ECE 318 Fundamentals of Optics

Lecture 7

ECE 318 Fundamentals of Optics

Lecture 7

Geometric Optics VI

• Spherical Mirrors

© Li Qian2 Lecture 7

Spherical MirrorsSpherical Mirrors

Sign Convention:- so is positive when O is to the left of V (i.e., real object)

- si is positive when I is to the left of V (i.e., real image)

- R is positive when C is to the left of V (i.e., concave)

- f is positive for a concave mirror (Note the sign convention for R and f is different from that defined in Pedrotti’s textbook.)

VCO Iφ

θiθr

sosi

Rlo

li

A

© Li Qian3 Lecture 7

Spherical Mirrors – Cont’dSpherical Mirrors – Cont’d

Use the same approach for solving refraction at a spherical interface, we can solve the imaging problem of spherical mirrors through geometric relationships:

Law of Sines applied to ∆OAC: ( )( )

( )RslRsl

o

io

i

oo

−=⇒

−=

−θφ

θφπsinsin

sinsin (1)

Law of Sines applied to ∆ACI:i

ri

r

ii

sRlsRl

−=⇒

−=

θφθφ

sinsinsinsin (2)

Law of Reflection: ri θθ = (3)

( ) ( )i

i

o

o

lRs

lRs −

−=− (4)(1)+(2)+(3):

where

( ) ( ) ( )φπ −−−+−= cos222 RRsRRsl ooo ( ) ( ) φcos222 RsRRsRl iii −−+−=

© Li Qian4 Lecture 7

Spherical Mirrors – Cont’dSpherical Mirrors – Cont’d

Eq. (4) is an implicit function of so, si and φ. Therefore, for a given so, si depends on φ, i.e. rays do not converge: A spherical reflecting surface is not a perfect imaging system.

However, if we again assume paraxial approximation, then cos φ ≈ 1, and we have:.; iioo slsl ≈≈ Equation (4) becomes:

fRss io

1211≡=+ (5)

Note Eq (5) has the same form as the Gaussian thin-lens formula (in air).

Q7-1: Does Eq (5) still hold if the medium is something other than air?

Q7-2: If a spherical mirror behaves like a thin lens, according to (5), what are the pros and cons of using a mirror instead of a lens for imaging?

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Spherical Mirrors – MT, MLSpherical Mirrors – MT, ML

VC

Sso

si

Pyoyi

Transverse Magnification:

RsRs

yyM

o

i

o

iT −

−=≡ (6)

VC

Sso

si

dso -dsi

Longitudinal Magnification:

2

2

o

i

o

iL s

sdsdsM −=≡ (7)

Q7-3: At what location will the source and image overlap? What are the MT and MLat this location?

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Mirrors vs. LensesMirrors vs. Lenses

• Mirrors can be made much larger in diameter than lenses– Mirrors have a lower requirement on the material’s optical quality– Only one high quality surface is required for a mirror, whereas

both surfaces of a lens has to be of high quality. – Better structural support can be provided to the mirror.

• Lenses have higher aberration due to chromatic dispersion.

• If real images are desired, mirrors form images on the same side as the incoming light, which makes placing image recording devices less convenient.

© Li Qian7 Lecture 7

Example - The Hubble Space TelescopeExample - The Hubble Space Telescope

The figure below shows the schematic of the Hubble Space Telescope (HST). The relevant parameters are given on the right.

Primary Mirror: Diameter = 2.4 mRadius of Curvature = 12.75 mHole Diameter = 0.60 m

Secondary Mirror: Diameter = 0.3 mRadius of Curvature = 5 mDistance to Primary Mirror = 4.6 m

1) Find the focal plane location relative to the primary mirror (see figure on next page).2) Find image location yi that corresponds to a ray angle of α (see figure on next page).3) If we were to use a refractive telescope objective for imaging, in order to achieve the

same scaling between α and yi, what would be the distance between the image plane and the objective?

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The Hubble Space Telescope –Simplified StructureThe Hubble Space Telescope –Simplified Structure

R1=12.75m

R2=-5m

?

yi

α

4.6 m

f1

-so2

si2

Final image plane

Focal Plane of R1

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The Hubble Space Telescope –Simplified CalculationsThe Hubble Space Telescope –Simplified Calculations1) Find the focal plane location relative to the primary mirror

If we ignored the secondary mirror, rays from a distant object would form an image on the focal plane of the primary mirror. Since f1=R1/2=6.375m, the image would be formed at 6.375m from the primary mirror. This image serves as the virtual object for the secondary mirror, and the object distance so2= –(6.375 – 4.6) = –1.775m. Using (5), we have

Therefore the image is formed at (6.12 – 4.6) = 1.52 m behind the primary mirror.

msR

sRss o

iio

12.6122111

222

222

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=⇒=+

−

2) Find image location yi that corresponds to a ray angle of α.Similar to part (1), if we ignored the secondary mirror, the image size on the focal plane of the primary mirror would be y1=αf1 under paraxial approximation. Next, using (6), we can find the final image size yi to be:

ααα 0.22375.65775.1

512.61

22

2212 =×

+−+

=−−

== fRsRsyMy

o

iTi

© Li Qian10 Lecture 7

The Hubble Space Telescope –Simplified Calculations (Cont’d)The Hubble Space Telescope –Simplified Calculations (Cont’d)3) If we were to use a refractive telescope objective for imaging, recall

where fobj is the focal length of the objective, and yi is the image size formed by the objective. Therefore,

which is a much larger dimension than the length of the HST.

obj

i

fy

=α

myf iobj 0.22==

α

Note: Here, we simplify the calculations considerably by assuming mirrors are spherical. The actual mirror shape of the HST is hyperbolodial, in order to reduce spherical aberrations.

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Homework – The Magic PigHomework – The Magic Pig

A pig figurine is placed at the bottom of a concave mirror with a radius of curvature R1. Another concave mirror with a hole in the centre and having a radius of curvature R2 is placed on top, as shown. As a result, two images are produced, mirroring each other, symmetrically located at the central hole location, as shown.

R2

R1

object

Image 1

Image 2

dFind:

(1) The relationship between R1, R2, and d, dbeing the central separation distance between the two mirrors.

(2) Explanations of how the two images form and whether the images are real or virtual.

(3) MT and ML of both images.

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Homework – Side-view MirrorHomework – Side-view MirrorA person sitting in the driver seat of a car is looking into the side-view mirror and finds a police car behind him. The following words are printed on the mirror: “Objects in mirror are closer than they appear”. Assuming that the mirror is spherical with a radius of curvature of 2 m, the person is 0.5 m away from the mirror, and that the police car is 1.5 m tall and 20 m away from the mirror, answer the following questions:(1) Is the side-view mirror a convex or concave one?(2) Where does the virtual image of the police car made by the mirror lie? (3) How tall is the virtual image of the police car? (4) What is the angle subtended by the height of the virtual image at the person’s eyes? (5) If the person turns back and looks at the police car directly, what is the angle

subtended by the height of the car at his eyes? (6) Based on the above calculations, interpret the meaning of “Objects in mirror are closer

than they appear”