/3 solve for q gives q=-6 cmNOTE q is negative which means it is on the same side of the lens as the objectb) M=-q/p=-(-6)/2=+3

The image is not inverted (M positive).The image is magnified (|M|>1)The image is virtual


question An object is placed in front of a converging (positive) lens

with the object distance larger than the focal distance. An image is created on a screen on the other side of the lens. Then, the lower half of the lens is covered with a piece of wood. Which of the following is true:

a) the image on the screen will become less bright only b) half of the image on the screen will disappear only c) half of the image will disappear and the remainder of the

image will become less bright.

O F

+

FI

screenrays of light can still make it to any point on the image,but there are less of them, so less bright (only)


NOT CORRECT


diverging lens p>|f|

O F

-

F


I


A virtual non-inverted image is created. The magnification |M|<1

1) A ray parallel to the central axis will be bend so that the ray passes through the focal point IN FRONT of the lens

3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis


example

an object is put 5 cm in front of a negative lens, with focal length of -3 cm. a) What is the image distance q? b) What is the magnification?

a) p=5 cm, f=-3 cm use 1/p + 1/q = 1/f1/5 + 1/q = -1/3 solve for q gives q=-1.88 cm

b) M=-q/p=-(-1.88)/5=+0.375

The image is non-inverted (M positive).The image is demagnified (|M|<1)


diverging lens p<|f|

OF

-

F


A virtual non-inverted image is created. The magnification |M|<1 similar to case with p>|f|

1) A ray parallel to the central axis will be bend so that the ray passes through the focal point IN FRONT of the lens

3) A ray aimed at the focal point on the other side of the lens will be bent parallel to the central axis

I



example

an object is put 2 cm in front of a negative lens, with focal length of -3 cm. a) What is the image distance q? b) What is the magnification?

a) p = 2 cm, f = -3 cm use 1/p + 1/q = 1/f1/2 + 1/q = -1/3 solve for q gives q = -1.2 cm

b) M = -q/p = -(-1.2)/2 = +0.6

The image is non-inverted (M positive).The image is demagnified (|M|<1)


lenses, an overview

type p? image image direction

M q f

converging

p>f real inverted |M|>0 M - + +

converging

p<f virtual not inverted

|M|>1 M + - +

diverging any p virtual not inverted

|M|<1 M + - -

mirror or lens equation 1/p + 1/q = 1/f mirror or lens magnification: M = -q/p lens makers equation: 1/f=(n-1)(1/R1-1/R2)


chromatic aberrations

Chromatic aberrations are due to light of different wavelengths having a different index of refractionCan be corrected by combining lenses/mirrors

If n varies with wavelength, the focal length fchanges with wavelength


two lenses

an object, 1 cm high, is placed 5 cm in front of a converging mirror with a focal length of 3 cm. This setup is placed in front of a diverging mirror with a focal length of –5 cm. The distance between the two lenses is 10 cm. Where is the image located, and what are its properties?

+ -

5 cm

3cm

15 cm

5cm


answer+ -

5 cm

3cm

15 cm

5cm

consider the converging lens first:1/p+1/q=1/f so 1/5+1/q=1/3 so q=7.5 cmM=-q/p=-7.5/5=-1.5, so 1.5x1cm=1.5 cm high A real inverted magnified image I1

5+7.5=12.5 cm

I1

O

consider the action of diverging lens on the image constructed above1/p+1/q=1/f so 1/(2.5) + 1/q = 1/(-5) q=-1.67 cm M=-q/p=-(-1.67)/2.5=0.67, so 0.67x 1.5=1 cm highA virtual non-inverted demagnified image I2 relative to I1

I2

15-1.67=13.33 cm

Mirrors and lenses PHY232 – Spring 2007 Jon Pumplin pumplin/PHY232 (Ppt courtesy of Remco Zegers)

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Transcript of Mirrors and lenses PHY232 – Spring 2007 Jon Pumplin pumplin/PHY232 (Ppt courtesy of Remco Zegers)