Lecture 22Wave Optics-3 Chapter 22

PHYSICS 270Dennis Papadopoulos

April 2, 2010

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R = A[cosωt + cos(ωt + φ) + cos(ωt + 2φ) + cos(ωt + 3φ) + ....]

φ = 2πd sinθ

λ

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O ˆ Q S = φ

A = 2rsin(φ /2)

O ˆ Q T = nφ

AR = 2rsin(nφ /2) = Asin(nφ /2)

sin(φ /2)

I = Io

sin2(nφ /2)

sin2(φ /2)

Io ≡ A2

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I = Io

sin2(nφ /2)

sin2(φ /2)

φ = 2πd sinθ

λ

Phase Arrays

Gratings

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I = Io

sin2(nφ /2)

sin2(φ /2)

φ = 2πd sinθ

λ

What happens when slit is too large to be considered a point source ?Huygens principle replace wave-front by a continuous series of point

sources

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d → 0;φ → 0

Take nφ the difference from one end to the other constant,say nφ = Φ

but send φ to zero. sinφ ≈ φ

I = n2Io

sin2(Φ /2)

Φ2= Imax

sin2(Φ /2)

Φ2

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w =2.44λL

D= D

Dc = 2.44λL

When to use ray optics and when wave optics

Actual double slit interference pattern (a<d and a> wavelength) –

Convolution of ideal double slit and single slit patterns

Raleigh Criterion

• Two objects are resolvable if min=1.22D, namely the angle of the first dark fringe of the diffraction pattern

• Objects not resolvable if min

• Objects marginally resolvable if min

Resolution limit -- Rayleigh’s criteriaFor circular aperture, slightly different:

The Resolution of Optical InstrumentsThe minimum spot size to which a lens can focus light of wavelength λ is

where D is the diameter of the circular aperture of the lens, and f is the focal length.In order to resolve two points, their angular separation must be greater than θmin, where

is called the angular resolution of the lens.

The same criterion applies to the focusing spot of mirrors if D is the diameter of the mirror

Raleigh Criterion

• Two objects are resolvable if min=1.22D, namely the angle of the first dark fringe of the diffraction pattern

• Objects not resolvable if min

• Objects marginally resolvable if min

EXAMPLE

The Hubble space telescope has a diameter 2.4 meters. It is used to photograph objects 30000 light years away ( 1 light year is 9.46x1015 meters). Assume that it uses red light with 650 nm wavelength. What is the distance between two stars that can be resolved?

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=1.22λ /D

s = Rθ = R(1.22λ /D) =1011 km

SECTIONS 24.3-24.4-24.5 The Eye-Angular Magnification- Resolution

Vision• The human eye is roughly spherical, about 2.4 cm in

diameter.• The transparent cornea and the lens are the eye’s

refractive elements. • The eye is filled with a clear, jellylike fluid called the

aqueous humor and the vitreous humor.• The indices of refraction of the aqueous and vitreous

humors are 1.34, only slightly different from water. • The lens has an average index of 1.44. • The pupil, a variable-diameter aperture in the iris,

automatically opens and closes to control the light intensity.

• The f-number varies from roughly f/3 to f/16, very similar to a camera.

f-number = f/D f/3 means that f-number is 3

Focusing and Accommodation• The eye focuses by changing the focal length of the lens

by using the ciliary muscles to change the curvature of the lens surface.

• Tensing the ciliary muscles causes accommodation, which decreases the lens’s radius of curvature and thus decreases its focal length.

• The farthest distance at which a relaxed eye can focus is called the eye’s far point (FP). The far point of a normal eye is infinity; that is, the eye can focus on objects extremely far away.

• The closest distance at which an eye can focus, using maximum accommodation, is the eye’s near point (NP).

Usually 25 cm

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≈h /s

θNP ≈ h /25cm

Cannot focus any closer than the near point of the eye ~ 25 cm

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sinθ ≈ θ ≈ h /s ≈ h / f

Angular Magnification

M = θ /θNP = 25cm / f

Angular Magnification

Exercise: Compare concepts and scaling of angular vs. lateral magnification